2014.05.29 21:58 *xzile* **Somerset, KY**

News and information about Somerset, KY and it's surrounding areas (Burnside, Eubank, Science Hill).

2015.02.28 14:13 *LynchMob_Lerry* **Kentuckiana Guns**

2022.08.16 00:23 *bobbyvette69* **SomersetShame**

A subreddit where the people of Pulaski County, KY can post how they really feel about the state of Somerset and surrounding areas without censorship. Nothing is off limits here. Anonymity encouraged. Put the methhead who stole your mower on blast. Shame your local dope dealer. Expose evil around your community. Safeguard those around you. These people have gotten away with this for far too long. It's time we make it harder for them.

2023.03.19 17:19 *SchlesingerMindy323* **[HIRING] 25 Jobs in KY Hiring Now!**

2023.03.18 17:23 *rrmdp* **π’ Comfort Keepers - Somerset - London - Corbin, KY is hiring a Caregiver!**

2023.03.15 01:30 *hApPiNeSsIsAmYtHH* **Got something really cool to share!**

If you've seen any of my other posts on here you'll know that I love history and have a small collection of antiques, one of which is this beautiful photograph of a baby girl.

Well as I was looking at my stuff I noticed that the sitter's name was listed on the back (written in pencil), another name was listed above the sitter's name (in pen), the photography company's/photographer's name was listed on the front, the location is listed below it, there's a date listed on the bottom of the photograph (only the year no month or day), and basically there's enough information here to at least try to look for this person so I did just that and I think found them!

The sitter's name is Ruth Downs and the woman's name is Ruth G Downs, the photo is listed as being taken in 1917 (once again no month or day is listed) and the sitter is clearly a baby and the woman was born on March 26 1916 (which depending on when the photo was taken in 1917 would match the sitter's age), the photograph was taken in Davenport Iowa and the woman was born died and looks like she spent her entire life in Davenport Iowa, the name listed above the sitter's name on the photograph is Stoltenberg and the woman's parent's last names are Stoltenberg (that's her dad's last name, her mom's married name, and her mom's maiden name is Oden), the woman passed away in 2007 at the age of 91, her husband passed away in 1997, going by the obituary it looks like they didn't have any children, that the only family they had (or at least were close enough with to be listed on the obituary) were her nephew, her nephew's wife, and her nephew's and his wife's two children, I found her 1933 high school senior yearbook picture, a picture of her husband (in his obituary), their marriage certificate, her parents were German immigrants, some documents have her birth year listed as 1917 but since her obituary (which is normally made by people who personally knew the person) and most documents have her birth year listed as 1916 i'm going by that, and a few censuses (she had an older sister and her husband had a sister too). Also if this is actually her then her name was probably written on the back of the photo sometime after she got married (she got married in 1943) since her married name is listed too not unless her parents could see into the future and knew who she was gonna marry haha.

I'm going to do some more digging to see if I can find anymore information, to quadruple check and confirm everything, and to see if there's any family members who would like to have this photograph but I just wanted to share because I thought this was really cool and to be able to give people attached to antiques their humanity back is a really humbling thing! :)

Here's a link showing the antique picture and her senior yearbook picture (unfortunately I couldn't find any other pictures of Ruth but if this is actually her it's still really cool to see how she grew up to look and that she lived a long life): https://imgur.com/a/QcuAWUN

Edit: Just did some more digging and managed to find an obituary showing a picture of Ruth when she was older, unfortunately the picture isn't clear since it requires a subscription with Newspapers (can't afford it right now) but just knowing I found a picture of her when she's older is pretty cool and hopefully will help point me to finding living family incase they want the photograph. Here's a link to the (very blurry) picture haha: https://imgur.com/a/gLBDeKy

submitted by hApPiNeSsIsAmYtHH to Genealogy [link] [comments]
Well as I was looking at my stuff I noticed that the sitter's name was listed on the back (written in pencil), another name was listed above the sitter's name (in pen), the photography company's/photographer's name was listed on the front, the location is listed below it, there's a date listed on the bottom of the photograph (only the year no month or day), and basically there's enough information here to at least try to look for this person so I did just that and I think found them!

The sitter's name is Ruth Downs and the woman's name is Ruth G Downs, the photo is listed as being taken in 1917 (once again no month or day is listed) and the sitter is clearly a baby and the woman was born on March 26 1916 (which depending on when the photo was taken in 1917 would match the sitter's age), the photograph was taken in Davenport Iowa and the woman was born died and looks like she spent her entire life in Davenport Iowa, the name listed above the sitter's name on the photograph is Stoltenberg and the woman's parent's last names are Stoltenberg (that's her dad's last name, her mom's married name, and her mom's maiden name is Oden), the woman passed away in 2007 at the age of 91, her husband passed away in 1997, going by the obituary it looks like they didn't have any children, that the only family they had (or at least were close enough with to be listed on the obituary) were her nephew, her nephew's wife, and her nephew's and his wife's two children, I found her 1933 high school senior yearbook picture, a picture of her husband (in his obituary), their marriage certificate, her parents were German immigrants, some documents have her birth year listed as 1917 but since her obituary (which is normally made by people who personally knew the person) and most documents have her birth year listed as 1916 i'm going by that, and a few censuses (she had an older sister and her husband had a sister too). Also if this is actually her then her name was probably written on the back of the photo sometime after she got married (she got married in 1943) since her married name is listed too not unless her parents could see into the future and knew who she was gonna marry haha.

I'm going to do some more digging to see if I can find anymore information, to quadruple check and confirm everything, and to see if there's any family members who would like to have this photograph but I just wanted to share because I thought this was really cool and to be able to give people attached to antiques their humanity back is a really humbling thing! :)

Here's a link showing the antique picture and her senior yearbook picture (unfortunately I couldn't find any other pictures of Ruth but if this is actually her it's still really cool to see how she grew up to look and that she lived a long life): https://imgur.com/a/QcuAWUN

Edit: Just did some more digging and managed to find an obituary showing a picture of Ruth when she was older, unfortunately the picture isn't clear since it requires a subscription with Newspapers (can't afford it right now) but just knowing I found a picture of her when she's older is pretty cool and hopefully will help point me to finding living family incase they want the photograph. Here's a link to the (very blurry) picture haha: https://imgur.com/a/gLBDeKy

2023.03.12 16:44 *SchlesingerMindy323* **[HIRING] 25 Jobs in KY Hiring Now!**

2023.03.10 19:25 *CorvusSchismaticus* **Remains Found in Lake Identified as Man Missing since 1990: Ruvil Hale of Pigeon Roost, KY.**

In March 2022 a Kentucky State Trooper was spending a leisurely day fishing on Dewey Lake in Floyd County, Kentucky, when his depth finder picked up something unusual in the lake; a submerged car. When the car, a 1988 Ford Tempo, was pulled from the lake, authorities also found human remains inside the car. The vehicle's description and license plate matched that of a vehicle that had been stolen on July 3,1990 from a restaurant in Paintsville which was near a nursing home called the Paintsville Healthcare Center. A patient at that nursing home, a 43 year old man named Ruvil Hale, had also disappeared from the nursing home that same day and hadn't been seen or heard from since. Investigators at the time surmised that Ruvil had stolen the car to make his escape, but could find no trace of him or the car. Nobody could even guess where he had planned to go. He likely had no more than $2 of cash on him and the stolen car had only a half tank of gas, according to the owner. Aerial and ground searches in a 20 mile radius, including searches of ponds and lakes, (including Dewey Lake) and back roadways turned up nothing.

For 32 years Ruvil Hale's family wondered what became of him. Ruvil, a divorced father of two and a former coal miner, had a long history of medical problems, including a stroke, severe headaches, seizures, memory loss and double vision. He had recently had surgery for a brain aneurysm , which is why he had been moved to the nursing home in Paintsville because of the round the clock nursing care he required due to his extensive medical issues. He had been housed in other healthcare facilities in the past, according to his family, and had tried to escape from those places as well. Ruvil hadn't driven in years, according to his son, Keith, due to his poor vision. They couldn't imagine he would have gotten far.

Ruvil's family hired a private investigator at one point as well, but no trace of Ruvil could be found and eventually he was declared legally dead in 1996. Now, 32 years later, some of the questions could finally be answered. In October of 2022 a DNA match confirmed that the remains in the submerged car were those of Ruvil Hale. The lake in which the car was found, Dewey Lake, was only about fifteen minutes away from the healthcare center.

On January 29,2023 Ruvil's family held a funeral and burial at the family cemetery in Pilgrim, KY.

Links:

https://www.jpinews.com/2023/01/10/father-of-former-glasgow-superintendent-found-after-32-years/

https://charleyproject.org/case/ruvil-hale

https://mountaincitizen.com/2022/03/23/remains-believed-to-be-hale-missing-32-years/

https://www.callahamfuneralhome.com/obituaries/Ruvil-Hale/#!/Obituary

https://mountaincitizen.com/2023/01/25/medical-examiner-confirms-remains-are-ruvil-hale/

submitted by CorvusSchismaticus to gratefuldoe [link] [comments]
For 32 years Ruvil Hale's family wondered what became of him. Ruvil, a divorced father of two and a former coal miner, had a long history of medical problems, including a stroke, severe headaches, seizures, memory loss and double vision. He had recently had surgery for a brain aneurysm , which is why he had been moved to the nursing home in Paintsville because of the round the clock nursing care he required due to his extensive medical issues. He had been housed in other healthcare facilities in the past, according to his family, and had tried to escape from those places as well. Ruvil hadn't driven in years, according to his son, Keith, due to his poor vision. They couldn't imagine he would have gotten far.

Ruvil's family hired a private investigator at one point as well, but no trace of Ruvil could be found and eventually he was declared legally dead in 1996. Now, 32 years later, some of the questions could finally be answered. In October of 2022 a DNA match confirmed that the remains in the submerged car were those of Ruvil Hale. The lake in which the car was found, Dewey Lake, was only about fifteen minutes away from the healthcare center.

On January 29,2023 Ruvil's family held a funeral and burial at the family cemetery in Pilgrim, KY.

Links:

https://www.jpinews.com/2023/01/10/father-of-former-glasgow-superintendent-found-after-32-years/

https://charleyproject.org/case/ruvil-hale

https://mountaincitizen.com/2022/03/23/remains-believed-to-be-hale-missing-32-years/

https://www.callahamfuneralhome.com/obituaries/Ruvil-Hale/#!/Obituary

https://mountaincitizen.com/2023/01/25/medical-examiner-confirms-remains-are-ruvil-hale/

2023.03.09 20:21 *rrmdp* **π’ Comfort Keepers - Somerset - London - Corbin, KY is hiring a Carer!**

2023.03.09 19:12 *SchlesingerMindy323* **[HIRING] 25 Jobs in KY Hiring Now!**

2023.03.09 16:04 *ornery_epidexipteryx* **Arrests made on the Don Franklin Dealership Hellcat theft in Somerset, KY**

submitted by ornery_epidexipteryx to True_Kentucky [link] [comments] |

2023.03.08 18:29 *SchlesingerMindy323* **[HIRING] 25 Jobs in KY Hiring Now!**

2023.03.04 16:00 *_call-me-al_* **[Sat, Mar 04 2023] TL;DR β This is what you missed in the last 24 hours on Reddit**

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2023.03.03 21:43 *tyzaginger* **Come check out our art located @ The Makerβs Mill in Somerset, Ky. Itβs a space where artists, artisans, and makers put their eye for excellence, passion, and craftmanship on display.**

submitted by tyzaginger to True_Kentucky [link] [comments] |

2023.03.03 20:33 *rrmdp* **π’ Comfort Keepers - Somerset - London - Corbin, KY is hiring a Caregiver: full-time!**

2023.03.03 19:15 *tyzaginger* **Come check out our art located @ The Makerβs Mill in Somerset, Ky. Itβs a space where artists, artisans, and makers put their eye for excellence, passion, and craftmanship on display.**

submitted by tyzaginger to Kentucky [link] [comments] |

2023.03.01 17:47 *rrmdp* **π’ Comfort Keepers - Somerset - London - Corbin, KY is hiring a Caregiver: full-time!**

2023.02.24 14:35 *johnwinston2* **https://snbc13.com/doug-schutte-louisville-ky-owner-of-the-bards-town-has-died-obituary/**

submitted by johnwinston2 to Louisville [link] [comments]

2023.02.21 18:43 *CooperVsBob* **Pennyroyal Season 2, Episode 4 OutlineβCreated by a Fan**

Apologies for the delay on this. Once I got my hands on a copy of *TlΓΆn, Uqbar, Orbis Tertius* and saw that it mentioned my hometown of Nashville, TN, I went down my own hyperstition rabbit hole.

"It looks like there aren't many great matches for your search." That was the theme of my citation process for this episode. The Pennyroyal crew dug*deep* for The Loop and found some amazing stuff. That said, some topics technically fall under the category of conspiracy theory, so some of the citations are sketchy (random and/or politicized blogs, etc.).

As always, all feedback is welcome. Please point out any and all oversights that may have slipped through. I apologize to anyone whose name I may have misspelled.

My intention for these outlines are to show the patterns of Pennyroyal via a visual medium. There is so much valuable information here it deserves to be preserved and studied. Enjoy!

**Season 1 Outline** **Season 2, Episodes 1-3 Outline**

# Episode 4: The Loop

Signs that appeared all over Somerset two weeks after Season 1 was released

Nathan: Definition and example of "tulpa"

βObserved informational structureβ concept: since Guterma was literally Mr. X before Pennyroyal crew discovered him, what if all of the new, conscious focus on him as a character created a new thought form, or tulpa?

Rowan Elizabeth Cabrales, University of Amsterdam: Experimental time theorist, occult scholar, and philosopher of art and metaphysics

Dark Enlightenment definition: βA disturbing philosophy that brings cyborgs into feudalism and merges Silicon Valleyβs startup ethos with the selective breeding originally proposed by Plato.β References:

**[Note: For the Chuck Hayes section, the events are presented as they appear in the episode, not chronological order]**

1992: Hayes testified regarding the Inslaw case; government portrayed him as a redneck schizophrenic who never worked for CIA; however, upon Hayes' testimony, they enact the National Security Act and seal the testimony

Nathan on FOIA Requests

Another sync: Feral House (James Shelby Downard; Richard Spence, etc.) also published The Octopus

J. Orlin Grabbe

submitted by CooperVsBob to PennyRoyalPodcast [link] [comments]
"It looks like there aren't many great matches for your search." That was the theme of my citation process for this episode. The Pennyroyal crew dug

As always, all feedback is welcome. Please point out any and all oversights that may have slipped through. I apologize to anyone whose name I may have misspelled.

My intention for these outlines are to show the patterns of Pennyroyal via a visual medium. There is so much valuable information here it deserves to be preserved and studied. Enjoy!

- One woman and two men were witnessed posting the signs
- βThis town is a cult; call the FBI; none of this is real; itβs The Truman Showβ
- Included references to local politicians and power figures, i.e. Lester Burns (see Season 1, Episode 3)
- A giant message was spray painted on the roof of an abandoned house on the main road entering downtown, that read βThis town is a cult; the cyber terrorists have taken over; nazis and MK Ultra; call the FBI;β the house was soon demolished
- It didnβt feel like these were fans of the podcast who left the messages, so what could explain the timing of them?

- Nathan acquired black-and-white photos of Alexander Guterma (see Season 1, Episode 6: Mr X)
- A married couple who lived near Mt. Victory mine saw the pictures, and the wife requested to talk to the Pennyroyal crew: she had been seeing Gutermaβs likeness in a recurring dream
- According to the woman, he was not human: "he had never been alive.β

Nathan: Definition and example of "tulpa"

βObserved informational structureβ concept: since Guterma was literally Mr. X before Pennyroyal crew discovered him, what if all of the new, conscious focus on him as a character created a new thought form, or tulpa?

