Transformation in PDE
Using the transformation $displaystyle u=frac{w}{y}$ in the PDE:
$$xfrac{du}{dx} = u + yfrac{du}{dy}$$
The transformed equation has a solution of the form
"$w=?$".
What is the method to solve such question?
I substituted $displaystyle u=frac{w}{y}$ in the equation and then tried to solve. I got:
$$-xfrac{w}{y^2} frac{dy}{dx} = u - frac{w}{y}$$
Is this correct?
pde partial-derivative
edited Dec 16 '16 at 11:48
Mythomorphic
5,3321733
asked Dec 16 '16 at 5:41
Rohit Gulabwani
4819
add a comment |
Using the transformation $displaystyle u=frac{w}{y}$ in the PDE:
$$xfrac{du}{dx} = u + yfrac{du}{dy}$$
The transformed equation has a solution of the form
"$w=?$".
What is the method to solve such question?
I substituted $displaystyle u=frac{w}{y}$ in the equation and then tried to solve. I got:
$$-xfrac{w}{y^2} frac{dy}{dx} = u - frac{w}{y}$$
Is this correct?
pde partial-derivative
edited Dec 16 '16 at 11:48
Mythomorphic
5,3321733
asked Dec 16 '16 at 5:41
Rohit Gulabwani
4819
add a comment |
Using the transformation $displaystyle u=frac{w}{y}$ in the PDE:
$$xfrac{du}{dx} = u + yfrac{du}{dy}$$
The transformed equation has a solution of the form
"$w=?$".
What is the method to solve such question?
I substituted $displaystyle u=frac{w}{y}$ in the equation and then tried to solve. I got:
$$-xfrac{w}{y^2} frac{dy}{dx} = u - frac{w}{y}$$
Is this correct?
pde partial-derivative
edited Dec 16 '16 at 11:48
Mythomorphic
5,3321733
asked Dec 16 '16 at 5:41
Rohit Gulabwani
4819
Using the transformation $displaystyle u=frac{w}{y}$ in the PDE:
$$xfrac{du}{dx} = u + yfrac{du}{dy}$$
The transformed equation has a solution of the form
"$w=?$".
What is the method to solve such question?
I substituted $displaystyle u=frac{w}{y}$ in the equation and then tried to solve. I got:
$$-xfrac{w}{y^2} frac{dy}{dx} = u - frac{w}{y}$$
Is this correct?
pde partial-derivative
pde partial-derivative
edited Dec 16 '16 at 11:48
Mythomorphic
5,3321733
asked Dec 16 '16 at 5:41
Rohit Gulabwani
4819
edited Dec 16 '16 at 11:48
Mythomorphic
5,3321733
asked Dec 16 '16 at 5:41
Rohit Gulabwani
4819
edited Dec 16 '16 at 11:48
Mythomorphic
5,3321733
edited Dec 16 '16 at 11:48
Mythomorphic
5,3321733
edited Dec 16 '16 at 11:48
Mythomorphic
5,3321733
5,3321733
asked Dec 16 '16 at 5:41
Rohit Gulabwani
4819
asked Dec 16 '16 at 5:41
Rohit Gulabwani
4819
asked Dec 16 '16 at 5:41
Rohit Gulabwani
4819
4819
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
What you got : $quad -xfrac{w}{y^2} frac{dy}{dx} = u - frac{w}{y}quad $
is not correct.
Since you didn't show the detail of your calculus, it's not possible to show you where is the mistake. I suppose that you can find the correct result by yourself. Or show us what you have done.
For this PDE the method of change of function is very good. Even simpler, use the change $u(x,y)=xv(x,y)$.
For information only, another method method is shown below (method of characteristics) :
The equations of characteristics curves are : $quad frac{dx}{x}=-frac{dy}{y}=frac{du}{u}quadtoquad begin{cases}xy=c_1 \ frac{u}{x}=c_2end{cases}$
This is valid with independent $c_1$ and $c_2$ on the characteristic curves only. Elsewhere $c_1$ and $c_2$ are dependent : $c_2=f(c_1)$. So, the general solution of the PDE is :
$$u=xf(xy)$$
with any differentiable function $f$.
answered Dec 16 '16 at 8:01
JJacquelin
42.5k21750
add a comment |
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
});
}
});
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%2f2060868%2ftransformation-in-pde%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
What you got : $quad -xfrac{w}{y^2} frac{dy}{dx} = u - frac{w}{y}quad $
is not correct.
