A simple looking DE












0












$begingroup$


While working on a physics problem, I have encountered the following DE.



$dx(y-x)+dy(k-y)=0$ where $k$ is a constant.



I have tried various approaches, like trying to concert into a linear DE form, and various other stuff in my knowledge, none if them seem to yield much.










share|cite|improve this question









$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/D%27Alembert%27s_equation
    $endgroup$
    – whpowell96
    Jan 15 at 17:46
















0












$begingroup$


While working on a physics problem, I have encountered the following DE.



$dx(y-x)+dy(k-y)=0$ where $k$ is a constant.



I have tried various approaches, like trying to concert into a linear DE form, and various other stuff in my knowledge, none if them seem to yield much.










share|cite|improve this question









$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/D%27Alembert%27s_equation
    $endgroup$
    – whpowell96
    Jan 15 at 17:46














0












0








0





$begingroup$


While working on a physics problem, I have encountered the following DE.



$dx(y-x)+dy(k-y)=0$ where $k$ is a constant.



I have tried various approaches, like trying to concert into a linear DE form, and various other stuff in my knowledge, none if them seem to yield much.










share|cite|improve this question









$endgroup$




While working on a physics problem, I have encountered the following DE.



$dx(y-x)+dy(k-y)=0$ where $k$ is a constant.



I have tried various approaches, like trying to concert into a linear DE form, and various other stuff in my knowledge, none if them seem to yield much.







ordinary-differential-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 15 at 17:37









Esha ManideepEsha Manideep

34




34












  • $begingroup$
    en.wikipedia.org/wiki/D%27Alembert%27s_equation
    $endgroup$
    – whpowell96
    Jan 15 at 17:46


















  • $begingroup$
    en.wikipedia.org/wiki/D%27Alembert%27s_equation
    $endgroup$
    – whpowell96
    Jan 15 at 17:46
















$begingroup$
en.wikipedia.org/wiki/D%27Alembert%27s_equation
$endgroup$
– whpowell96
Jan 15 at 17:46




$begingroup$
en.wikipedia.org/wiki/D%27Alembert%27s_equation
$endgroup$
– whpowell96
Jan 15 at 17:46










1 Answer
1






active

oldest

votes


















0












$begingroup$

Write your equation in the form $$frac{dy}{dx}=frac{x-y(x)}{k-y(x)}$$ now we substitute $$x=k+t,y=k+v$$ then we get
$$frac{dv(t)}{dt}=frac{t-v(t)}{v(t)}$$ now we substitute $$v(t)=tu(t)$$ and we get
$$tfrac{du(t)}{dt}+u(t)=-frac{t-tu(t)}{tu(t)}$$ and this is $$frac{du(t)}{dt}=frac{-u(t)^2+u(t)-1}{tu(t)}$$ and this can be written as
$$intfrac{frac{du(t)}{dt}u(t)}{-u(t)^2+u(t)-1}=intfrac{1}{t}dt$$
Can you finish?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    In your fifth mathematical equation, RHS's numerator should contain $tu(t)$ right ?
    $endgroup$
    – Esha Manideep
    Jan 15 at 18:19














Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074689%2fa-simple-looking-de%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Write your equation in the form $$frac{dy}{dx}=frac{x-y(x)}{k-y(x)}$$ now we substitute $$x=k+t,y=k+v$$ then we get
$$frac{dv(t)}{dt}=frac{t-v(t)}{v(t)}$$ now we substitute $$v(t)=tu(t)$$ and we get
$$tfrac{du(t)}{dt}+u(t)=-frac{t-tu(t)}{tu(t)}$$ and this is $$frac{du(t)}{dt}=frac{-u(t)^2+u(t)-1}{tu(t)}$$ and this can be written as
$$intfrac{frac{du(t)}{dt}u(t)}{-u(t)^2+u(t)-1}=intfrac{1}{t}dt$$
Can you finish?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    In your fifth mathematical equation, RHS's numerator should contain $tu(t)$ right ?
    $endgroup$
    – Esha Manideep
    Jan 15 at 18:19


















0












$begingroup$

Write your equation in the form $$frac{dy}{dx}=frac{x-y(x)}{k-y(x)}$$ now we substitute $$x=k+t,y=k+v$$ then we get
$$frac{dv(t)}{dt}=frac{t-v(t)}{v(t)}$$ now we substitute $$v(t)=tu(t)$$ and we get
$$tfrac{du(t)}{dt}+u(t)=-frac{t-tu(t)}{tu(t)}$$ and this is $$frac{du(t)}{dt}=frac{-u(t)^2+u(t)-1}{tu(t)}$$ and this can be written as
$$intfrac{frac{du(t)}{dt}u(t)}{-u(t)^2+u(t)-1}=intfrac{1}{t}dt$$
Can you finish?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    In your fifth mathematical equation, RHS's numerator should contain $tu(t)$ right ?
    $endgroup$
    – Esha Manideep
    Jan 15 at 18:19
















0












0








0





$begingroup$

Write your equation in the form $$frac{dy}{dx}=frac{x-y(x)}{k-y(x)}$$ now we substitute $$x=k+t,y=k+v$$ then we get
$$frac{dv(t)}{dt}=frac{t-v(t)}{v(t)}$$ now we substitute $$v(t)=tu(t)$$ and we get
$$tfrac{du(t)}{dt}+u(t)=-frac{t-tu(t)}{tu(t)}$$ and this is $$frac{du(t)}{dt}=frac{-u(t)^2+u(t)-1}{tu(t)}$$ and this can be written as
$$intfrac{frac{du(t)}{dt}u(t)}{-u(t)^2+u(t)-1}=intfrac{1}{t}dt$$
Can you finish?






share|cite|improve this answer











$endgroup$



Write your equation in the form $$frac{dy}{dx}=frac{x-y(x)}{k-y(x)}$$ now we substitute $$x=k+t,y=k+v$$ then we get
$$frac{dv(t)}{dt}=frac{t-v(t)}{v(t)}$$ now we substitute $$v(t)=tu(t)$$ and we get
$$tfrac{du(t)}{dt}+u(t)=-frac{t-tu(t)}{tu(t)}$$ and this is $$frac{du(t)}{dt}=frac{-u(t)^2+u(t)-1}{tu(t)}$$ and this can be written as
$$intfrac{frac{du(t)}{dt}u(t)}{-u(t)^2+u(t)-1}=intfrac{1}{t}dt$$
Can you finish?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 16 at 8:08

























answered Jan 15 at 17:55









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867




78.4k42867








  • 1




    $begingroup$
    In your fifth mathematical equation, RHS's numerator should contain $tu(t)$ right ?
    $endgroup$
    – Esha Manideep
    Jan 15 at 18:19
















  • 1




    $begingroup$
    In your fifth mathematical equation, RHS's numerator should contain $tu(t)$ right ?
    $endgroup$
    – Esha Manideep
    Jan 15 at 18:19










1




1




$begingroup$
In your fifth mathematical equation, RHS's numerator should contain $tu(t)$ right ?
$endgroup$
– Esha Manideep
Jan 15 at 18:19






$begingroup$
In your fifth mathematical equation, RHS's numerator should contain $tu(t)$ right ?
$endgroup$
– Esha Manideep
Jan 15 at 18:19




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074689%2fa-simple-looking-de%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Human spaceflight

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

張江高科駅