Rowan Elizabeth Cabrales, University of Amsterdam: Experimental time theorist, occult scholar, and philosopher of art and metaphysics

- Hyperstition is a fiction that makes itself real; it does this through nonlinear time dynamics
- The idea of hyperstition is itself hyperstitional: CCRU (Cybernetic culture research unit) discovered the idea of hyperstition in the 1990s
- Since discovering this term, the term itself has started to spiral backwards and forwards, undulating through time, finding instances of itself in the past, where it always
*was*, just not*yet* - TlΓΆn, Uqbar, Orbis Tertius was
*always*a hyperstition once it*was* - Itβs not a causal history of a tradition getting developed progressivelyβit is an acausal, chaotic process: "Iβve tracked it back to Nietzsche but it likely goes much earlier."
- If you can think hyperstitionally, you can spread hyperstition through time: itβs memetic, itβs a time virus, and itβs able to locate other time viruses
- Thereβs a refractive element of a) it and b) its doing
*as*it

Nathan: James Shelby Downard and the Dayton Witch (pp.143-145 of Carnivals of Life and Death, set in 1931) - Tomb of Dr. Simon Pendleton Kramer Ft. Thomas, KY
- Kramer's son hired him to break into the tomb to retrieve items
- Body was ritually beheaded
- Boxes of grave goods included several books with Downardβs name, accompanied by machine with a metal nameplate: βDayton Witchβ and paper tag: βBrunel Universityβ (founded in 1966)

- Land used Lovecraftian Cthulhu mythos incorporated into his academic folklore and framework about human consciousness, time, and belief
- CCRUβs tenets of hyperstition include that there is no difference (in principal) between a universe, a religion, and a hoax
- Reference: Simon OβSullivanβs βAccelerationism, Hyperstition, and Myth Scienceβ
- CCRU's ingredients for hyperstition:

- An element of effective culture that makes itself real
- A fictional quantity functional as a time traveling device
- Coincidence intensifier
- A call to the Old Ones
- Hellier and Pennyroyal both fit this description

- Origin: Sadie Plant and Nick Land meet at 1990s Warwick universityβs Virtual Futures conferences, researching overlap between cybernetics, VR, tech, and how that impacts various disciplines
- Sadie begins CCRU with Landβs help: recruited artists (Kodwo Eshun, jungle music; Orphan Drift, other experimental scholars)
- Begins researching numogrammatics and brings Lovecraft and magick in
- Collective writings of CCRU contains details/historical narratives of where they got their info
- Includes Lemurian Time Wars, Cthulhu club, and leads to work of hyperstition: Cyclonopedia
- Land leans into performative narrative too hard: full experimental/method; inviting entities to channel through him; becomes cult of personality; asked to leave the university; CCRU goes underground
- Eventually gets involved with Mencius Moldbug: Silicon Valley alt right, trans/post humanism, with the intention of βfixingβ society via a form of accelerationism: "finding whatβs wrong with society and creating positive feedback loops until it breaks itself"
- This accelerationism, in turn, leads to something "flipping over," and white supremacism becomes the face of the beast (*but itβs too soon to tell if this elementβwhite supremacyβis still there and/or if it was always there)

Dark Enlightenment definition: βA disturbing philosophy that brings cyborgs into feudalism and merges Silicon Valleyβs startup ethos with the selective breeding originally proposed by Plato.β References:

- Jessica Klein βHereβs the Dark Enlightenment Explainer You Never Wantedβ
- Benjamin Noys, University of Chichester, author of βMalign Velocities: Accelerationism and Capitalismβ
- David Golumbia, professor of New Media at Virginia Commonwealth University

- Accelerationism and Nick Land
- The state of capitalism and inevitable change
- Modern communication theory and Claude Shannon
- Sun Ra combining esoteric mysticism with sound and comm technology (Cosmic Tones for Mental Health; Lanquidity)
- Hyperstition was a response to Sun Raβs myth science (references DJ Spookyβs rhythm science)

- Charles/Chuck Hayes (from Pulaski county, Somerset area: Nancy, KY) was involved in the largest gemstone seizure in American history (1985) smuggling uncut gemstones from Brazil: got busted
- Hayes had a lakeside club in Pulaski County and also owned the Beckett hotel, running honeypot schemes on politicians, an operation started by his mother
- 1975: busted for smuggling rocket launchers and military weapons

- Bought government surplus auctioned computers in Lexington, but government immediately wanted them back after purchase: Hayes refuses
- Government raids him; he sues and wins
- The Promis (Prosecutors Management Information System) software was on the computersβthe government failed to successfully delete it before selling the machines
- Hayes knew about the Promis Software because he used to sell it as a contractor for the CIA

1992: Hayes testified regarding the Inslaw case; government portrayed him as a redneck schizophrenic who never worked for CIA; however, upon Hayes' testimony, they enact the National Security Act and seal the testimony

Nathan on FOIA Requests

- The Pennyroyal Crew filed FOIA requests for Guterma and were successful
- But FOIA requests for Chuck Hayes received a Glomar Response ("we can neither confirm nor deny") from the FBI
- The CIA was happy to comply, but only upon certificate of Hayes' death (Pennyroyal crew have not been able to find any record whatsoever of Hayesβ current status, whether dead or alive)

- After Hayes contracted with the CIA selling The Promis Software, he joined/formed a collective of programmers and hackers called The Fifth Column (unconfirmed?)
- The Fifth Column used The Promis Software to get back at the CIA and blackmail corrupt politicians
- This led to a record amount of voluntary resignations of US politicians
- David L. Baker, journalist, Lexington Herald-Leader, wrote a series of articles in 1990 (need links)
- "Computer Records Accidentally Sold," Lexington Herald-Leader, September 1, 1990.
- "Buyer Says Agents Didn't Find Computer With Secrets," Lexington Herald-Leader, September 5, 1990.
- "Buyer of US Computer Files To Be Disclosed," Lexington Herald-Leader, September 6, 1990.
- "US Says Pulaski Man Tried To Sell Secrets," Lexington Herald-Leader, September 22, 1990

- James Norman, a journalist who had written about Fifth Column and senior editor at Forbes, had been tipped off about the matter by Bill Hamilton, president of Inslaw
- Norman visited Chuck Hayes at Beckett Hotel, where Hayes learned about the connection between Promis, NSA, and Systematics (owned by Jackson Stephens)
- Hayes passed this info along to Danny Casolaro
- Unverifiable myth: Hayes built a supercomputer, linked it to COMSAT, and stored it in a semi truck trailer to drive it around and avoid detection, which was the inspiration for X-Files S5E11 "Kill Switch)," (directed by William Gibson)
- The verifiable truth: Hayes was set up by the FBI for murder for hire; Hayes claimed he knew it was a setup and was joking when he ordered the hit; he further appealed the charge because the FBI informant/double agent who busted him was already on trial for setting up a politician in the same manner (source?)
- Nothing that Danny Casolaro ever did would have ever happened without Hayesβ involvement

- Details on the historical context re. investigative journalism around the time of Octopus
- What The Octopus is, how itβs thriving today and the contributions Casolaro made
- Key names
- The blonde ghost
- Robert Booth Nichols (last person who saw Casolaro? Widely held view)
- Kenn Thomas and Jim Keith, authors of The Octopus
- (Sidenote: Jim Keith, Kerry Thornley, and Ron Bonds all wrote for Illuminet Press and randomly died around the same time/late 90s) [Note: T. Allen Greenfield of Hellier fame wrote for Illuminet at the same time, and is on record saying he slept with a gun under his pillow out of fear. -CooperVsBob]
- Gary Webb, author of Dark Alliance (see Season 1, Episode 6: Mr X)

Another sync: Feral House (James Shelby Downard; Richard Spence, etc.) also published The Octopus

J. Orlin Grabbe

- Responsible fodesigned modern derivatives/futures market
- Prosecutors actually accused Grabbe of creating Chuck Hayes as a fictional entity (I couldn't find anything on this, but I did find a piece asserting that the government accused Hayes of creating Grabbe's online presence as a pen name for himself. - CooperVsBob)
- One of the originators of the concept of cryptocurrency
- Ph.D. in Economics, Harvard, 1981
- Assistant professor of economics at Wharton university
- Author of International Financial Markets, 1986
- Founder of FX Systems, Inc.
- The End of Ordinary Money
- Digital Cash and the Future of Money
- Obituary quotes Grabbe: "A false dichotomy had grown up between order and chaos, 'where induced fear of the horrors of total chaos is the favorite tool of those who seek power to enforce their own order on others.'" Based on this, and his relationship with Hayes, itβs speculative/likely he was a foundemember of The Fifth Column?

- White Paper: Bitcoin: A Peer to Peer Electronic Cash System; Satoshi Nakamoto, 10/31/2008
- Grabbe died on 3/15/2008,
**230**days earlier **230**-day contracts are a common factor in Grabbeβs derivatives strategy (needs source)- Eric Engle: Bitcoin: Digital Finance Law offers a semantic analysis of Satoshiβs white paper and Grabbeβs white paper, and determined they were written by the same author

2023.01.19 08:42 *vladislavsd* **[2023/01/18] Somerset man won't face charges in intruder shooting case (Somerset, KY)**

submitted by vladislavsd to dgu [link] [comments]

2023.01.16 20:58 *Fontrill* **Recommendation after new eps**

After the latest eps Iβd like to recommend Hellier to everyone. Iβm from KY and the things mentioned in these 2 episodes are things old folk talked about when I was a kid.

If yβall donβt wanna watch Hellier at least look at Quartz (the Crystal), Ley-lines and EMF charts related to Somerset, KY.

Old school satanic shit up in those woods and itβs been known for decades in that community.

submitted by Fontrill to Spudmode [link] [comments]
If yβall donβt wanna watch Hellier at least look at Quartz (the Crystal), Ley-lines and EMF charts related to Somerset, KY.

Old school satanic shit up in those woods and itβs been known for decades in that community.

2023.01.16 10:34 *rajusingh79* **DIFFERENTIAL EQUATION**

https://docs.google.com/document/d/1iiDTnS3EU92uTkmO6KyCwqT8FRC1u-10/edit?usp=share_link&ouid=109474854956598892099&rtpof=true&sd=true

## DIFFERENTIAL EQUATION

## BASIC DEFINITION

An equation containing an independent variable, dependent variable and differential coefficients of dependent variable with respect to independent variable is called a differential equation. An ordinary differential equation is one is which there is only one independent variable.

**For Example :**

π· .........(1)

π· .........(2)

π· ..........(3)

π· .........(4)

π· .........(5)

π· .........(6)

## Order and Degree of a Differential Equation:

The **order** of differential equation is the order of highest order derivative appearing in the equation.

For Example :

Orders of differential equations (1), (2), (3), (4), (5) and (6) are 1, 2, 1, 3, 2 and 2 respectively.

The**degree** of differential equation is the degree of the highest order derivative involved in it, when the differential coefficients are free from radicals and fractions (i.e. write differential equations as polynomial in derivatives)

For Example :

Degrees of differential equations (1), (2), (3), (4), (5) and (6) are 1, 1, 1, 4, 3 and 2 respectively.

**Illustration 1:** **Find the order and degree (if defined) of the following differential equations :**

**(i)** **π·** **(ii)** **π·**

**Solution :** (i) The given differential equation can be re-written as

π·. Hence its order is 3 and degree 2.

(ii)**π·****.** Hence its order is 2 and degree 1.

**Formation of differential Equation:**

We know y2 = 4ax is a parabola whose vertex is origin and axis as the x-axis . If a is a parameter, it will represent a family of parabola with the vertex at (0, 0) and axis as y = 0 .

Differentiating y2 = 4ax . . (1)

π· . . (2)

From (1) and (2), y2 = 2yxπ· β y = 2xπ·

This is a differential equation for all the members of the family and it does not contain any parameter ( arbitrary constant).

(i) The differential equation of a family of curves of one parameter is a differential equation of the first order, obtained by eliminating the parameter by differentiation.

(ii) The differential equation of a family of curves of two parameter is a differential equation of the second order, obtained by eliminating the parameter by differentiating the algebraic equation twice. Similar procedure is used to find differential equation of a family of curves of three or more parameter.

**Example -1:** **Find the differential equation of the family of curves y = Aex + Be3x for different values of A and B.**

**Solution:** y = A ex + Be3x . . . . (1)

y1 = Aex + 3Be3x . . . (2) π·

y2 = Aex + 9B3x . . . (3) π·

Eliminating A and B from the above three, we get

π· = 0 β ex e3x π· = 0

β 3y + 4y1 β y2 = 0 β 3y + 4π·

#### SOLUTION OF DIFFERENTIAL EQUATION

The general solution of a differential equation is the relation in the variables x, y obtained by integrating (removing derivatives ) where the relation contains as many arbitrary constants as the order of the equation. The general solution of a differential equation of the first order contains one arbitrary constant while that of the second order contains two arbitrary constants. In the general solution, if particular values of the arbitrary constant are put , we get a particular solution which will give one member of the family of curves.

To solve differential equation of the first order and the first degree:

Simple standard form of differential equation of the first order and first degree are as follows:

#### (i) Variable Separable

Form f(x) dx + Ο(y) dy = 0

**Method:** Integrate it i.e., find β« f(x) dx + β« Ο(y)dy = c

**Example -2:** **Solve** **π·****.**

**Solution:** Given π· β π·

β π·

Integrating, we get ln y β ex =c

#### (ii) Reducible into Variable Separable

**Method:** Sometimes differential equation of the first order cannot be solved directly by variable separation but by some substitution we can reduce it to a differential equation with separable variables.

A differential equation of the π· is solved by writing ax + by + c = t

**Example -3:** **Solve (x β y)2** **π·****.**

**Solution:** Put z = x βy β π· β π·

Now z2π· β π·

β dx = π·, which is in the form of variable separable

Now integrating, we get x = z + π·

β Solution is x =(x β y) + π·

#### (iii) Homogeneous Equation

When π· is equal to a fraction whose numerator and the denominator both are homogeneous functions of x and y of the same degree then the differential equation is said to be homogeneous equation.

i.e. when π·, where f(km, ky) = f(x, y) then this differential equation is said to be homogeneous differential equation .

**Method:** Put y = vx

**Example -4:** **Solve** **π·****.**

**Solution:** π· (homogeneous ) . Put y = vx

β΄ π·

β π·, Integrate

C + lnx = - ln(1 βv2)

β lnkx + ln(1 βv2) =0

β kx(1- v2) = 1 β k(x2 β y2) = x .

#### (iv) Non-homogeneous Differential Equation

Form π·

**Method:** If π·, put x = X + h , y = Y + k such that a1h +b1k + c1 = 0, a2h + b2k +c2 = 0 . In this way the equation becomes homogeneous in X, Y. then use the method for homogeneous equation.

If π·. Put a1x+b1y=v or a2x + b2y = v. The equation in the form of variable separable in x, v.

**Example -5:** **π·****.**

**Solution:** Here π·

Hence we put x- y = v

π· β π·

or, 1 β π· β π· = dx or, π· Integrate

2v + ln (v +2) = x + C, Put the value of v

β΄ x β 2y + ln (x β y +2) = C

**(v)** **Linear Equation**

Form π·, where P(x) and Q(x) are functions of x

**Method:** Multiplying the equation by eβ«P(x)dx, called integrating factor. Then the equation becomes π·

Integrating π·

**(vi)** **Reducible into Linear Equation**

Form R(y)π· + P(x) S(y) = Q(x) , such that π·

**Method:** Put S(y) =z then π·

β΄ The equation becomes π·, which is in the linear form

**Example -6:** **π·****.**

**Solution:** π·

Multiplying both sides by I.F. and integrating

π·

Put π·

β π·

β 2 y π·

**(vii)** **Exact Differential Equations**

Mdx + Ndy = 0, where M and N are functions of x and y. If π·, then the equation is exact and its solution is given by β« Mdx + β« N dy = c

To find the solution of an exact differential equation Mdx + N dy = 0, integrate π· as if y were constant. Also integrate the terms of N that do not contain x w.r.t y. Equate the sum of these integrals to a constant.

**Example -7:** **(x2 βay)dx + (y2 βax)dy = 0.**

**Solution:** Here M = x2 βay

N = y2 βax

π·

π· β π·

β΄ equation is exact

solution is π· = c

π· β ayx + π· = c

or x3 β3axy + y3 = 3c.

**Integrating Factor:**

A factor, which when multiplied to a non exact, differential equation makes it exact, is known as integrating factor e.g. the non exact equation y dx βx dy = 0 can be made exact on multiplying by the factor π·. Hence π· is the integrating factor for this equation.

**Notes:**

In general such a factor exist but except in certain special cases, it is likely to be difficult to determine.

The number of integrating factor for equation M dx + N dy = 0 is infinite.

**Some Useful Results:**

d(xy) = xdy + ydx

π·

π·

π· = π·

d tan-1 π· = π·

π·

d(sin-1 xy) = π·

**Example -8:** **Solve x dy β y dx =** **π·****.**

**Solution:** π· β π·

β π· Integrating

β lnπ· β ln π· = 2 ln x + ln k

y + π· = kx2.

**(viii)** **Linear Differential Equation with constant coefficient**

Differential equation of the form π·, aI β R for I = 0, 1 , 2, 3, . . . , n is called a linear differential equation with constant coefficients.

In order to solve this differential equation, take the auxiliary equation as a0Dn + a1Dn-1 +β¦+ an =0

Find the roots of this equation and then solution of the given differential equation will be as given in the following table

Roots of the auxiliary equation

Corresponding complementary function

1

One real root Ξ±1

C1π·

2.

Two real and differential root Ξ±1 and Ξ±2

C1π· + C2π·

3.

Two real and equal roots Ξ±1 and Ξ±2

(C1 + C2x) π·

4.

Three real and equal roots Ξ±1, Ξ±2, Ξ±3

(C1 + C2x + C3x2 )π·

5.

One pair of imaginary roots Ξ± Β± iΞ²

(C1 cosΞ²x + C2 sinΞ²x) π·

6.