Since you didn't show the detail of your calculus, it's not possible to show you where is the mistake. I suppose that you can find the correct result by yourself. Or show us what you have done.
For this PDE the method of change of function is very good. Even simpler, use the change $u(x,y)=xv(x,y)$.
For information only, another method method is shown below (method of characteristics) :
The equations of characteristics curves are : $quad frac{dx}{x}=-frac{dy}{y}=frac{du}{u}quadtoquad begin{cases}xy=c_1 \ frac{u}{x}=c_2end{cases}$
This is valid with independent $c_1$ and $c_2$ on the characteristic curves only. Elsewhere $c_1$ and $c_2$ are dependent : $c_2=f(c_1)$. So, the general solution of the PDE is :
$$u=xf(xy)$$
with any differentiable function $f$.
answered Dec 16 '16 at 8:01
JJacquelin
42.5k21750
add a comment |
What you got : $quad -xfrac{w}{y^2} frac{dy}{dx} = u - frac{w}{y}quad $
is not correct.
Since you didn't show the detail of your calculus, it's not possible to show you where is the mistake. I suppose that you can find the correct result by yourself. Or show us what you have done.
For this PDE the method of change of function is very good. Even simpler, use the change $u(x,y)=xv(x,y)$.
For information only, another method method is shown below (method of characteristics) :
The equations of characteristics curves are : $quad frac{dx}{x}=-frac{dy}{y}=frac{du}{u}quadtoquad begin{cases}xy=c_1 \ frac{u}{x}=c_2end{cases}$
This is valid with independent $c_1$ and $c_2$ on the characteristic curves only. Elsewhere $c_1$ and $c_2$ are dependent : $c_2=f(c_1)$. So, the general solution of the PDE is :
$$u=xf(xy)$$
with any differentiable function $f$.
answered Dec 16 '16 at 8:01
JJacquelin
42.5k21750
add a comment |
What you got : $quad -xfrac{w}{y^2} frac{dy}{dx} = u - frac{w}{y}quad $
is not correct.
Since you didn't show the detail of your calculus, it's not possible to show you where is the mistake. I suppose that you can find the correct result by yourself. Or show us what you have done.
For this PDE the method of change of function is very good. Even simpler, use the change $u(x,y)=xv(x,y)$.
For information only, another method method is shown below (method of characteristics) :
The equations of characteristics curves are : $quad frac{dx}{x}=-frac{dy}{y}=frac{du}{u}quadtoquad begin{cases}xy=c_1 \ frac{u}{x}=c_2end{cases}$
This is valid with independent $c_1$ and $c_2$ on the characteristic curves only. Elsewhere $c_1$ and $c_2$ are dependent : $c_2=f(c_1)$. So, the general solution of the PDE is :
$$u=xf(xy)$$
with any differentiable function $f$.
answered Dec 16 '16 at 8:01
JJacquelin
42.5k21750
What you got : $quad -xfrac{w}{y^2} frac{dy}{dx} = u - frac{w}{y}quad $
is not correct.
Since you didn't show the detail of your calculus, it's not possible to show you where is the mistake. I suppose that you can find the correct result by yourself. Or show us what you have done.
For this PDE the method of change of function is very good. Even simpler, use the change $u(x,y)=xv(x,y)$.
For information only, another method method is shown below (method of characteristics) :
The equations of characteristics curves are : $quad frac{dx}{x}=-frac{dy}{y}=frac{du}{u}quadtoquad begin{cases}xy=c_1 \ frac{u}{x}=c_2end{cases}$
This is valid with independent $c_1$ and $c_2$ on the characteristic curves only. Elsewhere $c_1$ and $c_2$ are dependent : $c_2=f(c_1)$. So, the general solution of the PDE is :
$$u=xf(xy)$$
with any differentiable function $f$.
answered Dec 16 '16 at 8:01
JJacquelin
42.5k21750
edited Dec 16 '16 at 12:12
answered Dec 16 '16 at 8:01
JJacquelin
42.5k21750
answered Dec 16 '16 at 8:01
JJacquelin
42.5k21750
answered Dec 16 '16 at 8:01
JJacquelin
42.5k21750
42.5k21750
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f2060868%2ftransformation-in-pde%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
