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 GulabwaniRohit 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 GulabwaniRohit 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 GulabwaniRohit 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 GulabwaniRohit Gulabwani
4819
edited Dec 16 '16 at 11:48
Mythomorphic
5,3321733
asked Dec 16 '16 at 5:41
Rohit GulabwaniRohit 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 GulabwaniRohit Gulabwani
4819
asked Dec 16 '16 at 5:41
Rohit GulabwaniRohit Gulabwani
4819
asked Dec 16 '16 at 5:41
Rohit GulabwaniRohit 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
JJacquelinJJacquelin
42.8k21750
add a comment |
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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
JJacquelinJJacquelin
42.8k21750
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
JJacquelinJJacquelin
42.8k21750
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
JJacquelinJJacquelin
42.8k21750
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
JJacquelinJJacquelin
42.8k21750
edited Dec 16 '16 at 12:12
answered Dec 16 '16 at 8:01
JJacquelinJJacquelin
42.8k21750
answered Dec 16 '16 at 8:01
JJacquelinJJacquelin
42.8k21750
answered Dec 16 '16 at 8:01
JJacquelinJJacquelin
42.8k21750
42.8k21750
add a comment |
add a comment |
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