Two Pair of equal imaginary roots Ξ± Β± iΞ² and Ξ± Β± iΞ²

[ (C1 + C2x) cosΞ² + (C1 + C2x)sinΞ²]π·

**Example -9:** **Solve** **π·****.**

**Solution:** Its auxiliary equation is D2 β3D + 2 = 0 β D = 1, D = 2

Hence its solution is y = C1ex + C2e2x

So far only linear differential equation with constant coefficients of form

a0 π· + a1 π·+ β¦.+ an y = 0, aI β R

for i = 0, 1, 2, β¦., n were considered. Now we consider the following form

a0 π· + a1 π·+ β¦.+ an y = X

where X is either constant or functions of x alone.

**Theorem:**

If y = f1(x) is the general solution of a0 π· + a1 π·+ β¦.+ an y = 0

and y = f2(x) is a solution of

a0 π· + a1 π·+ β¦.+ an y = X

Then y = f1(x) + f2(x) is the general solution of a0 π· + a1 π·+ β¦.+ any = X

Expression f1(x) is known as complementary function and the expression f2(x) is known as particular integral. These are denotes by C.F. and P.I. respectively.

The nth derivative of y will be denoted Dny where D stands for π· and n denotes the order of derivative.

If we take Differential Equation:

π· + P1π·+ P2π· + β¦. + Pny = X

then we can write this differential equation in a symbolic form as

Dny + P1Dnβ 1y + P2Dnβ 2y + β¦.+ Pny = X

(Dn + P1Dnβ 1 + P2Dnβ 2 + β¦.+ Pn)y = X

The operator Dn + P1Dnβ 1 + P2Dnβ 2 + β¦.+ Pn is denoted by f(D) so that the equation takes the form f(D)y = X

y = π·

**Methods of finding P.I. :**

**Notes:**

If both m1 and m2 are constants, the expressions (D β m1)(D β m2)y and (Dβ m2)(D β m1)y are equivalent i.e. the expression is independent of the order of operational factors.

π·

We will explain the method with the help of following

**Example -10:** **Solve** **π·****.**

**Solution:** The equation can be written as (D2 β 5D + 6)y = e3x

(D β 3) (D β 2)y = e3x

C.F. = c1 e3x + c2e2x

And P.I. = π·= π·= π·

= e3x π· = x. e3x

β΄ y = c1 e3x + c2e2x + x e3x

P.I. can be found by resolving π·

Into partial functions

π·

β΄ P.I. = π·= π·

= π· = x e3x β e3x .

Second term can be neglected as it is included as it inclined in the first term of a C.F.

**Short Method of Finding P.I. :**

In certain cases, the P.I. can be obtained by methods shorter than the general method.

(i). To find P.I. when X = eax in f(D) y = X, where a is constant

y = π·

π· if f(a) β 0

π· if f(a) = 0 , where f(D) = (D β a)r Ο(D)

**Example -11:** **Solve (D3 β 5D2 + 7D β 3)y = e3x.**

**Solution:** (D β 1)2 (D β 3) y = e3x

C.F. = aex + bx ex + ce3x

And P.I. = π·e3x = π·

β΄ y = aex + bx ex + ce3x + π·

(ii). To find P.I. when X = cosax or sinax

f(D) y = X

y = π· sinax

If f( β a2) β 0 then π· = π·

If f(β a2) = 0 then (D2 + a2) is atleast one factor of f(D2)

Let f (D2) = (D2 + a2)r Ο (D2)

Where Ο(β a2) β 0

β΄ π· = π· = π·

when r = 1 π· sin ax = βπ·

Similarly If f(β a)2 β 0 then π·cos ax = π·cosax

and π·

**Example -12:** **Solve (D2 β 5D + 6)y = sin3x.**

**Solution:** (D β 2) (D β 3)y = sin3x

C.F. = ae2x + be3x

P.I.= π· sin3x = π·

= π· = β (5D β 3). π·sin3x = π·

β΄ ae2x + be3x + π·

(iii). To find the P.I. when X = xm where m β N

f(D) y = xm

y = π·

we will explain the method by taking an example

**Example -13:** **Find P.I. of (D3 + 3D2 + 2D)y = x2.**

**Solution:** P.I. = π· =π· = π·

= π·

= π· = π·

= π· = π·

(iv). To find the value of π·eaxV where βaβ is a constant and V is a function of x

π·

**Example -14:** **Solve (D2 + 2)y = x2 e3x.**

**Solution:** C.F. = a cos π· x + b sinπ·x

P.I. = π· x2 e3x = e3x π·

= π·π· = π·x2

= π·x2 = π·x2

= π·

β΄ a cosπ·x + b sinπ·x + π·

(v). To find π· where V is a function of x

π·

**Example -15:** **Solve (D2 + 4) y = x sin2x.**

**Solution:** C.F. = c1 cos 2x + c2 sin2x

P.I. = π· = π·

= π· = β π·

= β π·= β π·

y = c1 cos2x + c2 sin2x β π·.

**Some Results on Tangents and Normals:**

(i) The equation of the tangent at P(x, y) to the curve y= f(x) is Y β y = π·

(ii) The equation of the normal at point P(x, y) to the curve y = f(x) is

Y β y = π·(X β x )

(iii) The length of the tangent = CP = π·

(iv) The length of the normal = PD =π·

π·

(v) The length of the cartesian subtangent = CA = π·

(vi) The length of the cartesian subnormal = AD = π·

(viii) The initial ordinate of the tangent =OB = y - xπ·

**PROBLEMS**

**SUBJECTIVE**

(C) x2 = 16y β 60 (D) x2 = -16y + 68

Solution: Let the curve be y = f(x). Subnormal at any point = π·

β yπ· = Β±8 β y dy = Β±8dx β π· = Β±8x + c

β y2 = 16 x+2c1, β c1= -8 or y2 = -16x +2c2, β c2= 24

Hence (A), (B) are correct answers.

(C) (y+2) = Ξ»(x+1) (D) (y+1) = Ξ»(x+2) where Ξ» β R.

Solution: Any line passing through the centre of the given circle would meet the circle orthogonally.

Hence (A) is the correct answer.

(C) m = 1, n = 2 (D) None

Solution: π· = c β the differential equation is ;

y = x.π· + π·-3. π·+2

Clearly its order is one and degree 4.

Hence (A) is the correct answer.

(C) π· = y (D) π· = y

Solution: π· = a cosx - b sinx β π· = -a sinx β bcosx = -y

Hence π· + y = 0 .

Hence (A) is the correct answer.

(C) equal to 2y2 (D) equal to x2

Solution: π· = π· for a differential equation

or π· = -1 π·π· = - π·π· = - π·

or π· + π·π· = 0.

Hence (A) is the correct answer.

(C) 3 (D) none of these

Solution: Since the equation is not a polynomial in all the differential coefficient so the degree of equation is not defined.

Hence (D) is the correct answer.

(C) 3, 2 (D) none of these

Solution: Equation of required parabola is of the form y2 = 4a(x βh)

Differentiating, we have 2yπ· = 4a β yπ· = 2a

Required differential equation π·.

Degree of the equation is 1 and order is 2.

Hence (B) is the correct answer.

(C) yy2 + xy12 βxy1 = 0 (D) none of these

Solution: Ellipse centred at origin are given by π· = 1 β¦β¦(1)

where a and b are unknown constants

π· β π·y1 = 0 β¦β¦(2)

Differentiating again, we get

π·(y12 + yy2) = 0 β¦β¦(3)

Multiplying (3) with x and then subtracting from (2) we get

π·( yy1 β xy12 βxyy2) = 0 β xyy2 + xy12 βyy1 = 0.

Hence (B) is the correct answer.

(C) y = 1 + π· (D) y = π· + 11π·

Solution: π· + (3x)y = x

I.F = π·

β΄ Solution of given equation is

yπ· + c = π· + c

If curve passes through (0, 4), then

4 βπ· = c β c = π·

y = π· β 3y = 1 + 11π·.

Hence (A) is the correct answer.

(C) x βy + c = = log (3x β4y β3) (D) x βy + c = log (3x β4y + 1)

Solution: Let 3x β4y = z

3 β4π· β π·

Therefore the given equation π·

β βπ·β βπ·dz = dx

β βz + 4 log (z + 1) = x + c β log (3x β4y + 1) = x βy + c.

Hence (B) is the correct answer.

(A) 5 (B) 4

(C) 3 (D) none of these

Solution: y = (c1 + c4) ex + c2 e2x + c3 e3x + c4π·

y = c1 ex + c2 e2x + c3 e3x + c4 π·

y = k1 ex + k2 e2x + k3 e3x + k4

Therefore 4 obituary constants

Hence (B) is the correct answer.

(C) ln xy (D) none of these

Solution: I.F. = π· = ln y

Hence (B) is the correct answer.

(A) π· (B) π·

(C) π· (D) π·

Solution: y = A e2x + b eβ2x β π· = 2 (A e2x β b eβ2x)

π· = 4 (A e2x + b eβ2x) = ln y

Hence (C) is the correct answer.

(C) log tan π· = c β 2 cos π· (D) none of these

Solution: π·= β 2 cosπ·

β π· β c β 2 sin π· = log tan π·

Hence (A) is the correct answer.

(C) tan π· = kx (D) none of these

Solution: π· = π·

put y = vx β v + x π· = v + tan v

cot v dv = π·

Integrating, we get ln sin v = ln x + ln k β sin π· = kx

Hence (A) is the correct answer.

(C) sinβ1 x = c sinβ1 y (D) (sinβ1 x) (sinβ1 y) = c

Solution: π· β sinβ1 y + sinβ1 x = c

Hence (B) is the correct answer.

(C) y = π·eβ2x + cx + d (D) y = eβ2x + cx2 + d

Solution: π· = eβ2x, π· + k1

Integrating, y = π· + k1 x + k2 β y = π· + cx + d

Hence (C) is the correct answer.

(C) xy = π·x4 + c (D) y β x = π·x4 + c

Solution: π· = x2

I.F. = π· = x

Therefore solution is xy = π·+ c

Hence (C) is the correct answer.

(C) π· + y = 0 (D) none of these

Solution: y = A e3x + B e5x

yβ² = 3 A e3x + 5 B e5x

yβ³ = 9 A e3x + 25 B e5x

therefore yβ³ β 8y + 15y = 0

Hence (A) is the correct answer.

(C) pair of straight line (D) none of these

Solution: π· β ln y = ln y2 + ln c β y2 = kx

it represents a family of parabola

Hence (A) is the correct answer.

(C) y tanx = x (D) none of these

Solution: π·β y tan x = sec x

I.F. = π· = cos x

y cos x = β« sec x cos x dx = x + c

y cos x = x + c

At (0, ΞΈ), y cos x = x Since c = 0

Hence (B) is the correct answer.

(B) a set of circles with centre on x-axis

(C) a set of ellipses

(D) none of these

Solution: y dy = (a β x) dx

π·= ax β π· β x2 + y2 = 2ax β x2 + 2ax + y2 = 0

It represents a set of sides with centre on xβaxis

Hence (B) is the correct answer.

(C) e6 + 1 (D) loge 6

Solution: e2y dy = dx β π· + c = x β c = 5 β π·

At. y = 3 π· = x β x = π·

Hence (B) is the correct answer.

(C) xy = x2 + x + 1 (D) xy = y + 1

Solution: π· β y = x + π· + c

At (2, 7/2) π· = 2 + π·+ c β c = 1

Therefore y = x + π· + 1 β xy = x2 + x + 1

Hence (C) is the correct answer.

(C) e (D) e2

Solution: π· = ex dx β ln y = ex + c

At x = 0, y = 1; c = 0

ln y = ex

Therefore At x = 1, y = ec

Hence (C) is the correct answer.

(C) 4 (D) None of these

Solutions : The general equation of all conics with centre at origin is ax2 + 2hxy + by2 = 1

Since it has three arbitrary constants, the differential equation is of order 3.

Hence (B) is the correct answer.

(C) 3 (D) 4

Solutions : The general equation of all conics with axes as coordinate axes is ax2 + by2 = 1.

Since it has two arbitrary constants, the differential equation is of order 2.

Hence (B) is the correct answer.

(C) π· (D) π·

Solutions : The general equation of non-vertical lines in a plane is ax + by = 1, b β 0

π· π· π·

Hence (C) is the correct answer.

(C) π· (D) π·

Solutions : ax = by = 0, a β 0 π·.

Hence (D) is the correct answer.

(C) 1, 3 (D) 1, 4

Solutions : Equation of any tangent to x2 = 4y is π·, where m is arbitrary constant.

π· π·

β΄ Putting this value of m in π·, we get

π· π·

Which is a differential equation of order 1 and degree 2.

Hence (A) is the correct answer.

(C) 3e2 (D) 2e3

Solutions : π· β log f(x) = x + c

Since f(1) = 2

β΄ log 2 = 1 + c

β΄ log f(x) = x + log 2 β 1

β΄ log f(3) = 3 + log 2 β 1 = 2 + log 2

β f(3) = e2+log2 = e2 . elog2 = 2e2.

Hence (B) is the correct answer.

(C) 6xy = 3x3 β 29x β 6 (D) None of these

Solutions : π· π·

It passes through (3, 9)

π· π·

π·

Hence (C) is the correct answer.

(C) y2 = cosx + 1 (D) None of these

Solutions : On dividing by sin x,

π·

Put y2 = v π·

I.F. = eβ«cotxdx = elog sinx = sin x

β΄ Solution is v. sin x β«sinx . (2 cos x) dx + c

β y2 sin x = sin2x + c

When π·, y = 1, then c = 0

β΄ y2 = sin x.

Hence (A) is the correct answer.

(C) π· (D) None of these

Solutions : π· β P = βx, Q = 1

I.F. = π·

β΄ Solution is π·

Hence (D) is the correct answer.

(C) ex(x β 1) + 1 = y (D) None of these

Solutions : π·

β eβy dy = (ex + eβx)dx β eβy = ex β eβx + c

Hence (D) is the correct answer.

(C) π· (D) π·

Solutions : y = eβx (A cos x + B sin x) ....(1)

π· (βA sin x + B cos x) β eβx (A cos x + B sin x)

π·(βA sin x + B cos x) β y .....(2)

π·(βA cos x β B sin x) β eβx (βA sin x + B cos x) π·

Using (1) and (2), we get

π·

Hence (C) is the correct answer.

(C) π· (D) π·

Solutions : π· .....(1)

Put y = vx π·

β΄ (1) becomes π·

π·

π· π·

π·.

Hence (A) is the correct answer.

(C) 1, 4 (D) 2, 4

Solutions : On simplification the equation becomes π·

Hence (D) is the correct answer.

(C) π· (D) 6

Solutions : Clearly the equation is of order 2.

Hence (A) is the correct answer.

(A) m = 3, n = 3 (B) m = 3, n = 2

(C) m = 3, n = 5 (D) m = 3, n = 1

Solutions : The given differential equation can be written as

π·

β m = 3, n = 2

Hence (B) is the correct answer.

(C) parabolas (D) ellipses

Solutions : π· π·

β 2log(y + 3) = logx + log c

β (y + 3)2 = ex, which represent parabolas.

Hence (C) is the correct answer.

(C) cos x tan y = c (D) cos x sin y = c

Solutions : Given equation can be written as :

π·

π·

π·

π·

π·

Hence (A) is the correct answer.

(C) y = emx + ceβax (D) (a + m) y = emx + ceβax

Solutions : I.F. = eβ«adx = eax

β΄ Solution is π·

π·

β (a + m)y = emx + c1 (a + m) eβax

β (a + m) y = emx + ceβax

where c = c1 (a + m)

Hence (D) is the correct answer.

(C) sec x (D) sin x

Solutions : π·

Here P = tan x, Q = sec x

β΄ I. F. = eβ«tan x dx = elog secx = sec x.

Hence (C) is the correct answer.

(B) straight line passing through origin

(C) parabola whose vertex is at origin

(D) circle whose centre is at origin

Solutions : x dy β y dx = 0 π·

β log y β log x = log c

π·

π· or y = cx which is a straight line.

Hence (B) is the correct answer.

(C) log(log x) (D) x

Solutions : π·

π· π·

π·

Hence (B) is the correct answer.

(C) (x β y)ex+y = 1 (D) (x β c)ex+y + 1 = 0

Solutions : π· π·

Put eβy = z π· π·

π· β΄ P = β1, Q = βex.

β΄ I.F. = eβ«β1dx = eβx

Solution is z . eβx = β« βex . eβx dx + c = βx + c

β eβy . eβx = c β x β eβ(x+y) = c β x

β (x β c) ex+y + 1 = 0

Hence (D) is the correct answer.

(C) 1 (D) None of these

Solutions : There is only one arbitrary constant.

So, degree of the differential equation is one.

Hence (C) is the correct answer.

(C) βx (D) ex

Solutions : I.F. π·

Hence (A) is the correct answer.

(C) π· (D) π·

Solutions : π·

π·

π·

Hence (D) is the correct answer.

(C) y = 4x (D) y + 4x + 1 = v

Solutions : y + 4x + 1 = v

Hence (D) is the correct answer.

(C) 1 (D) π·

Solutions : π·

β slope at (1, 0) = π·

Hence (C) is the correct answer.

