Transformation in PDE

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Using the transformation $displaystyle u=fracwy$ in the PDE:
$$xfracdudx = u + yfracdudy$$
The transformed equation has a solution of the form
"$w=?$".

What is the method to solve such question?

I substituted $displaystyle u=fracwy$ in the equation and then tried to solve. I got:
$$-xfracwy^2 fracdydx = u - fracwy$$
Is this correct?

edited Dec 16 '16 at 11:48

Mythomorphic

5,1811733

asked Dec 16 '16 at 5:41

Rohit Gulabwani

489

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up vote
0
down vote

favorite

Using the transformation $displaystyle u=fracwy$ in the PDE:
$$xfracdudx = u + yfracdudy$$
The transformed equation has a solution of the form
"$w=?$".

What is the method to solve such question?

I substituted $displaystyle u=fracwy$ in the equation and then tried to solve. I got:
$$-xfracwy^2 fracdydx = u - fracwy$$
Is this correct?

edited Dec 16 '16 at 11:48

Mythomorphic

5,1811733

asked Dec 16 '16 at 5:41

Rohit Gulabwani

489

add a commentÂ |Â

up vote
0
down vote

favorite

Using the transformation $displaystyle u=fracwy$ in the PDE:
$$xfracdudx = u + yfracdudy$$
The transformed equation has a solution of the form
"$w=?$".

What is the method to solve such question?

I substituted $displaystyle u=fracwy$ in the equation and then tried to solve. I got:
$$-xfracwy^2 fracdydx = u - fracwy$$
Is this correct?

edited Dec 16 '16 at 11:48

Mythomorphic

5,1811733

asked Dec 16 '16 at 5:41

Rohit Gulabwani

489

Using the transformation $displaystyle u=fracwy$ in the PDE:
$$xfracdudx = u + yfracdudy$$
The transformed equation has a solution of the form
"$w=?$".

What is the method to solve such question?

I substituted $displaystyle u=fracwy$ in the equation and then tried to solve. I got:
$$-xfracwy^2 fracdydx = u - fracwy$$
Is this correct?

edited Dec 16 '16 at 11:48

Mythomorphic

5,1811733

asked Dec 16 '16 at 5:41

Rohit Gulabwani

489

edited Dec 16 '16 at 11:48

Mythomorphic

5,1811733

edited Dec 16 '16 at 11:48

Mythomorphic

5,1811733

edited Dec 16 '16 at 11:48

Mythomorphic

5,1811733

asked Dec 16 '16 at 5:41

Rohit Gulabwani

489

asked Dec 16 '16 at 5:41

Rohit Gulabwani

489

asked Dec 16 '16 at 5:41

Rohit Gulabwani

489

add a commentÂ |Â

1 Answer
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What you got : $quad -xfracwy^2 fracdydx = u - fracwyquad $
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 : $quad fracdxx=-fracdyy=fracduuquadtoquad begincasesxy=c_1 \ fracux=c_2endcases$

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 $f$.

edited Dec 16 '16 at 12:12

answered Dec 16 '16 at 8:01

JJacquelin

40.6k21650

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1 Answer
1

active

oldest

votes

up vote
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down vote

What you got : $quad -xfracwy^2 fracdydx = u - fracwyquad $
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 : $quad fracdxx=-fracdyy=fracduuquadtoquad begincasesxy=c_1 \ fracux=c_2endcases$

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 $f$.

edited Dec 16 '16 at 12:12

answered Dec 16 '16 at 8:01

JJacquelin

40.6k21650

add a commentÂ |Â

up vote
0
down vote

What you got : $quad -xfracwy^2 fracdydx = u - fracwyquad $
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 : $quad fracdxx=-fracdyy=fracduuquadtoquad begincasesxy=c_1 \ fracux=c_2endcases$

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 $f$.

edited Dec 16 '16 at 12:12

answered Dec 16 '16 at 8:01

JJacquelin

40.6k21650

add a commentÂ |Â

up vote
0
down vote

What you got : $quad -xfracwy^2 fracdydx = u - fracwyquad $
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 : $quad fracdxx=-fracdyy=fracduuquadtoquad begincasesxy=c_1 \ fracux=c_2endcases$

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 $f$.

edited Dec 16 '16 at 12:12

answered Dec 16 '16 at 8:01

JJacquelin

40.6k21650

What you got : $quad -xfracwy^2 fracdydx = u - fracwyquad $
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 : $quad fracdxx=-fracdyy=fracduuquadtoquad begincasesxy=c_1 \ fracux=c_2endcases$

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 $f$.

edited Dec 16 '16 at 12:12

answered Dec 16 '16 at 8:01

JJacquelin

40.6k21650

edited Dec 16 '16 at 12:12

answered Dec 16 '16 at 8:01

JJacquelin

40.6k21650

answered Dec 16 '16 at 8:01

JJacquelin

40.6k21650

answered Dec 16 '16 at 8:01

JJacquelin

40.6k21650

add a commentÂ |Â

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