Solve differential equation $xyy'=x^4+y^4$
$begingroup$
How to find general solution to this differential equation (if it exists):
$$ xyy'=x^4+y^4 ?$$
I do not know how to even approach it since I never dealt with nonlinear equations. Only thing that I notice is that $x$ and $y$ are symetric in equation: $$ y'=dfrac{x^4+y^4}{xy}$$
so I expect (maybe) some kind of symetric function for y(x).
Thanks for any help.
ordinary-differential-equations analysis nonlinear-analysis
asked Jan 12 at 17:01
ThomThom
521111
$endgroup$
add a comment |
$begingroup$
How to find general solution to this differential equation (if it exists):
$$ xyy'=x^4+y^4 ?$$
I do not know how to even approach it since I never dealt with nonlinear equations. Only thing that I notice is that $x$ and $y$ are symetric in equation: $$ y'=dfrac{x^4+y^4}{xy}$$
so I expect (maybe) some kind of symetric function for y(x).
Thanks for any help.
ordinary-differential-equations analysis nonlinear-analysis
asked Jan 12 at 17:01
ThomThom
521111
$endgroup$
$begingroup$
Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck
$endgroup$
– poetasis
Jan 12 at 17:19
$begingroup$
@poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$?
$endgroup$
– Thom
Jan 12 at 17:23
$begingroup$
I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more.
$endgroup$
– poetasis
Jan 12 at 17:25
$begingroup$
Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4),dx-xy,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants.
$endgroup$
– Adrian Keister
Jan 12 at 17:45
3
$begingroup$
The definition of a new dependent variable $z := frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
$endgroup$
– Christoph
Jan 12 at 17:49
add a comment |
$begingroup$
How to find general solution to this differential equation (if it exists):
$$ xyy'=x^4+y^4 ?$$
I do not know how to even approach it since I never dealt with nonlinear equations. Only thing that I notice is that $x$ and $y$ are symetric in equation: $$ y'=dfrac{x^4+y^4}{xy}$$
so I expect (maybe) some kind of symetric function for y(x).
Thanks for any help.
ordinary-differential-equations analysis nonlinear-analysis
asked Jan 12 at 17:01
ThomThom
521111
$endgroup$
How to find general solution to this differential equation (if it exists):
$$ xyy'=x^4+y^4 ?$$
I do not know how to even approach it since I never dealt with nonlinear equations. Only thing that I notice is that $x$ and $y$ are symetric in equation: $$ y'=dfrac{x^4+y^4}{xy}$$
so I expect (maybe) some kind of symetric function for y(x).
Thanks for any help.
ordinary-differential-equations analysis nonlinear-analysis
ordinary-differential-equations analysis nonlinear-analysis
asked Jan 12 at 17:01
ThomThom
521111
asked Jan 12 at 17:01
ThomThom
521111
asked Jan 12 at 17:01
ThomThom
521111
asked Jan 12 at 17:01
ThomThom
521111
asked Jan 12 at 17:01
ThomThom
521111
521111
$begingroup$
Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck
$endgroup$
– poetasis
Jan 12 at 17:19
$begingroup$
@poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$?
$endgroup$
– Thom
Jan 12 at 17:23
$begingroup$
I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more.
$endgroup$
– poetasis
Jan 12 at 17:25
$begingroup$
Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4),dx-xy,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants.
$endgroup$
– Adrian Keister
Jan 12 at 17:45
3
$begingroup$
The definition of a new dependent variable $z := frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
$endgroup$
– Christoph
Jan 12 at 17:49
add a comment |
$begingroup$
Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck
$endgroup$
– poetasis
Jan 12 at 17:19
$begingroup$
@poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$?
$endgroup$
– Thom
Jan 12 at 17:23
$begingroup$
I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more.
$endgroup$
– poetasis
Jan 12 at 17:25
$begingroup$
Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4),dx-xy,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants.
$endgroup$
– Adrian Keister
Jan 12 at 17:45
3
$begingroup$
The definition of a new dependent variable $z := frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
$endgroup$
– Christoph
Jan 12 at 17:49
$begingroup$
Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck
$endgroup$
– poetasis
Jan 12 at 17:19
$begingroup$
Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck
$endgroup$
– poetasis
Jan 12 at 17:19
$begingroup$
@poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$?