(C) y = 2x β 4 (D) y = 2x2 β 4

Solutions : Ans is (C)

(C) xy2 = 2y5 + c (D) x(y2 + xy) = 0

Solutions : π· β β2x + 10y3 β π·

π·

β΄ I.F. = π·

β΄ Solution is π·

π·.

Hence (C) is the correct answer.

(C) 2 log10 e (D) π·

Solutions : π· π·

β log(x + 1) = t + c ....(1)

Initially, when t = 0, x = 0

β΄ c = 0

β΄ log (x + 1) = t

When x = 99, then t = loge (100) = 2 loge 10.

Hence (B) is the correct answer.

(C) circle (D) parabola

Solutions : π· β ydy = xdx

π· β y2 β x2 = 2c

which is a rectangular hyperbola

Hence (B) is the correct answer.

(C) x = π· (D) None of these

Solution: The given equation is dx β x (ydx+ xdy )= x5y4(ydx + xdy)

π· β lnx = xy + π·

β x = π·.

Hence (A) is the correct answer.

(C) β13, 12 (D) 12, β13

Solution: On differentiatingπ·, w.r.t x we get

π·π·

Putting these values in π·, we get

π·.

On solving we get A= β13 and B = β12

Hence (B) is the correct answer.

(C) π· (D) π·

Solutions : If (0, a) is centre on y-axis, then its radius is a because it passes through origin.

β΄ Equation of circle is x2 + (y β a)2 = a2

β x2 + y2 β 2ay = 0 ....(1)

π· .....(2)

Using (1) in (2), π·

π·

π·

Hence (A) is the correct answer.

(C) π· (D) π·

Solutions : π·

Put y = vx β π·

π·

π·

β log(log v) = logx + logc = logcx

β log v = cx π·.

Hence (D) is the correct answer.

(A) 2 (B) 3

(C) 4 (D) None of these

Solution: Putting x = tanΞΈ and y = tanΟ. Then equation becomes

secΞΈ + secΟ = A (tanΞΈ secΟ β tanΟsecΞΈ).

β π·βπ·

βπ·β π· βtan β1x β tan β1y = 2cot β1A

Differentiating, we get π·

Which is a differential equation of degree 1.

Hence (D) is the correct answer.

submitted by rajusingh79 to u/rajusingh79 [link] [comments]
π· .........(1)

π· .........(2)

π· ..........(3)

π· .........(4)

π· .........(5)

π· .........(6)

For Example :

Orders of differential equations (1), (2), (3), (4), (5) and (6) are 1, 2, 1, 3, 2 and 2 respectively.

The

For Example :

Degrees of differential equations (1), (2), (3), (4), (5) and (6) are 1, 1, 1, 4, 3 and 2 respectively.

π·. Hence its order is 3 and degree 2.

(ii)

We know y2 = 4ax is a parabola whose vertex is origin and axis as the x-axis . If a is a parameter, it will represent a family of parabola with the vertex at (0, 0) and axis as y = 0 .

Differentiating y2 = 4ax . . (1)

π· . . (2)

From (1) and (2), y2 = 2yxπ· β y = 2xπ·

This is a differential equation for all the members of the family and it does not contain any parameter ( arbitrary constant).

(i) The differential equation of a family of curves of one parameter is a differential equation of the first order, obtained by eliminating the parameter by differentiation.

(ii) The differential equation of a family of curves of two parameter is a differential equation of the second order, obtained by eliminating the parameter by differentiating the algebraic equation twice. Similar procedure is used to find differential equation of a family of curves of three or more parameter.

y1 = Aex + 3Be3x . . . (2) π·

y2 = Aex + 9B3x . . . (3) π·

Eliminating A and B from the above three, we get

π· = 0 β ex e3x π· = 0

β 3y + 4y1 β y2 = 0 β 3y + 4π·

To solve differential equation of the first order and the first degree:

Simple standard form of differential equation of the first order and first degree are as follows:

β π·

Integrating, we get ln y β ex =c

A differential equation of the π· is solved by writing ax + by + c = t

Now z2π· β π·

β dx = π·, which is in the form of variable separable

Now integrating, we get x = z + π·

β Solution is x =(x β y) + π·

i.e. when π·, where f(km, ky) = f(x, y) then this differential equation is said to be homogeneous differential equation .

β΄ π·

β π·, Integrate

C + lnx = - ln(1 βv2)

β lnkx + ln(1 βv2) =0

β kx(1- v2) = 1 β k(x2 β y2) = x .

If π·. Put a1x+b1y=v or a2x + b2y = v. The equation in the form of variable separable in x, v.

Hence we put x- y = v

π· β π·

or, 1 β π· β π· = dx or, π· Integrate

2v + ln (v +2) = x + C, Put the value of v

β΄ x β 2y + ln (x β y +2) = C

Form π·, where P(x) and Q(x) are functions of x

Integrating π·

Form R(y)π· + P(x) S(y) = Q(x) , such that π·

β΄ The equation becomes π·, which is in the linear form

Multiplying both sides by I.F. and integrating

π·

Put π·

β π·

β 2 y π·

Mdx + Ndy = 0, where M and N are functions of x and y. If π·, then the equation is exact and its solution is given by β« Mdx + β« N dy = c

To find the solution of an exact differential equation Mdx + N dy = 0, integrate π· as if y were constant. Also integrate the terms of N that do not contain x w.r.t y. Equate the sum of these integrals to a constant.

N = y2 βax

π·

π· β π·

β΄ equation is exact

solution is π· = c

π· β ayx + π· = c

or x3 β3axy + y3 = 3c.

A factor, which when multiplied to a non exact, differential equation makes it exact, is known as integrating factor e.g. the non exact equation y dx βx dy = 0 can be made exact on multiplying by the factor π·. Hence π· is the integrating factor for this equation.

In general such a factor exist but except in certain special cases, it is likely to be difficult to determine.

The number of integrating factor for equation M dx + N dy = 0 is infinite.

d(xy) = xdy + ydx

π·

π·

π· = π·

d tan-1 π· = π·

π·

d(sin-1 xy) = π·

β π· Integrating

β lnπ· β ln π· = 2 ln x + ln k

y + π· = kx2.

Differential equation of the form π·, aI β R for I = 0, 1 , 2, 3, . . . , n is called a linear differential equation with constant coefficients.

In order to solve this differential equation, take the auxiliary equation as a0Dn + a1Dn-1 +β¦+ an =0

Find the roots of this equation and then solution of the given differential equation will be as given in the following table

Roots of the auxiliary equation

Corresponding complementary function

1

One real root Ξ±1

C1π·

2.

Two real and differential root Ξ±1 and Ξ±2

C1π· + C2π·

3.

Two real and equal roots Ξ±1 and Ξ±2

(C1 + C2x) π·

4.

Three real and equal roots Ξ±1, Ξ±2, Ξ±3

(C1 + C2x + C3x2 )π·

5.

One pair of imaginary roots Ξ± Β± iΞ²

(C1 cosΞ²x + C2 sinΞ²x) π·

6.

Two Pair of equal imaginary roots Ξ± Β± iΞ² and Ξ± Β± iΞ²

[ (C1 + C2x) cosΞ² + (C1 + C2x)sinΞ²]π·

Hence its solution is y = C1ex + C2e2x

So far only linear differential equation with constant coefficients of form

a0 π· + a1 π·+ β¦.+ an y = 0, aI β R

for i = 0, 1, 2, β¦., n were considered. Now we consider the following form

a0 π· + a1 π·+ β¦.+ an y = X

where X is either constant or functions of x alone.

If y = f1(x) is the general solution of a0 π· + a1 π·+ β¦.+ an y = 0

and y = f2(x) is a solution of

a0 π· + a1 π·+ β¦.+ an y = X

Then y = f1(x) + f2(x) is the general solution of a0 π· + a1 π·+ β¦.+ any = X

Expression f1(x) is known as complementary function and the expression f2(x) is known as particular integral. These are denotes by C.F. and P.I. respectively.

The nth derivative of y will be denoted Dny where D stands for π· and n denotes the order of derivative.

If we take Differential Equation:

π· + P1π·+ P2π· + β¦. + Pny = X

then we can write this differential equation in a symbolic form as

Dny + P1Dnβ 1y + P2Dnβ 2y + β¦.+ Pny = X

(Dn + P1Dnβ 1 + P2Dnβ 2 + β¦.+ Pn)y = X

The operator Dn + P1Dnβ 1 + P2Dnβ 2 + β¦.+ Pn is denoted by f(D) so that the equation takes the form f(D)y = X

y = π·

If both m1 and m2 are constants, the expressions (D β m1)(D β m2)y and (Dβ m2)(D β m1)y are equivalent i.e. the expression is independent of the order of operational factors.

π·

We will explain the method with the help of following

(D β 3) (D β 2)y = e3x

C.F. = c1 e3x + c2e2x

And P.I. = π·= π·= π·

= e3x π· = x. e3x

β΄ y = c1 e3x + c2e2x + x e3x

P.I. can be found by resolving π·

Into partial functions

π·

β΄ P.I. = π·= π·

= π· = x e3x β e3x .

Second term can be neglected as it is included as it inclined in the first term of a C.F.

In certain cases, the P.I. can be obtained by methods shorter than the general method.

(i). To find P.I. when X = eax in f(D) y = X, where a is constant

y = π·

π· if f(a) β 0

π· if f(a) = 0 , where f(D) = (D β a)r Ο(D)

C.F. = aex + bx ex + ce3x

And P.I. = π·e3x = π·

β΄ y = aex + bx ex + ce3x + π·

(ii). To find P.I. when X = cosax or sinax

f(D) y = X

y = π· sinax

If f( β a2) β 0 then π· = π·

If f(β a2) = 0 then (D2 + a2) is atleast one factor of f(D2)

Let f (D2) = (D2 + a2)r Ο (D2)

Where Ο(β a2) β 0

β΄ π· = π· = π·

when r = 1 π· sin ax = βπ·

Similarly If f(β a)2 β 0 then π·cos ax = π·cosax

and π·

C.F. = ae2x + be3x

P.I.= π· sin3x = π·

= π· = β (5D β 3). π·sin3x = π·

β΄ ae2x + be3x + π·

(iii). To find the P.I. when X = xm where m β N

f(D) y = xm

y = π·

we will explain the method by taking an example

= π·

= π· = π·

= π· = π·

(iv). To find the value of π·eaxV where βaβ is a constant and V is a function of x

π·

P.I. = π· x2 e3x = e3x π·

= π·π· = π·x2

= π·x2 = π·x2

= π·

β΄ a cosπ·x + b sinπ·x + π·

(v). To find π· where V is a function of x

π·

P.I. = π· = π·

= π· = β π·

= β π·= β π·

y = c1 cos2x + c2 sin2x β π·.

(i) The equation of the tangent at P(x, y) to the curve y= f(x) is Y β y = π·

(ii) The equation of the normal at point P(x, y) to the curve y = f(x) is

Y β y = π·(X β x )

(iii) The length of the tangent = CP = π·

(iv) The length of the normal = PD =π·

π·

(v) The length of the cartesian subtangent = CA = π·

(vi) The length of the cartesian subnormal = AD = π·

(viii) The initial ordinate of the tangent =OB = y - xπ·

- A curve that passes through (2, 4) and having subnormal of constant length of 8 units can be;

(C) x2 = 16y β 60 (D) x2 = -16y + 68

Solution: Let the curve be y = f(x). Subnormal at any point = π·

β yπ· = Β±8 β y dy = Β±8dx β π· = Β±8x + c

β y2 = 16 x+2c1, β c1= -8 or y2 = -16x +2c2, β c2= 24

Hence (A), (B) are correct answers.

- Equation of a curve that would cut x2 + y2 - 2x - 4y - 15 = 0 orthogonally can be ;

(C) (y+2) = Ξ»(x+1) (D) (y+1) = Ξ»(x+2) where Ξ» β R.

Solution: Any line passing through the centre of the given circle would meet the circle orthogonally.

Hence (A) is the correct answer.

- Let m and n be the order and the degree of the differential equation whose solution is y = cx +c2- 3c3/2 +2, where c is a parameter. Then,

(C) m = 1, n = 2 (D) None

Solution: π· = c β the differential equation is ;

y = x.π· + π·-3. π·+2

Clearly its order is one and degree 4.

Hence (A) is the correct answer.

- y = a sinx + b cosx is the solution of differential equation :

(C) π· = y (D) π· = y

Solution: π· = a cosx - b sinx β π· = -a sinx β bcosx = -y

Hence π· + y = 0 .

Hence (A) is the correct answer.

- For any differential function y = f(x), the value of π· + π·π· is :

(C) equal to 2y2 (D) equal to x2

Solution: π· = π· for a differential equation

or π· = -1 π·π· = - π·π· = - π·

or π· + π·π· = 0.

Hence (A) is the correct answer.

- The degree of differential equation π· is :

(C) 3 (D) none of these

Solution: Since the equation is not a polynomial in all the differential coefficient so the degree of equation is not defined.

Hence (D) is the correct answer.

- The degree and order of the differential equation of all parabolas whose axis is x-axis are :

(C) 3, 2 (D) none of these

Solution: Equation of required parabola is of the form y2 = 4a(x βh)

Differentiating, we have 2yπ· = 4a β yπ· = 2a

Required differential equation π·.

Degree of the equation is 1 and order is 2.

Hence (B) is the correct answer.

- The differential equation of all ellipses centred at origin is :

(C) yy2 + xy12 βxy1 = 0 (D) none of these

Solution: Ellipse centred at origin are given by π· = 1 β¦β¦(1)

where a and b are unknown constants

π· β π·y1 = 0 β¦β¦(2)

Differentiating again, we get

π·(y12 + yy2) = 0 β¦β¦(3)

Multiplying (3) with x and then subtracting from (2) we get

π·( yy1 β xy12 βxyy2) = 0 β xyy2 + xy12 βyy1 = 0.

Hence (B) is the correct answer.

- Particular solution of y1 + 3xy = x which passes through (0, 4) is :

(C) y = 1 + π· (D) y = π· + 11π·

Solution: π· + (3x)y = x

I.F = π·

β΄ Solution of given equation is

yπ· + c = π· + c

If curve passes through (0, 4), then

4 βπ· = c β c = π·

y = π· β 3y = 1 + 11π·.

Hence (A) is the correct answer.

- Solution of equation π· is :

(C) x βy + c = = log (3x β4y β3) (D) x βy + c = log (3x β4y + 1)

Solution: Let 3x β4y = z

3 β4π· β π·

Therefore the given equation π·

β βπ·β βπ·dz = dx

β βz + 4 log (z + 1) = x + c β log (3x β4y + 1) = x βy + c.

Hence (B) is the correct answer.

- The order of the differential equation, whose general solution is

(A) 5 (B) 4

(C) 3 (D) none of these

Solution: y = (c1 + c4) ex + c2 e2x + c3 e3x + c4π·

y = c1 ex + c2 e2x + c3 e3x + c4 π·

y = k1 ex + k2 e2x + k3 e3x + k4

Therefore 4 obituary constants

Hence (B) is the correct answer.

- I.F. for y ln π· is :

(C) ln xy (D) none of these

Solution: I.F. = π· = ln y

Hence (B) is the correct answer.

- Which one of the following is a differential equation of the family of curves

(A) π· (B) π·

(C) π· (D) π·

Solution: y = A e2x + b eβ2x β π· = 2 (A e2x β b eβ2x)

π· = 4 (A e2x + b eβ2x) = ln y

Hence (C) is the correct answer.

- Solution of π· is :

(C) log tan π· = c β 2 cos π· (D) none of these

Solution: π·= β 2 cosπ·

β π· β c β 2 sin π· = log tan π·

Hence (A) is the correct answer.

- Solution of π· is :

(C) tan π· = kx (D) none of these

Solution: π· = π·

put y = vx β v + x π· = v + tan v

cot v dv = π·

Integrating, we get ln sin v = ln x + ln k β sin π· = kx

Hence (A) is the correct answer.

- Solution of π· = 0 is :

(C) sinβ1 x = c sinβ1 y (D) (sinβ1 x) (sinβ1 y) = c

Solution: π· β sinβ1 y + sinβ1 x = c

Hence (B) is the correct answer.

- General solution of π· = eβ2x is :

(C) y = π·eβ2x + cx + d (D) y = eβ2x + cx2 + d

Solution: π· = eβ2x, π· + k1

Integrating, y = π· + k1 x + k2 β y = π· + cx + d

Hence (C) is the correct answer.

- Solution of π· = x2 is :

(C) xy = π·x4 + c (D) y β x = π·x4 + c

Solution: π· = x2

I.F. = π· = x

Therefore solution is xy = π·+ c

Hence (C) is the correct answer.

- Differential equation associated with primitive y = Ae3x + Be5x :

(C) π· + y = 0 (D) none of these

Solution: y = A e3x + B e5x

yβ² = 3 A e3x + 5 B e5x

yβ³ = 9 A e3x + 25 B e5x

therefore yβ³ β 8y + 15y = 0

Hence (A) is the correct answer.

