Fdtxjr

Triggering an ultrasonic sensor

Why do early math courses focus on the cross sections of a cone and not on other 3D objects?

Has negative voting ever been officially implemented in elections, or seriously proposed, or even studied?

Using audio cues to encourage good posture

How to dry out epoxy resin faster than usual?

Should I use a zero-interest credit card for a large one-time purchase?

Take 2! Is this homebrew Lady of Pain warlock patron balanced?

Did Mueller's report provide an evidentiary basis for the claim of Russian govt election interference via social media?

Sum letters are not two different

How to compare two different files line by line in unix?

Time to Settle Down!

Did any compiler fully use 80-bit floating point?

What initially awakened the Balrog?

Why do we bend a book to keep it straight?

How fail-safe is nr as stop bytes?

What is the meaning of 'breadth' in breadth first search?

What order were files/directories outputted in dir?

How would a mousetrap for use in space work?

Find 108 by using 3,4,6

Project Euler #1 in C++

What is "gratricide"?

How to run automated tests after each commit?

MLE of the unknown radius

Can a new player join a group only when a new campaign starts?

Solving an initial value problem involving the second-order nonlinear ordinary differential equation $y''(x) = y(x) cdot y'(x) + (y'(x))^2$

1

$begingroup$

I have the following equation $y''(x) = y(x) cdot y'(x) + (y'(x))^2$ with the initial values $y(1) = 0$ and $y'(1) = -1$.

I am seeking some guidance for how to best tackle this particular problem.

this article suggests that a v substitution may reduce the problem in some capacity. Thus I let $v = y'(x)$ as follows:

beginalign*
y''(x) &= y(x) cdot y'(x) + (y'(x))^2 & (1) \
y''(x) &= y(x) cdot v + v^2 & (2) \
endalign*

When I enter (2) into wolfram, the engine says that the equation is a second-order linear ordinary differential equation.

Can the substitution be leveraged from here to solve the problem? Or is there a better way of doing this?

ordinary-differential-equations initial-value-problems

share|cite|improve this question

asked Mar 27 at 19:42

OscarOscar

358111

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think the solution to this ODE has a closed form.
  $endgroup$
  – Peter Foreman
  Mar 27 at 20:01
  

  
  
  
  

add a comment | 

1

$begingroup$

I have the following equation $y''(x) = y(x) cdot y'(x) + (y'(x))^2$ with the initial values $y(1) = 0$ and $y'(1) = -1$.

I am seeking some guidance for how to best tackle this particular problem.

this article suggests that a v substitution may reduce the problem in some capacity. Thus I let $v = y'(x)$ as follows:

beginalign*
y''(x) &= y(x) cdot y'(x) + (y'(x))^2 & (1) \
y''(x) &= y(x) cdot v + v^2 & (2) \
endalign*

When I enter (2) into wolfram, the engine says that the equation is a second-order linear ordinary differential equation.

Can the substitution be leveraged from here to solve the problem? Or is there a better way of doing this?

ordinary-differential-equations initial-value-problems

share|cite|improve this question

asked Mar 27 at 19:42

OscarOscar

358111

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think the solution to this ODE has a closed form.
  $endgroup$
  – Peter Foreman
  Mar 27 at 20:01
  

  
  
  
  

add a comment | 

$begingroup$

I have the following equation $y''(x) = y(x) cdot y'(x) + (y'(x))^2$ with the initial values $y(1) = 0$ and $y'(1) = -1$.

I am seeking some guidance for how to best tackle this particular problem.

this article suggests that a v substitution may reduce the problem in some capacity. Thus I let $v = y'(x)$ as follows:

beginalign*
y''(x) &= y(x) cdot y'(x) + (y'(x))^2 & (1) \
y''(x) &= y(x) cdot v + v^2 & (2) \
endalign*

When I enter (2) into wolfram, the engine says that the equation is a second-order linear ordinary differential equation.

Can the substitution be leveraged from here to solve the problem? Or is there a better way of doing this?