$endgroup$
– Thom
Jan 12 at 17:23
$begingroup$
@poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$?
$endgroup$
– Thom
Jan 12 at 17:23
$begingroup$
I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more.
$endgroup$
– poetasis
Jan 12 at 17:25
$begingroup$
I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more.
$endgroup$
– poetasis
Jan 12 at 17:25
$begingroup$
Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4),dx-xy,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants.
$endgroup$
– Adrian Keister
Jan 12 at 17:45
$begingroup$
Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4),dx-xy,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants.
$endgroup$
– Adrian Keister
Jan 12 at 17:45
3
3
$begingroup$
The definition of a new dependent variable $z := frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
$endgroup$
– Christoph
Jan 12 at 17:49
$begingroup$
The definition of a new dependent variable $z := frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
$endgroup$
– Christoph
Jan 12 at 17:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Observing that $y y' = left(frac{1}{2}y^2right)'$ we define the new dependent variable $z := frac{1}{2} y^2$, $y^4 = 4 z^2$. We then obtain a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing
begin{equation}
z = - frac{x}{4} frac{u'}{u}, quad z^2 = frac{x^2}{16} frac{(u')^2}{u^2}, quad z' = - frac{1}{4} frac{u'}{u} - frac{x}{4} frac{u'' u - (u')^2}{u^2},
end{equation}
which yields $x^2 u'' + x u' + 4 x^4 u = 0$. With the definition of a new independent variable
begin{equation}
xi := x^2, quad frac{d}{dx} = 2 xi^{1/2} frac{d}{dxi}, quad frac{d^2}{dx^2} = 2 frac{d}{dxi} + 4 xi frac{d^2}{dxi^2},
end{equation}
we obtain the Bessel differential equation
begin{equation}
xi^2 frac{d^2 u}{d xi^2} + xi frac{d u}{dxi} + xi^2 u = 0,
end{equation}
with fundamental solutions $J_0(xi)$, $Y_0(xi)$ (zeroth-order Bessel functions).
answered Jan 12 at 19:15
ChristophChristoph
59616
$endgroup$
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071118%2fsolve-differential-equation-xyy-x4y4%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
$begingroup$
Observing that $y y' = left(frac{1}{2}y^2right)'$ we define the new dependent variable $z := frac{1}{2} y^2$, $y^4 = 4 z^2$. We then obtain a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing
begin{equation}
z = - frac{x}{4} frac{u'}{u}, quad z^2 = frac{x^2}{16} frac{(u')^2}{u^2}, quad z' = - frac{1}{4} frac{u'}{u} - frac{x}{4} frac{u'' u - (u')^2}{u^2},
end{equation}
which yields $x^2 u'' + x u' + 4 x^4 u = 0$. With the definition of a new independent variable
begin{equation}
xi := x^2, quad frac{d}{dx} = 2 xi^{1/2} frac{d}{dxi}, quad frac{d^2}{dx^2} = 2 frac{d}{dxi} + 4 xi frac{d^2}{dxi^2},
end{equation}
we obtain the Bessel differential equation
begin{equation}
xi^2 frac{d^2 u}{d xi^2} + xi frac{d u}{dxi} + xi^2 u = 0,
end{equation}
with fundamental solutions $J_0(xi)$, $Y_0(xi)$ (zeroth-order Bessel functions).