- The curve satisfying y = 2x π· is a :

(C) pair of straight line (D) none of these

Solution: π· β ln y = ln y2 + ln c β y2 = kx

it represents a family of parabola

Hence (A) is the correct answer.

- The equation of the curve through the origin satisfying the equation dy = (sec x + y tanx) dx is :

(C) y tanx = x (D) none of these

Solution: π·β y tan x = sec x

I.F. = π· = cos x

y cos x = β« sec x cos x dx = x + c

y cos x = x + c

At (0, ΞΈ), y cos x = x Since c = 0

Hence (B) is the correct answer.

- The differential equation yπ· + x = a (where βaβ is a constant) represents :

(B) a set of circles with centre on x-axis

(C) a set of ellipses

(D) none of these

Solution: y dy = (a β x) dx

π·= ax β π· β x2 + y2 = 2ax β x2 + 2ax + y2 = 0

It represents a set of sides with centre on xβaxis

Hence (B) is the correct answer.

- If π· = eβ2y and y = 0 when x = 5, then value of x where y = 3 is given by :

(C) e6 + 1 (D) loge 6

Solution: e2y dy = dx β π· + c = x β c = 5 β π·

At. y = 3 π· = x β x = π·

Hence (B) is the correct answer.

- The equation of curve passing through π· and having slope 1 β π· at (x, y) is :

(C) xy = x2 + x + 1 (D) xy = y + 1

Solution: π· β y = x + π· + c

At (2, 7/2) π· = 2 + π·+ c β c = 1

Therefore y = x + π· + 1 β xy = x2 + x + 1

Hence (C) is the correct answer.

- Given that π· = yex such that x = 0, y = e. The value of y (y > 0) when x = 1 will be :

(C) e (D) e2

Solution: π· = ex dx β ln y = ex + c

At x = 0, y = 1; c = 0

ln y = ex

Therefore At x = 1, y = ec

Hence (C) is the correct answer.

- The differential equation of all conics with centre at origin is of order

(C) 4 (D) None of these

Solutions : The general equation of all conics with centre at origin is ax2 + 2hxy + by2 = 1

Since it has three arbitrary constants, the differential equation is of order 3.

Hence (B) is the correct answer.

- The differential equation of all conics with the coordinate axes, is of order

(C) 3 (D) 4

Solutions : The general equation of all conics with axes as coordinate axes is ax2 + by2 = 1.

Since it has two arbitrary constants, the differential equation is of order 2.

Hence (B) is the correct answer.

- The differential equation of all non-vertical lines in a plane is :

(C) π· (D) π·

Solutions : The general equation of non-vertical lines in a plane is ax + by = 1, b β 0

π· π· π·

Hence (C) is the correct answer.

- The differential equation of all non-horizontal lines in a plane is :

(C) π· (D) π·

Solutions : ax = by = 0, a β 0 π·.

Hence (D) is the correct answer.

- The degree and order of the differential equation of all tangent lines to the parabola x2 = 4y is :

(C) 1, 3 (D) 1, 4

Solutions : Equation of any tangent to x2 = 4y is π·, where m is arbitrary constant.

π· π·

β΄ Putting this value of m in π·, we get

π· π·

Which is a differential equation of order 1 and degree 2.

Hence (A) is the correct answer.

- If f(x) = f'(x) and f(1) = 2, then f(3) =

(C) 3e2 (D) 2e3

Solutions : π· β log f(x) = x + c

Since f(1) = 2

β΄ log 2 = 1 + c

β΄ log f(x) = x + log 2 β 1

β΄ log f(3) = 3 + log 2 β 1 = 2 + log 2

β f(3) = e2+log2 = e2 . elog2 = 2e2.

Hence (B) is the correct answer.

- Equation of the curve passing through (3, 9) which satisfies the differential equation π· is :

(C) 6xy = 3x3 β 29x β 6 (D) None of these

Solutions : π· π·

It passes through (3, 9)

π· π·

π·

Hence (C) is the correct answer.

- Solution of π·, y = 1 is given by

(C) y2 = cosx + 1 (D) None of these

Solutions : On dividing by sin x,

π·

Put y2 = v π·

I.F. = eβ«cotxdx = elog sinx = sin x

β΄ Solution is v. sin x β«sinx . (2 cos x) dx + c

β y2 sin x = sin2x + c

When π·, y = 1, then c = 0

β΄ y2 = sin x.

Hence (A) is the correct answer.

- The general solution of the equation π· is :

(C) π· (D) None of these

Solutions : π· β P = βx, Q = 1

I.F. = π·

β΄ Solution is π·

Hence (D) is the correct answer.

- Solution of π· is :

(C) ex(x β 1) + 1 = y (D) None of these

Solutions : π·

β eβy dy = (ex + eβx)dx β eβy = ex β eβx + c

Hence (D) is the correct answer.

- If y = eβx (A cosx + B sin x), then y satisfies

(C) π· (D) π·

Solutions : y = eβx (A cos x + B sin x) ....(1)

π· (βA sin x + B cos x) β eβx (A cos x + B sin x)

π·(βA sin x + B cos x) β y .....(2)

π·(βA cos x β B sin x) β eβx (βA sin x + B cos x) π·

Using (1) and (2), we get

π·

Hence (C) is the correct answer.

- The solution of π· is :

(C) π· (D) π·

Solutions : π· .....(1)

Put y = vx π·

β΄ (1) becomes π·

π·

π· π·

π·.

Hence (A) is the correct answer.

- Order and degree of differential equation π· are :

(C) 1, 4 (D) 2, 4

Solutions : On simplification the equation becomes π·

Hence (D) is the correct answer.

- The order of differential equation π· is :

(C) π· (D) 6

Solutions : Clearly the equation is of order 2.

Hence (A) is the correct answer.

- If m and n are order and degree of the equation

(A) m = 3, n = 3 (B) m = 3, n = 2

(C) m = 3, n = 5 (D) m = 3, n = 1

Solutions : The given differential equation can be written as

π·

β m = 3, n = 2

Hence (B) is the correct answer.

- The solution of the differential equation π· represent

(C) parabolas (D) ellipses

Solutions : π· π·

β 2log(y + 3) = logx + log c

β (y + 3)2 = ex, which represent parabolas.

Hence (C) is the correct answer.

- Solution of differential equation dy β sinx siny dx = 0 is :

(C) cos x tan y = c (D) cos x sin y = c

Solutions : Given equation can be written as :

π·

π·

π·

π·

π·

Hence (A) is the correct answer.

- Solution of differential equation π· is :

(C) y = emx + ceβax (D) (a + m) y = emx + ceβax

Solutions : I.F. = eβ«adx = eax

β΄ Solution is π·

π·

β (a + m)y = emx + c1 (a + m) eβax

β (a + m) y = emx + ceβax

where c = c1 (a + m)

Hence (D) is the correct answer.

- Integrating factor of the differential equation π· is :

(C) sec x (D) sin x

Solutions : π·

Here P = tan x, Q = sec x

β΄ I. F. = eβ«tan x dx = elog secx = sec x.

Hence (C) is the correct answer.

- Solution of differential equation xdy β ydx = 0 represents

(B) straight line passing through origin

(C) parabola whose vertex is at origin

(D) circle whose centre is at origin

Solutions : x dy β y dx = 0 π·

β log y β log x = log c

π·

π· or y = cx which is a straight line.

Hence (B) is the correct answer.

- The integrating factor of the differential equation π· is given by

(C) log(log x) (D) x

Solutions : π·

π· π·

π·

Hence (B) is the correct answer.

- The solution of π· is :

(C) (x β y)ex+y = 1 (D) (x β c)ex+y + 1 = 0

Solutions : π· π·

Put eβy = z π· π·

π· β΄ P = β1, Q = βex.

β΄ I.F. = eβ«β1dx = eβx

Solution is z . eβx = β« βex . eβx dx + c = βx + c

β eβy . eβx = c β x β eβ(x+y) = c β x

β (x β c) ex+y + 1 = 0

Hence (D) is the correct answer.

- Family y = Ax + A3 of curves is represented by the differential equation of degree

(C) 1 (D) None of these

Solutions : There is only one arbitrary constant.

So, degree of the differential equation is one.

Hence (C) is the correct answer.

- Integrating factor of π·; (x > 0) is :

(C) βx (D) ex

Solutions : I.F. π·

Hence (A) is the correct answer.

- If integrating factor of x(1 β x2) dy + (2x2y β y β ax3) dx = 0 is eβ«Pdx, then P is :

(C) π· (D) π·

Solutions : π·

π·

π·

Hence (D) is the correct answer.

- For solving π·, suitable substitution is :

(C) y = 4x (D) y + 4x + 1 = v

Solutions : y + 4x + 1 = v

Hence (D) is the correct answer.

- Integral curve satisfying π· has the slope at the point (1, 0) of the curve, equal to :

(C) 1 (D) π·

Solutions : π·

β slope at (1, 0) = π·

Hence (C) is the correct answer.

- A solution of the differential equation π· is :

(C) y = 2x β 4 (D) y = 2x2 β 4

Solutions : Ans is (C)

- The solution of π· is :

(C) xy2 = 2y5 + c (D) x(y2 + xy) = 0

Solutions : π· β β2x + 10y3 β π·

π·

β΄ I.F. = π·

β΄ Solution is π·

π·.

Hence (C) is the correct answer.

- A particle moves in a straight line with velocity given by π· (x being the distance described). The time taken by the particle to describe 99 metres is :

(C) 2 log10 e (D) π·

Solutions : π· π·

β log(x + 1) = t + c ....(1)

Initially, when t = 0, x = 0

β΄ c = 0

β΄ log (x + 1) = t

When x = 99, then t = loge (100) = 2 loge 10.

Hence (B) is the correct answer.

- The curve for which slope of the tangent at any point equals the ratio of the abscissa to the ordinate of the point is a/an

(C) circle (D) parabola

Solutions : π· β ydy = xdx

π· β y2 β x2 = 2c

which is a rectangular hyperbola

Hence (B) is the correct answer.

- The solution of the differential equation (1 β xy β x5y5) dx β x2(x4y4 + 1)dy = 0 is given by

(C) x = π· (D) None of these

Solution: The given equation is dx β x (ydx+ xdy )= x5y4(ydx + xdy)

π· β lnx = xy + π·

β x = π·.

Hence (A) is the correct answer.

- If π·satisfies the relation π·then value of A and B respectively are:

(C) β13, 12 (D) 12, β13

Solution: On differentiatingπ·, w.r.t x we get

π·π·

Putting these values in π·, we get

π·.

On solving we get A= β13 and B = β12

Hence (B) is the correct answer.

- The differential equation of all circles which pass through the origin and whose centres lie on y-axis is :

(C) π· (D) π·

Solutions : If (0, a) is centre on y-axis, then its radius is a because it passes through origin.

β΄ Equation of circle is x2 + (y β a)2 = a2

β x2 + y2 β 2ay = 0 ....(1)

π· .....(2)

Using (1) in (2), π·

π·

π·

Hence (A) is the correct answer.

- If π·(logy β logx + 1) then the solution of the equation is :

(C) π· (D) π·

Solutions : π·

Put y = vx β π·

π·

π·

β log(log v) = logx + logc = logcx

β log v = cx π·.

Hence (D) is the correct answer.

- The degree of the differential equation satisfying

(A) 2 (B) 3

(C) 4 (D) None of these

Solution: Putting x = tanΞΈ and y = tanΟ. Then equation becomes

secΞΈ + secΟ = A (tanΞΈ secΟ β tanΟsecΞΈ).

β π·βπ·

βπ·β π· βtan β1x β tan β1y = 2cot β1A

Differentiating, we get π·

Which is a differential equation of degree 1.

Hence (D) is the correct answer.

2023.01.16 10:29 *rajusingh79* **DIFFERENTIAL EQUATION**

https://docs.google.com/document/d/1iiDTnS3EU92uTkmO6KyCwqT8FRC1u-10/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true

## DIFFERENTIAL EQUATION

## BASIC DEFINITION

An equation containing an independent variable, dependent variable and differential coefficients of dependent variable with respect to independent variable is called a differential equation. An ordinary differential equation is one is which there is only one independent variable.

**For Example :**

π· .........(1)

π· .........(2)

π· ..........(3)

π· .........(4)

π· .........(5)

π· .........(6)

## Order and Degree of a Differential Equation:

The **order** of differential equation is the order of highest order derivative appearing in the equation.

For Example :

Orders of differential equations (1), (2), (3), (4), (5) and (6) are 1, 2, 1, 3, 2 and 2 respectively.

The**degree** of differential equation is the degree of the highest order derivative involved in it, when the differential coefficients are free from radicals and fractions (i.e. write differential equations as polynomial in derivatives)

For Example :

Degrees of differential equations (1), (2), (3), (4), (5) and (6) are 1, 1, 1, 4, 3 and 2 respectively.

**Illustration 1:** **Find the order and degree (if defined) of the following differential equations :**

**(i)** **π·** **(ii)** **π·**

**Solution :** (i) The given differential equation can be re-written as

π·. Hence its order is 3 and degree 2.

(ii)**π·****.** Hence its order is 2 and degree 1.

**Formation of differential Equation:**

We know y2 = 4ax is a parabola whose vertex is origin and axis as the x-axis . If a is a parameter, it will represent a family of parabola with the vertex at (0, 0) and axis as y = 0 .

Differentiating y2 = 4ax . . (1)

π· . . (2)

From (1) and (2), y2 = 2yxπ· β y = 2xπ·

This is a differential equation for all the members of the family and it does not contain any parameter ( arbitrary constant).

(i) The differential equation of a family of curves of one parameter is a differential equation of the first order, obtained by eliminating the parameter by differentiation.

(ii) The differential equation of a family of curves of two parameter is a differential equation of the second order, obtained by eliminating the parameter by differentiating the algebraic equation twice. Similar procedure is used to find differential equation of a family of curves of three or more parameter.

**Example -1:** **Find the differential equation of the family of curves y = Aex + Be3x for different values of A and B.**

**Solution:** y = A ex + Be3x . . . . (1)

y1 = Aex + 3Be3x . . . (2) π·

y2 = Aex + 9B3x . . . (3) π·

Eliminating A and B from the above three, we get

π· = 0 β ex e3x π· = 0

β 3y + 4y1 β y2 = 0 β 3y + 4π·

#### SOLUTION OF DIFFERENTIAL EQUATION

The general solution of a differential equation is the relation in the variables x, y obtained by integrating (removing derivatives ) where the relation contains as many arbitrary constants as the order of the equation. The general solution of a differential equation of the first order contains one arbitrary constant while that of the second order contains two arbitrary constants. In the general solution, if particular values of the arbitrary constant are put , we get a particular solution which will give one member of the family of curves.

To solve differential equation of the first order and the first degree:

Simple standard form of differential equation of the first order and first degree are as follows:

#### (i) Variable Separable

Form f(x) dx + Ο(y) dy = 0

**Method:** Integrate it i.e., find β« f(x) dx + β« Ο(y)dy = c

**Example -2:** **Solve** **π·****.**

**Solution:** Given π· β π·

β π·

Integrating, we get ln y β ex =c

#### (ii) Reducible into Variable Separable

**Method:** Sometimes differential equation of the first order cannot be solved directly by variable separation but by some substitution we can reduce it to a differential equation with separable variables.

A differential equation of the π· is solved by writing ax + by + c = t

**Example -3:** **Solve (x β y)2** **π·****.**

**Solution:** Put z = x βy β π· β π·

Now z2π· β π·

β dx = π·, which is in the form of variable separable

Now integrating, we get x = z + π·

β Solution is x =(x β y) + π·

#### (iii) Homogeneous Equation

When π· is equal to a fraction whose numerator and the denominator both are homogeneous functions of x and y of the same degree then the differential equation is said to be homogeneous equation.

i.e. when π·, where f(km, ky) = f(x, y) then this differential equation is said to be homogeneous differential equation .

**Method:** Put y = vx

**Example -4:** **Solve** **π·****.**

**Solution:** π· (homogeneous ) . Put y = vx

β΄ π·

β π·, Integrate

C + lnx = - ln(1 βv2)

β lnkx + ln(1 βv2) =0

β kx(1- v2) = 1 β k(x2 β y2) = x .

#### (iv) Non-homogeneous Differential Equation

Form π·

**Method:** If π·, put x = X + h , y = Y + k such that a1h +b1k + c1 = 0, a2h + b2k +c2 = 0 . In this way the equation becomes homogeneous in X, Y. then use the method for homogeneous equation.

If π·. Put a1x+b1y=v or a2x + b2y = v. The equation in the form of variable separable in x, v.