ordinary-differential-equations initial-value-problems

share|cite|improve this question

asked Mar 27 at 19:42

OscarOscar

358111

$endgroup$

share|cite|improve this question

asked Mar 27 at 19:42

OscarOscar

358111

share|cite|improve this question

asked Mar 27 at 19:42

OscarOscar

358111

asked Mar 27 at 19:42

OscarOscar

358111

asked Mar 27 at 19:42

OscarOscar

358111

2 Answers
2

active

oldest

votes

3

$begingroup$

The idea that is proposed is to try to find an equation for a function $v$ so that $y'(x)=v(y(x))$. By the chain rule, $y''=v'v$ and thus the equation reads as
$$
vv'=yv+v^2implies v=0lor v'=y+v.
$$
The first case gives constant solutions, but contradicts the initicial conditions. In the second case, $v(y)=Ce^y-y-1$. With $v(0)=-1$ one finds $C=0$. For this exceptional case one can now also solve the resulting first order equation
$$
y'(x)=-y(x)-1implies y(x)=e^1-x-1.
$$

share|cite|improve this answer

answered Mar 27 at 20:11

LutzLLutzL

60.8k42057

$endgroup$

add a comment | 

0

$begingroup$

I think Wolfram fails to realize $y$ and $v$ are related. You should end up with the system of equations
$$
left{
beginmatrix
v &= y' &\
v' &= yv &+ v^2
endmatrix
right|
$$

share|cite|improve this answer

answered Mar 27 at 19:48

gt6989bgt6989b

36k22557

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3165032%2fsolving-an-initial-value-problem-involving-the-second-order-nonlinear-ordinary-d%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

The idea that is proposed is to try to find an equation for a function $v$ so that $y'(x)=v(y(x))$. By the chain rule, $y''=v'v$ and thus the equation reads as
$$
vv'=yv+v^2implies v=0lor v'=y+v.
$$
The first case gives constant solutions, but contradicts the initicial conditions. In the second case, $v(y)=Ce^y-y-1$. With $v(0)=-1$ one finds $C=0$. For this exceptional case one can now also solve the resulting first order equation
$$
y'(x)=-y(x)-1implies y(x)=e^1-x-1.
$$

share|cite|improve this answer

answered Mar 27 at 20:11

LutzLLutzL

60.8k42057

$endgroup$

add a comment | 

3

$begingroup$

The idea that is proposed is to try to find an equation for a function $v$ so that $y'(x)=v(y(x))$. By the chain rule, $y''=v'v$ and thus the equation reads as
$$
vv'=yv+v^2implies v=0lor v'=y+v.
$$
The first case gives constant solutions, but contradicts the initicial conditions. In the second case, $v(y)=Ce^y-y-1$. With $v(0)=-1$ one finds $C=0$. For this exceptional case one can now also solve the resulting first order equation
$$
y'(x)=-y(x)-1implies y(x)=e^1-x-1.
$$

share|cite|improve this answer

answered Mar 27 at 20:11

LutzLLutzL

60.8k42057

$endgroup$

add a comment | 

$begingroup$

The idea that is proposed is to try to find an equation for a function $v$ so that $y'(x)=v(y(x))$. By the chain rule, $y''=v'v$ and thus the equation reads as
$$
vv'=yv+v^2implies v=0lor v'=y+v.
$$
The first case gives constant solutions, but contradicts the initicial conditions. In the second case, $v(y)=Ce^y-y-1$. With $v(0)=-1$ one finds $C=0$. For this exceptional case one can now also solve the resulting first order equation
$$
y'(x)=-y(x)-1implies y(x)=e^1-x-1.
$$

share|cite|improve this answer

answered Mar 27 at 20:11

LutzLLutzL

60.8k42057

$endgroup$

share|cite|improve this answer

answered Mar 27 at 20:11

LutzLLutzL

60.8k42057

answered Mar 27 at 20:11

LutzLLutzL

60.8k42057

answered Mar 27 at 20:11

LutzLLutzL

60.8k42057

0

$begingroup$

I think Wolfram fails to realize $y$ and $v$ are related. You should end up with the system of equations
$$
left{
beginmatrix
v &= y' &\
v' &= yv &+ v^2
endmatrix
right|
$$

share|cite|improve this answer

answered Mar 27 at 19:48

gt6989bgt6989b

36k22557

$endgroup$

add a comment | 

0

$begingroup$

I think Wolfram fails to realize $y$ and $v$ are related. You should end up with the system of equations
$$
left{
beginmatrix
v &= y' &\
v' &= yv &+ v^2
endmatrix
right|
$$

share|cite|improve this answer

answered Mar 27 at 19:48

gt6989bgt6989b

36k22557

$endgroup$

add a comment | 

$begingroup$

I think Wolfram fails to realize $y$ and $v$ are related. You should end up with the system of equations
$$
left{
beginmatrix
v &= y' &\
v' &= yv &+ v^2
endmatrix
right|
$$

share|cite|improve this answer

answered Mar 27 at 19:48

gt6989bgt6989b

36k22557

$endgroup$

share|cite|improve this answer

answered Mar 27 at 19:48

gt6989bgt6989b

36k22557

answered Mar 27 at 19:48

gt6989bgt6989b

36k22557

answered Mar 27 at 19:48

gt6989bgt6989b

36k22557