answered Jan 12 at 19:15
ChristophChristoph
59616
$endgroup$
add a comment |
$begingroup$
Observing that $y y' = left(frac{1}{2}y^2right)'$ we define the new dependent variable $z := frac{1}{2} y^2$, $y^4 = 4 z^2$. We then obtain a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing
begin{equation}
z = - frac{x}{4} frac{u'}{u}, quad z^2 = frac{x^2}{16} frac{(u')^2}{u^2}, quad z' = - frac{1}{4} frac{u'}{u} - frac{x}{4} frac{u'' u - (u')^2}{u^2},
end{equation}
which yields $x^2 u'' + x u' + 4 x^4 u = 0$. With the definition of a new independent variable
begin{equation}
xi := x^2, quad frac{d}{dx} = 2 xi^{1/2} frac{d}{dxi}, quad frac{d^2}{dx^2} = 2 frac{d}{dxi} + 4 xi frac{d^2}{dxi^2},
end{equation}
we obtain the Bessel differential equation
begin{equation}
xi^2 frac{d^2 u}{d xi^2} + xi frac{d u}{dxi} + xi^2 u = 0,
end{equation}
with fundamental solutions $J_0(xi)$, $Y_0(xi)$ (zeroth-order Bessel functions).
answered Jan 12 at 19:15
ChristophChristoph
59616
$endgroup$
add a comment |
$begingroup$
Observing that $y y' = left(frac{1}{2}y^2right)'$ we define the new dependent variable $z := frac{1}{2} y^2$, $y^4 = 4 z^2$. We then obtain a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing
begin{equation}
z = - frac{x}{4} frac{u'}{u}, quad z^2 = frac{x^2}{16} frac{(u')^2}{u^2}, quad z' = - frac{1}{4} frac{u'}{u} - frac{x}{4} frac{u'' u - (u')^2}{u^2},
end{equation}
which yields $x^2 u'' + x u' + 4 x^4 u = 0$. With the definition of a new independent variable
begin{equation}
xi := x^2, quad frac{d}{dx} = 2 xi^{1/2} frac{d}{dxi}, quad frac{d^2}{dx^2} = 2 frac{d}{dxi} + 4 xi frac{d^2}{dxi^2},
end{equation}
we obtain the Bessel differential equation
begin{equation}
xi^2 frac{d^2 u}{d xi^2} + xi frac{d u}{dxi} + xi^2 u = 0,
end{equation}
with fundamental solutions $J_0(xi)$, $Y_0(xi)$ (zeroth-order Bessel functions).
answered Jan 12 at 19:15
ChristophChristoph
59616
$endgroup$
Observing that $y y' = left(frac{1}{2}y^2right)'$ we define the new dependent variable $z := frac{1}{2} y^2$, $y^4 = 4 z^2$. We then obtain a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing
begin{equation}
z = - frac{x}{4} frac{u'}{u}, quad z^2 = frac{x^2}{16} frac{(u')^2}{u^2}, quad z' = - frac{1}{4} frac{u'}{u} - frac{x}{4} frac{u'' u - (u')^2}{u^2},
end{equation}
which yields $x^2 u'' + x u' + 4 x^4 u = 0$. With the definition of a new independent variable
begin{equation}
xi := x^2, quad frac{d}{dx} = 2 xi^{1/2} frac{d}{dxi}, quad frac{d^2}{dx^2} = 2 frac{d}{dxi} + 4 xi frac{d^2}{dxi^2},
end{equation}
we obtain the Bessel differential equation
begin{equation}
xi^2 frac{d^2 u}{d xi^2} + xi frac{d u}{dxi} + xi^2 u = 0,
end{equation}
with fundamental solutions $J_0(xi)$, $Y_0(xi)$ (zeroth-order Bessel functions).
answered Jan 12 at 19:15
ChristophChristoph
59616
answered Jan 12 at 19:15
ChristophChristoph
59616
answered Jan 12 at 19:15
ChristophChristoph
59616
answered Jan 12 at 19:15
ChristophChristoph
59616
59616
add a comment |
add a comment |
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.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071118%2fsolve-differential-equation-xyy-x4y4%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck
$endgroup$
– poetasis
Jan 12 at 17:19
$begingroup$
@poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$?
$endgroup$
– Thom
Jan 12 at 17:23
$begingroup$
I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more.
$endgroup$
– poetasis
Jan 12 at 17:25
$begingroup$
Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4),dx-xy,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants.
$endgroup$
– Adrian Keister
Jan 12 at 17:45
3
$begingroup$
The definition of a new dependent variable $z := frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + frac{4}{x} z^2$.
$endgroup$
– Christoph
Jan 12 at 17:49