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Find the parametric solution of the PDE

$$xu_x -xyu_y - u=0$$

which follows the side condition $u(s^2, s)=s^3$

The solution uses another method rather than finding the general solution using parametrization of $x,y$ into $s,t$.

It says $X(s,t)=Ae^t$ and $Y(s,t)=Be^{-xt}$ and $U(s,t)=Ce^t$.

(I understand that they found $X(s,t)$ using $dx/dt$. I can't seem to solve $Y(s,t)$ and $U(s,t)$.

I also don't undestand how to find the constans A,B and C after.

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What's $y_y$? I suppose it should be $u_y$. –  Kaster Feb 18 '13 at 0:09
    
yes thank u for the correction @kaster –  Yenny Chen Feb 18 '13 at 0:45
    
$u(x=s^2,y=s)=s^3$ or $u(y=s^2,x=s)=s^3$ ? –  doraemonpaul Apr 21 '13 at 1:39

1 Answer 1

$xu_x-xyu_y-u=0$

$xu_x-xyu_y=u$

$u_x-yu_y=\dfrac{u}{x}$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$

$\dfrac{dy}{dt}=-y$ , letting $y(0)=y_0$ , we have $y=y_0e^{-t}=y_0e^{-x}$

$\dfrac{du}{dt}=\dfrac{u}{x}=\dfrac{u}{t}$ , we have $u(x,y)=tf(y_0)=xf(ye^x)$

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