**Example -5:** **π·****.**

**Solution:** Here π·

Hence we put x- y = v

π· β π·

or, 1 β π· β π· = dx or, π· Integrate

2v + ln (v +2) = x + C, Put the value of v

β΄ x β 2y + ln (x β y +2) = C

**(v)** **Linear Equation**

Form π·, where P(x) and Q(x) are functions of x

**Method:** Multiplying the equation by eβ«P(x)dx, called integrating factor. Then the equation becomes π·

Integrating π·

**(vi)** **Reducible into Linear Equation**

Form R(y)π· + P(x) S(y) = Q(x) , such that π·

**Method:** Put S(y) =z then π·

β΄ The equation becomes π·, which is in the linear form

**Example -6:** **π·****.**

**Solution:** π·

Multiplying both sides by I.F. and integrating

π·

Put π·

β π·

β 2 y π·

**(vii)** **Exact Differential Equations**

Mdx + Ndy = 0, where M and N are functions of x and y. If π·, then the equation is exact and its solution is given by β« Mdx + β« N dy = c

To find the solution of an exact differential equation Mdx + N dy = 0, integrate π· as if y were constant. Also integrate the terms of N that do not contain x w.r.t y. Equate the sum of these integrals to a constant.

**Example -7:** **(x2 βay)dx + (y2 βax)dy = 0.**

**Solution:** Here M = x2 βay

N = y2 βax

π·

π· β π·

β΄ equation is exact

solution is π· = c

π· β ayx + π· = c

or x3 β3axy + y3 = 3c.

**Integrating Factor:**

A factor, which when multiplied to a non exact, differential equation makes it exact, is known as integrating factor e.g. the non exact equation y dx βx dy = 0 can be made exact on multiplying by the factor π·. Hence π· is the integrating factor for this equation.

**Notes:**

In general such a factor exist but except in certain special cases, it is likely to be difficult to determine.

The number of integrating factor for equation M dx + N dy = 0 is infinite.

**Some Useful Results:**

d(xy) = xdy + ydx

π·

π·

π· = π·

d tan-1 π· = π·

π·

d(sin-1 xy) = π·

**Example -8:** **Solve x dy β y dx =** **π·****.**

**Solution:** π· β π·

β π· Integrating

β lnπ· β ln π· = 2 ln x + ln k

y + π· = kx2.

**(viii)** **Linear Differential Equation with constant coefficient**

Differential equation of the form π·, aI β R for I = 0, 1 , 2, 3, . . . , n is called a linear differential equation with constant coefficients.

In order to solve this differential equation, take the auxiliary equation as a0Dn + a1Dn-1 +β¦+ an =0

Find the roots of this equation and then solution of the given differential equation will be as given in the following table

Roots of the auxiliary equation

Corresponding complementary function

1

One real root Ξ±1

C1π·

2.

Two real and differential root Ξ±1 and Ξ±2

C1π· + C2π·

3.

Two real and equal roots Ξ±1 and Ξ±2

(C1 + C2x) π·

4.

Three real and equal roots Ξ±1, Ξ±2, Ξ±3

(C1 + C2x + C3x2 )π·

5.

One pair of imaginary roots Ξ± Β± iΞ²

(C1 cosΞ²x + C2 sinΞ²x) π·

6.

Two Pair of equal imaginary roots Ξ± Β± iΞ² and Ξ± Β± iΞ²

[ (C1 + C2x) cosΞ² + (C1 + C2x)sinΞ²]π·

**Example -9:** **Solve** **π·****.**

**Solution:** Its auxiliary equation is D2 β3D + 2 = 0 β D = 1, D = 2

Hence its solution is y = C1ex + C2e2x

So far only linear differential equation with constant coefficients of form

a0 π· + a1 π·+ β¦.+ an y = 0, aI β R

for i = 0, 1, 2, β¦., n were considered. Now we consider the following form

a0 π· + a1 π·+ β¦.+ an y = X

where X is either constant or functions of x alone.

**Theorem:**

If y = f1(x) is the general solution of a0 π· + a1 π·+ β¦.+ an y = 0

and y = f2(x) is a solution of

a0 π· + a1 π·+ β¦.+ an y = X

Then y = f1(x) + f2(x) is the general solution of a0 π· + a1 π·+ β¦.+ any = X

Expression f1(x) is known as complementary function and the expression f2(x) is known as particular integral. These are denotes by C.F. and P.I. respectively.

The nth derivative of y will be denoted Dny where D stands for π· and n denotes the order of derivative.

If we take Differential Equation:

π· + P1π·+ P2π· + β¦. + Pny = X

then we can write this differential equation in a symbolic form as

Dny + P1Dnβ 1y + P2Dnβ 2y + β¦.+ Pny = X

(Dn + P1Dnβ 1 + P2Dnβ 2 + β¦.+ Pn)y = X

The operator Dn + P1Dnβ 1 + P2Dnβ 2 + β¦.+ Pn is denoted by f(D) so that the equation takes the form f(D)y = X

y = π·

**Methods of finding P.I. :**

**Notes:**

If both m1 and m2 are constants, the expressions (D β m1)(D β m2)y and (Dβ m2)(D β m1)y are equivalent i.e. the expression is independent of the order of operational factors.

π·

We will explain the method with the help of following

**Example -10:** **Solve** **π·****.**

**Solution:** The equation can be written as (D2 β 5D + 6)y = e3x

(D β 3) (D β 2)y = e3x

C.F. = c1 e3x + c2e2x

And P.I. = π·= π·= π·

= e3x π· = x. e3x

β΄ y = c1 e3x + c2e2x + x e3x

P.I. can be found by resolving π·

Into partial functions

π·

β΄ P.I. = π·= π·

= π· = x e3x β e3x .

Second term can be neglected as it is included as it inclined in the first term of a C.F.

**Short Method of Finding P.I. :**

In certain cases, the P.I. can be obtained by methods shorter than the general method.

(i). To find P.I. when X = eax in f(D) y = X, where a is constant

y = π·

π· if f(a) β 0

π· if f(a) = 0 , where f(D) = (D β a)r Ο(D)

**Example -11:** **Solve (D3 β 5D2 + 7D β 3)y = e3x.**

**Solution:** (D β 1)2 (D β 3) y = e3x

C.F. = aex + bx ex + ce3x

And P.I. = π·e3x = π·

β΄ y = aex + bx ex + ce3x + π·

(ii). To find P.I. when X = cosax or sinax

f(D) y = X

y = π· sinax

If f( β a2) β 0 then π· = π·

If f(β a2) = 0 then (D2 + a2) is atleast one factor of f(D2)

Let f (D2) = (D2 + a2)r Ο (D2)

Where Ο(β a2) β 0

β΄ π· = π· = π·

when r = 1 π· sin ax = βπ·

Similarly If f(β a)2 β 0 then π·cos ax = π·cosax

and π·

**Example -12:** **Solve (D2 β 5D + 6)y = sin3x.**

**Solution:** (D β 2) (D β 3)y = sin3x

C.F. = ae2x + be3x

P.I.= π· sin3x = π·

= π· = β (5D β 3). π·sin3x = π·

β΄ ae2x + be3x + π·

(iii). To find the P.I. when X = xm where m β N

f(D) y = xm

y = π·

we will explain the method by taking an example

**Example -13:** **Find P.I. of (D3 + 3D2 + 2D)y = x2.**

**Solution:** P.I. = π· =π· = π·

= π·

= π· = π·

= π· = π·

(iv). To find the value of π·eaxV where βaβ is a constant and V is a function of x

π·

**Example -14:** **Solve (D2 + 2)y = x2 e3x.**

**Solution:** C.F. = a cos π· x + b sinπ·x

P.I. = π· x2 e3x = e3x π·

= π·π· = π·x2

= π·x2 = π·x2

= π·

β΄ a cosπ·x + b sinπ·x + π·

(v). To find π· where V is a function of x

π·

**Example -15:** **Solve (D2 + 4) y = x sin2x.**

**Solution:** C.F. = c1 cos 2x + c2 sin2x

P.I. = π· = π·

= π· = β π·

= β π·= β π·

y = c1 cos2x + c2 sin2x β π·.

**Some Results on Tangents and Normals:**

(i) The equation of the tangent at P(x, y) to the curve y= f(x) is Y β y = π·

(ii) The equation of the normal at point P(x, y) to the curve y = f(x) is

Y β y = π·(X β x )

(iii) The length of the tangent = CP = π·

(iv) The length of the normal = PD =π·

π·

(v) The length of the cartesian subtangent = CA = π·

(vi) The length of the cartesian subnormal = AD = π·

(viii) The initial ordinate of the tangent =OB = y - xπ·

**PROBLEMS**

**SUBJECTIVE**

(C) x2 = 16y β 60 (D) x2 = -16y + 68

Solution: Let the curve be y = f(x). Subnormal at any point = π·

β yπ· = Β±8 β y dy = Β±8dx β π· = Β±8x + c

β y2 = 16 x+2c1, β c1= -8 or y2 = -16x +2c2, β c2= 24

Hence (A), (B) are correct answers.

(C) (y+2) = Ξ»(x+1) (D) (y+1) = Ξ»(x+2) where Ξ» β R.

Solution: Any line passing through the centre of the given circle would meet the circle orthogonally.

Hence (A) is the correct answer.

(C) m = 1, n = 2 (D) None

Solution: π· = c β the differential equation is ;

y = x.π· + π·-3. π·+2

Clearly its order is one and degree 4.

Hence (A) is the correct answer.

(C) π· = y (D) π· = y

Solution: π· = a cosx - b sinx β π· = -a sinx β bcosx = -y

Hence π· + y = 0 .

Hence (A) is the correct answer.

(C) equal to 2y2 (D) equal to x2

Solution: π· = π· for a differential equation

or π· = -1 π·π· = - π·π· = - π·

or π· + π·π· = 0.

Hence (A) is the correct answer.

(C) 3 (D) none of these

Solution: Since the equation is not a polynomial in all the differential coefficient so the degree of equation is not defined.

Hence (D) is the correct answer.

(C) 3, 2 (D) none of these

Solution: Equation of required parabola is of the form y2 = 4a(x βh)

Differentiating, we have 2yπ· = 4a β yπ· = 2a

Required differential equation π·.

Degree of the equation is 1 and order is 2.

Hence (B) is the correct answer.

(C) yy2 + xy12 βxy1 = 0 (D) none of these

Solution: Ellipse centred at origin are given by π· = 1 β¦β¦(1)

where a and b are unknown constants

π· β π·y1 = 0 β¦β¦(2)

Differentiating again, we get

π·(y12 + yy2) = 0 β¦β¦(3)

Multiplying (3) with x and then subtracting from (2) we get

π·( yy1 β xy12 βxyy2) = 0 β xyy2 + xy12 βyy1 = 0.

Hence (B) is the correct answer.

(C) y = 1 + π· (D) y = π· + 11π·

Solution: π· + (3x)y = x

I.F = π·

β΄ Solution of given equation is

yπ· + c = π· + c

If curve passes through (0, 4), then

4 βπ· = c β c = π·

y = π· β 3y = 1 + 11π·.

Hence (A) is the correct answer.

(C) x βy + c = = log (3x β4y β3) (D) x βy + c = log (3x β4y + 1)

Solution: Let 3x β4y = z

3 β4π· β π·

Therefore the given equation π·

β βπ·β βπ·dz = dx

β βz + 4 log (z + 1) = x + c β log (3x β4y + 1) = x βy + c.

Hence (B) is the correct answer.

(A) 5 (B) 4

(C) 3 (D) none of these

Solution: y = (c1 + c4) ex + c2 e2x + c3 e3x + c4π·

y = c1 ex + c2 e2x + c3 e3x + c4 π·

y = k1 ex + k2 e2x + k3 e3x + k4

Therefore 4 obituary constants

Hence (B) is the correct answer.

(C) ln xy (D) none of these

Solution: I.F. = π· = ln y

Hence (B) is the correct answer.

(A) π· (B) π·

(C) π· (D) π·

Solution: y = A e2x + b eβ2x β π· = 2 (A e2x β b eβ2x)

π· = 4 (A e2x + b eβ2x) = ln y

Hence (C) is the correct answer.

(C) log tan π· = c β 2 cos π· (D) none of these

Solution: π·= β 2 cosπ·

β π· β c β 2 sin π· = log tan π·

Hence (A) is the correct answer.

(C) tan π· = kx (D) none of these

Solution: π· = π·

put y = vx β v + x π· = v + tan v

cot v dv = π·

Integrating, we get ln sin v = ln x + ln k β sin π· = kx

Hence (A) is the correct answer.

(C) sinβ1 x = c sinβ1 y (D) (sinβ1 x) (sinβ1 y) = c

Solution: π· β sinβ1 y + sinβ1 x = c

Hence (B) is the correct answer.

(C) y = π·eβ2x + cx + d (D) y = eβ2x + cx2 + d

Solution: π· = eβ2x, π· + k1

Integrating, y = π· + k1 x + k2 β y = π· + cx + d

Hence (C) is the correct answer.

(C) xy = π·x4 + c (D) y β x = π·x4 + c

Solution: π· = x2

I.F. = π· = x

Therefore solution is xy = π·+ c

Hence (C) is the correct answer.

(C) π· + y = 0 (D) none of these

Solution: y = A e3x + B e5x

yβ² = 3 A e3x + 5 B e5x

yβ³ = 9 A e3x + 25 B e5x

therefore yβ³ β 8y + 15y = 0

Hence (A) is the correct answer.

(C) pair of straight line (D) none of these

Solution: π· β ln y = ln y2 + ln c β y2 = kx

it represents a family of parabola

Hence (A) is the correct answer.

(C) y tanx = x (D) none of these

Solution: π·β y tan x = sec x

I.F. = π· = cos x

y cos x = β« sec x cos x dx = x + c

y cos x = x + c

At (0, ΞΈ), y cos x = x Since c = 0

Hence (B) is the correct answer.

(B) a set of circles with centre on x-axis

(C) a set of ellipses

(D) none of these

Solution: y dy = (a β x) dx

π·= ax β π· β x2 + y2 = 2ax β x2 + 2ax + y2 = 0

It represents a set of sides with centre on xβaxis

Hence (B) is the correct answer.

(C) e6 + 1 (D) loge 6

Solution: e2y dy = dx β π· + c = x β c = 5 β π·

At. y = 3 π· = x β x = π·

Hence (B) is the correct answer.

(C) xy = x2 + x + 1 (D) xy = y + 1

Solution: π· β y = x + π· + c

At (2, 7/2) π· = 2 + π·+ c β c = 1

Therefore y = x + π· + 1 β xy = x2 + x + 1

Hence (C) is the correct answer.

(C) e (D) e2

Solution: π· = ex dx β ln y = ex + c

At x = 0, y = 1; c = 0

ln y = ex

Therefore At x = 1, y = ec

Hence (C) is the correct answer.

(C) 4 (D) None of these

Solutions : The general equation of all conics with centre at origin is ax2 + 2hxy + by2 = 1

Since it has three arbitrary constants, the differential equation is of order 3.

Hence (B) is the correct answer.

(C) 3 (D) 4

Solutions : The general equation of all conics with axes as coordinate axes is ax2 + by2 = 1.

Since it has two arbitrary constants, the differential equation is of order 2.

Hence (B) is the correct answer.

(C) π· (D) π·

Solutions : The general equation of non-vertical lines in a plane is ax + by = 1, b β 0

π· π· π·

Hence (C) is the correct answer.

(C) π· (D) π·

Solutions : ax = by = 0, a β 0 π·.

Hence (D) is the correct answer.

(C) 1, 3 (D) 1, 4

Solutions : Equation of any tangent to x2 = 4y is π·, where m is arbitrary constant.

π· π·

β΄ Putting this value of m in π·, we get

π· π·

Which is a differential equation of order 1 and degree 2.

Hence (A) is the correct answer.

(C) 3e2 (D) 2e3

Solutions : π· β log f(x) = x + c

Since f(1) = 2

β΄ log 2 = 1 + c

β΄ log f(x) = x + log 2 β 1

β΄ log f(3) = 3 + log 2 β 1 = 2 + log 2

β f(3) = e2+log2 = e2 . elog2 = 2e2.

Hence (B) is the correct answer.

(C) 6xy = 3x3 β 29x β 6 (D) None of these

Solutions : π· π·

It passes through (3, 9)

π· π·

π·

Hence (C) is the correct answer.

(C) y2 = cosx + 1 (D) None of these

Solutions : On dividing by sin x,

π·

Put y2 = v π·

I.F. = eβ«cotxdx = elog sinx = sin x

β΄ Solution is v. sin x β«sinx . (2 cos x) dx + c

β y2 sin x = sin2x + c

When π·, y = 1, then c = 0

β΄ y2 = sin x.

Hence (A) is the correct answer.

(C) π· (D) None of these

Solutions : π· β P = βx, Q = 1

I.F. = π·

β΄ Solution is π·

Hence (D) is the correct answer.

(C) ex(x β 1) + 1 = y (D) None of these

Solutions : π·

β eβy dy = (ex + eβx)dx β eβy = ex β eβx + c

Hence (D) is the correct answer.

(C) π· (D) π·

Solutions : y = eβx (A cos x + B sin x) ....(1)

π· (βA sin x + B cos x) β eβx (A cos x + B sin x)

π·(βA sin x + B cos x) β y .....(2)

π·(βA cos x β B sin x) β eβx (βA sin x + B cos x) π·

Using (1) and (2), we get

π·

Hence (C) is the correct answer.

(C) π· (D) π·

Solutions : π· .....(1)

Put y = vx π·

β΄ (1) becomes π·

π·

π· π·

π·.

Hence (A) is the correct answer.

(C) 1, 4 (D) 2, 4

Solutions : On simplification the equation becomes π·

Hence (D) is the correct answer.

(C) π· (D) 6

Solutions : Clearly the equation is of order 2.

Hence (A) is the correct answer.

(A) m = 3, n = 3 (B) m = 3, n = 2

(C) m = 3, n = 5 (D) m = 3, n = 1

Solutions : The given differential equation can be written as

π·

β m = 3, n = 2

Hence (B) is the correct answer.

(C) parabolas (D) ellipses

Solutions : π· π·

β 2log(y + 3) = logx + log c

β (y + 3)2 = ex, which represent parabolas.

Hence (C) is the correct answer.

(C) cos x tan y = c (D) cos x sin y = c

Solutions : Given equation can be written as :

π·

π·

π·

π·

π·

Hence (A) is the correct answer.

(C) y = emx + ceβax (D) (a + m) y = emx + ceβax

Solutions : I.F. = eβ«adx = eax

β΄ Solution is π·

π·

β (a + m)y = emx + c1 (a + m) eβax

β (a + m) y = emx + ceβax

where c = c1 (a + m)

Hence (D) is the correct answer.

(C) sec x (D) sin x

Solutions : π·

Here P = tan x, Q = sec x

β΄ I. F. = eβ«tan x dx = elog secx = sec x.

Hence (C) is the correct answer.

(B) straight line passing through origin

(C) parabola whose vertex is at origin

(D) circle whose centre is at origin

Solutions : x dy β y dx = 0 π·

β log y β log x = log c

π·

π· or y = cx which is a straight line.

Hence (B) is the correct answer.

(C) log(log x) (D) x

Solutions : π·

π· π·

π·

Hence (B) is the correct answer.

(C) (x β y)ex+y = 1 (D) (x β c)ex+y + 1 = 0

Solutions : π· π·

Put eβy = z π· π·

π· β΄ P = β1, Q = βex.

β΄ I.F. = eβ«β1dx = eβx

Solution is z . eβx = β« βex . eβx dx + c = βx + c

β eβy . eβx = c β x β eβ(x+y) = c β x

β (x β c) ex+y + 1 = 0

Hence (D) is the correct answer.

(C) 1 (D) None of these

Solutions : There is only one arbitrary constant.

So, degree of the differential equation is one.

Hence (C) is the correct answer.

(C) βx (D) ex

Solutions : I.F. π·

Hence (A) is the correct answer.

(C) π· (D) π·

Solutions : π·

π·

π·

Hence (D) is the correct answer.

(C) y = 4x (D) y + 4x + 1 = v

Solutions : y + 4x + 1 = v

Hence (D) is the correct answer.

(C) 1 (D) π·

Solutions : π·

β slope at (1, 0) = π·

Hence (C) is the correct answer.

(C) y = 2x β 4 (D) y = 2x2 β 4

Solutions : Ans is (C)

(C) xy2 = 2y5 + c (D) x(y2 + xy) = 0

Solutions : π· β β2x + 10y3 β π·

π·

β΄ I.F. = π·

β΄ Solution is π·

π·.

Hence (C) is the correct answer.

(C) 2 log10 e (D) π·

Solutions : π· π·

β log(x + 1) = t + c ....(1)

Initially, when t = 0, x = 0

β΄ c = 0

β΄ log (x + 1) = t

When x = 99, then t = loge (100) = 2 loge 10.

Hence (B) is the correct answer.

(C) circle (D) parabola

Solutions : π· β ydy = xdx

π· β y2 β x2 = 2c

which is a rectangular hyperbola

Hence (B) is the correct answer.

(C) x = π· (D) None of these

Solution: The given equation is dx β x (ydx+ xdy )= x5y4(ydx + xdy)

π· β lnx = xy + π·

β x = π·.

Hence (A) is the correct answer.

(C) β13, 12 (D) 12, β13

Solution: On differentiatingπ·, w.r.t x we get

π·π·

Putting these values in π·, we get

π·.

On solving we get A= β13 and B = β12

Hence (B) is the correct answer.

(C) π· (D) π·

Solutions : If (0, a) is centre on y-axis, then its radius is a because it passes through origin.

β΄ Equation of circle is x2 + (y β a)2 = a2

β x2 + y2 β 2ay = 0 ....(1)

π· .....(2)

Using (1) in (2), π·

π·

π·

Hence (A) is the correct answer.

(C) π· (D) π·

Solutions : π·

Put y = vx β π·

π·

π·

β log(log v) = logx + logc = logcx

β log v = cx π·.

Hence (D) is the correct answer.

(A) 2 (B) 3

(C) 4 (D) None of these

Solution: Putting x = tanΞΈ and y = tanΟ. Then equation becomes

secΞΈ + secΟ = A (tanΞΈ secΟ β tanΟsecΞΈ).

β π·βπ·

βπ·β π· βtan β1x β tan β1y = 2cot β1A

Differentiating, we get π·

Which is a differential equation of degree 1.

Hence (D) is the correct answer.

submitted by rajusingh79 to u/rajusingh79 [link] [comments]
π· .........(1)

π· .........(2)

π· ..........(3)

π· .........(4)

π· .........(5)

π· .........(6)

For Example :

Orders of differential equations (1), (2), (3), (4), (5) and (6) are 1, 2, 1, 3, 2 and 2 respectively.

The

For Example :

Degrees of differential equations (1), (2), (3), (4), (5) and (6) are 1, 1, 1, 4, 3 and 2 respectively.

π·. Hence its order is 3 and degree 2.

(ii)

We know y2 = 4ax is a parabola whose vertex is origin and axis as the x-axis . If a is a parameter, it will represent a family of parabola with the vertex at (0, 0) and axis as y = 0 .

Differentiating y2 = 4ax . . (1)

π· . . (2)

From (1) and (2), y2 = 2yxπ· β y = 2xπ·

This is a differential equation for all the members of the family and it does not contain any parameter ( arbitrary constant).

(i) The differential equation of a family of curves of one parameter is a differential equation of the first order, obtained by eliminating the parameter by differentiation.

(ii) The differential equation of a family of curves of two parameter is a differential equation of the second order, obtained by eliminating the parameter by differentiating the algebraic equation twice. Similar procedure is used to find differential equation of a family of curves of three or more parameter.

y1 = Aex + 3Be3x . . . (2) π·

y2 = Aex + 9B3x . . . (3) π·

Eliminating A and B from the above three, we get

π· = 0 β ex e3x π· = 0

β 3y + 4y1 β y2 = 0 β 3y + 4π·

To solve differential equation of the first order and the first degree:

Simple standard form of differential equation of the first order and first degree are as follows:

β π·

Integrating, we get ln y β ex =c

A differential equation of the π· is solved by writing ax + by + c = t

Now z2π· β π·

β dx = π·, which is in the form of variable separable

Now integrating, we get x = z + π·

β Solution is x =(x β y) + π·

i.e. when π·, where f(km, ky) = f(x, y) then this differential equation is said to be homogeneous differential equation .

β΄ π·

β π·, Integrate

C + lnx = - ln(1 βv2)

β lnkx + ln(1 βv2) =0

β kx(1- v2) = 1 β k(x2 β y2) = x .

If π·. Put a1x+b1y=v or a2x + b2y = v. The equation in the form of variable separable in x, v.

Hence we put x- y = v

π· β π·

or, 1 β π· β π· = dx or, π· Integrate

2v + ln (v +2) = x + C, Put the value of v

β΄ x β 2y + ln (x β y +2) = C

Form π·, where P(x) and Q(x) are functions of x

Integrating π·

Form R(y)π· + P(x) S(y) = Q(x) , such that π·

β΄ The equation becomes π·, which is in the linear form

Multiplying both sides by I.F. and integrating

π·

Put π·

β π·

β 2 y π·

Mdx + Ndy = 0, where M and N are functions of x and y. If π·, then the equation is exact and its solution is given by β« Mdx + β« N dy = c

To find the solution of an exact differential equation Mdx + N dy = 0, integrate π· as if y were constant. Also integrate the terms of N that do not contain x w.r.t y. Equate the sum of these integrals to a constant.

N = y2 βax

π·

π· β π·

β΄ equation is exact

solution is π· = c

π· β ayx + π· = c

or x3 β3axy + y3 = 3c.

A factor, which when multiplied to a non exact, differential equation makes it exact, is known as integrating factor e.g. the non exact equation y dx βx dy = 0 can be made exact on multiplying by the factor π·. Hence π· is the integrating factor for this equation.

In general such a factor exist but except in certain special cases, it is likely to be difficult to determine.

The number of integrating factor for equation M dx + N dy = 0 is infinite.

d(xy) = xdy + ydx

π·

π·

π· = π·

d tan-1 π· = π·

π·

d(sin-1 xy) = π·

β π· Integrating

β lnπ· β ln π· = 2 ln x + ln k

y + π· = kx2.

Differential equation of the form π·, aI β R for I = 0, 1 , 2, 3, . . . , n is called a linear differential equation with constant coefficients.

In order to solve this differential equation, take the auxiliary equation as a0Dn + a1Dn-1 +β¦+ an =0

Find the roots of this equation and then solution of the given differential equation will be as given in the following table

Roots of the auxiliary equation

Corresponding complementary function

1

One real root Ξ±1

C1π·

2.

Two real and differential root Ξ±1 and Ξ±2

C1π· + C2π·

3.

Two real and equal roots Ξ±1 and Ξ±2

(C1 + C2x) π·

4.

Three real and equal roots Ξ±1, Ξ±2, Ξ±3

(C1 + C2x + C3x2 )π·

5.

One pair of imaginary roots Ξ± Β± iΞ²

(C1 cosΞ²x + C2 sinΞ²x) π·

6.

Two Pair of equal imaginary roots Ξ± Β± iΞ² and Ξ± Β± iΞ²

[ (C1 + C2x) cosΞ² + (C1 + C2x)sinΞ²]π·

Hence its solution is y = C1ex + C2e2x

So far only linear differential equation with constant coefficients of form

a0 π· + a1 π·+ β¦.+ an y = 0, aI β R

for i = 0, 1, 2, β¦., n were considered. Now we consider the following form

a0 π· + a1 π·+ β¦.+ an y = X

where X is either constant or functions of x alone.

If y = f1(x) is the general solution of a0 π· + a1 π·+ β¦.+ an y = 0

and y = f2(x) is a solution of

a0 π· + a1 π·+ β¦.+ an y = X

Then y = f1(x) + f2(x) is the general solution of a0 π· + a1 π·+ β¦.+ any = X

Expression f1(x) is known as complementary function and the expression f2(x) is known as particular integral. These are denotes by C.F. and P.I. respectively.

The nth derivative of y will be denoted Dny where D stands for π· and n denotes the order of derivative.

If we take Differential Equation:

π· + P1π·+ P2π· + β¦. + Pny = X

then we can write this differential equation in a symbolic form as

Dny + P1Dnβ 1y + P2Dnβ 2y + β¦.+ Pny = X

(Dn + P1Dnβ 1 + P2Dnβ 2 + β¦.+ Pn)y = X

The operator Dn + P1Dnβ 1 + P2Dnβ 2 + β¦.+ Pn is denoted by f(D) so that the equation takes the form f(D)y = X

y = π·

If both m1 and m2 are constants, the expressions (D β m1)(D β m2)y and (Dβ m2)(D β m1)y are equivalent i.e. the expression is independent of the order of operational factors.

π·

We will explain the method with the help of following

(D β 3) (D β 2)y = e3x

C.F. = c1 e3x + c2e2x

And P.I. = π·= π·= π·

= e3x π· = x. e3x

β΄ y = c1 e3x + c2e2x + x e3x

P.I. can be found by resolving π·

Into partial functions

π·

β΄ P.I. = π·= π·

= π· = x e3x β e3x .

Second term can be neglected as it is included as it inclined in the first term of a C.F.

In certain cases, the P.I. can be obtained by methods shorter than the general method.

(i). To find P.I. when X = eax in f(D) y = X, where a is constant

y = π·

π· if f(a) β 0

π· if f(a) = 0 , where f(D) = (D β a)r Ο(D)

C.F. = aex + bx ex + ce3x

And P.I. = π·e3x = π·

β΄ y = aex + bx ex + ce3x + π·

(ii). To find P.I. when X = cosax or sinax

f(D) y = X

y = π· sinax

If f( β a2) β 0 then π· = π·

If f(β a2) = 0 then (D2 + a2) is atleast one factor of f(D2)

Let f (D2) = (D2 + a2)r Ο (D2)

Where Ο(β a2) β 0

β΄ π· = π· = π·

when r = 1 π· sin ax = βπ·

Similarly If f(β a)2 β 0 then π·cos ax = π·cosax

and π·

C.F. = ae2x + be3x

P.I.= π· sin3x = π·

= π· = β (5D β 3). π·sin3x = π·

β΄ ae2x + be3x + π·

(iii). To find the P.I. when X = xm where m β N

f(D) y = xm

y = π·

we will explain the method by taking an example

= π·

= π· = π·

= π· = π·

(iv). To find the value of π·eaxV where βaβ is a constant and V is a function of x

π·

P.I. = π· x2 e3x = e3x π·

= π·π· = π·x2

= π·x2 = π·x2

= π·

β΄ a cosπ·x + b sinπ·x + π·

(v). To find π· where V is a function of x

π·

P.I. = π· = π·

= π· = β π·

= β π·= β π·

y = c1 cos2x + c2 sin2x β π·.

(i) The equation of the tangent at P(x, y) to the curve y= f(x) is Y β y = π·

(ii) The equation of the normal at point P(x, y) to the curve y = f(x) is

Y β y = π·(X β x )

(iii) The length of the tangent = CP = π·

(iv) The length of the normal = PD =π·

π·

(v) The length of the cartesian subtangent = CA = π·

(vi) The length of the cartesian subnormal = AD = π·

(viii) The initial ordinate of the tangent =OB = y - xπ·

- A curve that passes through (2, 4) and having subnormal of constant length of 8 units can be;

(C) x2 = 16y β 60 (D) x2 = -16y + 68

Solution: Let the curve be y = f(x). Subnormal at any point = π·

β yπ· = Β±8 β y dy = Β±8dx β π· = Β±8x + c

β y2 = 16 x+2c1, β c1= -8 or y2 = -16x +2c2, β c2= 24

Hence (A), (B) are correct answers.

- Equation of a curve that would cut x2 + y2 - 2x - 4y - 15 = 0 orthogonally can be ;

(C) (y+2) = Ξ»(x+1) (D) (y+1) = Ξ»(x+2) where Ξ» β R.

Solution: Any line passing through the centre of the given circle would meet the circle orthogonally.

Hence (A) is the correct answer.

- Let m and n be the order and the degree of the differential equation whose solution is y = cx +c2- 3c3/2 +2, where c is a parameter. Then,

(C) m = 1, n = 2 (D) None

Solution: π· = c β the differential equation is ;

y = x.π· + π·-3. π·+2

Clearly its order is one and degree 4.

Hence (A) is the correct answer.

- y = a sinx + b cosx is the solution of differential equation :

(C) π· = y (D) π· = y

Solution: π· = a cosx - b sinx β π· = -a sinx β bcosx = -y

Hence π· + y = 0 .

Hence (A) is the correct answer.

- For any differential function y = f(x), the value of π· + π·π· is :

(C) equal to 2y2 (D) equal to x2

Solution: π· = π· for a differential equation

or π· = -1 π·π· = - π·π· = - π·

or π· + π·π· = 0.

Hence (A) is the correct answer.

- The degree of differential equation π· is :

(C) 3 (D) none of these

Solution: Since the equation is not a polynomial in all the differential coefficient so the degree of equation is not defined.

Hence (D) is the correct answer.

- The degree and order of the differential equation of all parabolas whose axis is x-axis are :

(C) 3, 2 (D) none of these

Solution: Equation of required parabola is of the form y2 = 4a(x βh)

Differentiating, we have 2yπ· = 4a β yπ· = 2a

Required differential equation π·.

Degree of the equation is 1 and order is 2.

Hence (B) is the correct answer.

- The differential equation of all ellipses centred at origin is :

(C) yy2 + xy12 βxy1 = 0 (D) none of these

Solution: Ellipse centred at origin are given by π· = 1 β¦β¦(1)

where a and b are unknown constants

π· β π·y1 = 0 β¦β¦(2)

Differentiating again, we get

π·(y12 + yy2) = 0 β¦β¦(3)

Multiplying (3) with x and then subtracting from (2) we get

π·( yy1 β xy12 βxyy2) = 0 β xyy2 + xy12 βyy1 = 0.

Hence (B) is the correct answer.

- Particular solution of y1 + 3xy = x which passes through (0, 4) is :

(C) y = 1 + π· (D) y = π· + 11π·

Solution: π· + (3x)y = x

I.F = π·

β΄ Solution of given equation is

yπ· + c = π· + c

If curve passes through (0, 4), then

4 βπ· = c β c = π·

y = π· β 3y = 1 + 11π·.

Hence (A) is the correct answer.

- Solution of equation π· is :

(C) x βy + c = = log (3x β4y β3) (D) x βy + c = log (3x β4y + 1)

Solution: Let 3x β4y = z

3 β4π· β π·

Therefore the given equation π·

β βπ·β βπ·dz = dx

β βz + 4 log (z + 1) = x + c β log (3x β4y + 1) = x βy + c.

Hence (B) is the correct answer.

- The order of the differential equation, whose general solution is

(A) 5 (B) 4

(C) 3 (D) none of these

Solution: y = (c1 + c4) ex + c2 e2x + c3 e3x + c4π·

y = c1 ex + c2 e2x + c3 e3x + c4 π·

y = k1 ex + k2 e2x + k3 e3x + k4

Therefore 4 obituary constants

Hence (B) is the correct answer.

- I.F. for y ln π· is :

(C) ln xy (D) none of these

Solution: I.F. = π· = ln y

Hence (B) is the correct answer.

- Which one of the following is a differential equation of the family of curves

(A) π· (B) π·

(C) π· (D) π·

Solution: y = A e2x + b eβ2x β π· = 2 (A e2x β b eβ2x)

π· = 4 (A e2x + b eβ2x) = ln y

Hence (C) is the correct answer.

- Solution of π· is :

(C) log tan π· = c β 2 cos π· (D) none of these

Solution: π·= β 2 cosπ·

β π· β c β 2 sin π· = log tan π·

Hence (A) is the correct answer.

- Solution of π· is :

(C) tan π· = kx (D) none of these

Solution: π· = π·

put y = vx β v + x π· = v + tan v

cot v dv = π·

Integrating, we get ln sin v = ln x + ln k β sin π· = kx

Hence (A) is the correct answer.

- Solution of π· = 0 is :

(C) sinβ1 x = c sinβ1 y (D) (sinβ1 x) (sinβ1 y) = c

Solution: π· β sinβ1 y + sinβ1 x = c

Hence (B) is the correct answer.

- General solution of π· = eβ2x is :

(C) y = π·eβ2x + cx + d (D) y = eβ2x + cx2 + d

Solution: π· = eβ2x, π· + k1

Integrating, y = π· + k1 x + k2 β y = π· + cx + d

Hence (C) is the correct answer.

- Solution of π· = x2 is :

(C) xy = π·x4 + c (D) y β x = π·x4 + c

Solution: π· = x2

I.F. = π· = x

Therefore solution is xy = π·+ c

Hence (C) is the correct answer.

- Differential equation associated with primitive y = Ae3x + Be5x :

(C) π· + y = 0 (D) none of these

Solution: y = A e3x + B e5x

yβ² = 3 A e3x + 5 B e5x

yβ³ = 9 A e3x + 25 B e5x

therefore yβ³ β 8y + 15y = 0

Hence (A) is the correct answer.

- The curve satisfying y = 2x π· is a :

(C) pair of straight line (D) none of these

Solution: π· β ln y = ln y2 + ln c β y2 = kx

it represents a family of parabola

Hence (A) is the correct answer.

- The equation of the curve through the origin satisfying the equation dy = (sec x + y tanx) dx is :

(C) y tanx = x (D) none of these

Solution: π·β y tan x = sec x

I.F. = π· = cos x

y cos x = β« sec x cos x dx = x + c

y cos x = x + c

At (0, ΞΈ), y cos x = x Since c = 0

Hence (B) is the correct answer.

- The differential equation yπ· + x = a (where βaβ is a constant) represents :

(B) a set of circles with centre on x-axis

(C) a set of ellipses

(D) none of these

Solution: y dy = (a β x) dx

π·= ax β π· β x2 + y2 = 2ax β x2 + 2ax + y2 = 0

It represents a set of sides with centre on xβaxis

Hence (B) is the correct answer.

- If π· = eβ2y and y = 0 when x = 5, then value of x where y = 3 is given by :

(C) e6 + 1 (D) loge 6

Solution: e2y dy = dx β π· + c = x β c = 5 β π·

At. y = 3 π· = x β x = π·

Hence (B) is the correct answer.

- The equation of curve passing through π· and having slope 1 β π· at (x, y) is :

(C) xy = x2 + x + 1 (D) xy = y + 1

Solution: π· β y = x + π· + c

At (2, 7/2) π· = 2 + π·+ c β c = 1

Therefore y = x + π· + 1 β xy = x2 + x + 1

Hence (C) is the correct answer.

- Given that π· = yex such that x = 0, y = e. The value of y (y > 0) when x = 1 will be :

(C) e (D) e2

Solution: π· = ex dx β ln y = ex + c

At x = 0, y = 1; c = 0

ln y = ex

Therefore At x = 1, y = ec

Hence (C) is the correct answer.

- The differential equation of all conics with centre at origin is of order

(C) 4 (D) None of these

Solutions : The general equation of all conics with centre at origin is ax2 + 2hxy + by2 = 1

Since it has three arbitrary constants, the differential equation is of order 3.

Hence (B) is the correct answer.

- The differential equation of all conics with the coordinate axes, is of order

(C) 3 (D) 4

Solutions : The general equation of all conics with axes as coordinate axes is ax2 + by2 = 1.

Since it has two arbitrary constants, the differential equation is of order 2.

Hence (B) is the correct answer.

- The differential equation of all non-vertical lines in a plane is :

(C) π· (D) π·

Solutions : The general equation of non-vertical lines in a plane is ax + by = 1, b β 0

π· π· π·

Hence (C) is the correct answer.

- The differential equation of all non-horizontal lines in a plane is :

(C) π· (D) π·

Solutions : ax = by = 0, a β 0 π·.

Hence (D) is the correct answer.

- The degree and order of the differential equation of all tangent lines to the parabola x2 = 4y is :

(C) 1, 3 (D) 1, 4

Solutions : Equation of any tangent to x2 = 4y is π·, where m is arbitrary constant.

π· π·

β΄ Putting this value of m in π·, we get

π· π·

Which is a differential equation of order 1 and degree 2.

Hence (A) is the correct answer.

- If f(x) = f'(x) and f(1) = 2, then f(3) =

(C) 3e2 (D) 2e3

Solutions : π· β log f(x) = x + c

Since f(1) = 2

β΄ log 2 = 1 + c

β΄ log f(x) = x + log 2 β 1

β΄ log f(3) = 3 + log 2 β 1 = 2 + log 2

β f(3) = e2+log2 = e2 . elog2 = 2e2.

Hence (B) is the correct answer.

- Equation of the curve passing through (3, 9) which satisfies the differential equation π· is :

(C) 6xy = 3x3 β 29x β 6 (D) None of these

Solutions : π· π·

It passes through (3, 9)

π· π·

π·

Hence (C) is the correct answer.

- Solution of π·, y = 1 is given by

(C) y2 = cosx + 1 (D) None of these

Solutions : On dividing by sin x,

π·

Put y2 = v π·

I.F. = eβ«cotxdx = elog sinx = sin x

β΄ Solution is v. sin x β«sinx . (2 cos x) dx + c

β y2 sin x = sin2x + c

When π·, y = 1, then c = 0

β΄ y2 = sin x.

Hence (A) is the correct answer.

- The general solution of the equation π· is :

(C) π· (D) None of these

Solutions : π· β P = βx, Q = 1

I.F. = π·

β΄ Solution is π·

Hence (D) is the correct answer.

- Solution of π· is :

(C) ex(x β 1) + 1 = y (D) None of these

Solutions : π·

β eβy dy = (ex + eβx)dx β eβy = ex β eβx + c

Hence (D) is the correct answer.

- If y = eβx (A cosx + B sin x), then y satisfies

(C) π· (D) π·

Solutions : y = eβx (A cos x + B sin x) ....(1)

π· (βA sin x + B cos x) β eβx (A cos x + B sin x)

π·(βA sin x + B cos x) β y .....(2)

π·(βA cos x β B sin x) β eβx (βA sin x + B cos x) π·

Using (1) and (2), we get

π·

Hence (C) is the correct answer.

- The solution of π· is :

(C) π· (D) π·

Solutions : π· .....(1)

Put y = vx π·

β΄ (1) becomes π·

π·

π· π·

π·.

Hence (A) is the correct answer.

- Order and degree of differential equation π· are :

(C) 1, 4 (D) 2, 4

Solutions : On simplification the equation becomes π·

Hence (D) is the correct answer.

- The order of differential equation π· is :

(C) π· (D) 6

Solutions : Clearly the equation is of order 2.

Hence (A) is the correct answer.

- If m and n are order and degree of the equation

(A) m = 3, n = 3 (B) m = 3, n = 2

(C) m = 3, n = 5 (D) m = 3, n = 1

Solutions : The given differential equation can be written as

π·

β m = 3, n = 2

Hence (B) is the correct answer.

- The solution of the differential equation π· represent

(C) parabolas (D) ellipses

Solutions : π· π·

β 2log(y + 3) = logx + log c

β (y + 3)2 = ex, which represent parabolas.

Hence (C) is the correct answer.

- Solution of differential equation dy β sinx siny dx = 0 is :

(C) cos x tan y = c (D) cos x sin y = c

Solutions : Given equation can be written as :

π·

π·

π·

π·

π·

Hence (A) is the correct answer.

- Solution of differential equation π· is :

(C) y = emx + ceβax (D) (a + m) y = emx + ceβax

Solutions : I.F. = eβ«adx = eax

β΄ Solution is π·

π·

β (a + m)y = emx + c1 (a + m) eβax

β (a + m) y = emx + ceβax

where c = c1 (a + m)

Hence (D) is the correct answer.

- Integrating factor of the differential equation π· is :

(C) sec x (D) sin x

Solutions : π·

Here P = tan x, Q = sec x

β΄ I. F. = eβ«tan x dx = elog secx = sec x.

Hence (C) is the correct answer.

- Solution of differential equation xdy β ydx = 0 represents

(B) straight line passing through origin

(C) parabola whose vertex is at origin

(D) circle whose centre is at origin

Solutions : x dy β y dx = 0 π·

β log y β log x = log c

π·

π· or y = cx which is a straight line.

Hence (B) is the correct answer.

- The integrating factor of the differential equation π· is given by

(C) log(log x) (D) x

Solutions : π·

π· π·

π·

Hence (B) is the correct answer.

- The solution of π· is :

(C) (x β y)ex+y = 1 (D) (x β c)ex+y + 1 = 0

Solutions : π· π·

Put eβy = z π· π·

π· β΄ P = β1, Q = βex.

β΄ I.F. = eβ«β1dx = eβx

Solution is z . eβx = β« βex . eβx dx + c = βx + c

β eβy . eβx = c β x β eβ(x+y) = c β x

β (x β c) ex+y + 1 = 0

Hence (D) is the correct answer.

- Family y = Ax + A3 of curves is represented by the differential equation of degree

(C) 1 (D) None of these

Solutions : There is only one arbitrary constant.

So, degree of the differential equation is one.

Hence (C) is the correct answer.

- Integrating factor of π·; (x > 0) is :

(C) βx (D) ex

Solutions : I.F. π·

Hence (A) is the correct answer.

- If integrating factor of x(1 β x2) dy + (2x2y β y β ax3) dx = 0 is eβ«Pdx, then P is :

(C) π· (D) π·

Solutions : π·

π·

π·

Hence (D) is the correct answer.

- For solving π·, suitable substitution is :

(C) y = 4x (D) y + 4x + 1 = v

Solutions : y + 4x + 1 = v

Hence (D) is the correct answer.

- Integral curve satisfying π· has the slope at the point (1, 0) of the curve, equal to :

(C) 1 (D) π·

Solutions : π·

β slope at (1, 0) = π·

Hence (C) is the correct answer.

- A solution of the differential equation π· is :

(C) y = 2x β 4 (D) y = 2x2 β 4

Solutions : Ans is (C)

- The solution of π· is :

(C) xy2 = 2y5 + c (D) x(y2 + xy) = 0

Solutions : π· β β2x + 10y3 β π·

π·

β΄ I.F. = π·

β΄ Solution is π·

π·.

Hence (C) is the correct answer.

- A particle moves in a straight line with velocity given by π· (x being the distance described). The time taken by the particle to describe 99 metres is :

(C) 2 log10 e (D) π·

Solutions : π· π·

β log(x + 1) = t + c ....(1)

Initially, when t = 0, x = 0

β΄ c = 0

β΄ log (x + 1) = t

When x = 99, then t = loge (100) = 2 loge 10.

Hence (B) is the correct answer.

- The curve for which slope of the tangent at any point equals the ratio of the abscissa to the ordinate of the point is a/an

(C) circle (D) parabola

Solutions : π· β ydy = xdx

π· β y2 β x2 = 2c

which is a rectangular hyperbola

Hence (B) is the correct answer.

- The solution of the differential equation (1 β xy β x5y5) dx β x2(x4y4 + 1)dy = 0 is given by

(C) x = π· (D) None of these

Solution: The given equation is dx β x (ydx+ xdy )= x5y4(ydx + xdy)

π· β lnx = xy + π·

β x = π·.

Hence (A) is the correct answer.

- If π·satisfies the relation π·then value of A and B respectively are:

(C) β13, 12 (D) 12, β13

Solution: On differentiatingπ·, w.r.t x we get

π·π·

Putting these values in π·, we get

π·.

On solving we get A= β13 and B = β12

Hence (B) is the correct answer.

- The differential equation of all circles which pass through the origin and whose centres lie on y-axis is :

(C) π· (D) π·

Solutions : If (0, a) is centre on y-axis, then its radius is a because it passes through origin.

β΄ Equation of circle is x2 + (y β a)2 = a2

β x2 + y2 β 2ay = 0 ....(1)

π· .....(2)

Using (1) in (2), π·

π·

π·

Hence (A) is the correct answer.

- If π·(logy β logx + 1) then the solution of the equation is :

(C) π· (D) π·

Solutions : π·

Put y = vx β π·

π·

π·

β log(log v) = logx + logc = logcx

β log v = cx π·.

Hence (D) is the correct answer.

- The degree of the differential equation satisfying

(A) 2 (B) 3

(C) 4 (D) None of these

Solution: Putting x = tanΞΈ and y = tanΟ. Then equation becomes

secΞΈ + secΟ = A (tanΞΈ secΟ β tanΟsecΞΈ).

β π·βπ·

βπ·β π· βtan β1x β tan β1y = 2cot β1A

Differentiating, we get π·

Which is a differential equation of degree 1.

Hence (D) is the correct answer.

2023.01.11 00:43 *InfiniteClient4631* **18 and homeless with a cat and my grandma**

So this is a long story and i honestly don't what to do or how to feel i am an 18 year old female from somerset ky and i have questions and don't know how to ask so let me just tell you the story and how it begin i was in foster for not attending school because of anxiety and at the time i did know what it was and just told people i don't know why i hate school so much and they all called me a trouble teen and i guess i was because i didn't go well shortly on my senior year i got placed in foster care and so little time to catch up and graduate but i did it but the only reason why i graduated was because of my grandma i wanted to live with her and take care of her and i did just that and on our way on getting food we came across a little kitten in the sewer she was scared but i eventually got her and tool her home and took care of her and named her priscilla aka prissy for short and well shortly after that our place that we was staying at told us we had only a month or two to get out or they'll take my grandma to court well my grandma and i tried everything to find a place to live but she only draws 1052 dollars a month everything is so expensive we needed enough money to have food but we don't know anybody in this town and we only keep to our selves we have contacted people to help and our family members barely help us and won't let us stay with them but the only reason we got thrown out was because of my grandma's hoarding problem and i am scared we will get kicked out again and where she was evicted and we lived in under housing authority goverment housing it's hard to find places anymore i need comfort and the good news is some guy said he was willing to rent to us but he hasn't given us a date to when the apartment is ready and hasn't called us or hasn't answered to our phone calls and people keep telling me to leave her she caused this on her own and is ruining my life and taking the opportunities and stealing my future and they also say get rid the cat you both will have somewhere to lay but priscilla has already went through that as a baby and she won't have a home and she will search for food i will not put her back to where she came from that is awful and cruel she gives me hope and happiness through the pain i know i can't fix because i also know nothing and don't know where to start and people close to me tell me to leave but i can't just leave the both of them alone to suffer yes i have choices but there is a path for all three of us i hope atleast and i read so many stories and reddit posts but never one like this i just need someone to tell me how to fix this or atleast tell me something good i am so scared of the world and feel so depressed and sad feel like i failed at taking care of the people i love

submitted by InfiniteClient4631 to homeless [link] [comments]
2023.01.08 19:12 *angibrown76* **Story suggestion - Somerset KY, early 80βs. Man murdered his identical twin, hid the body in the bedroom, and pretended to be both brothers for a couple of months.**

submitted by angibrown76 to mrballen [link] [comments] |