Find the general solution of the PDE $ xu_x-xyu_y-y=0 $
for all $u(x,y)$
and find the parametric form of the solution of the PDE which follows the side condition $ **u(s^2,s)=s^3** $
I got part (a) of the solution. The general solution is $ u(x,y)=-xf(ye^x) $
I have the solution that the parametric form of the PDE is $ x(s,t)=s^2e^t $ but i am not sure on how to solve it.
$\frac{dx}{x}=-\frac{dy}{-xy}=\frac{du}{y}$
From this we have $dx=-\frac{dy}{y}$ and so $x=-\ln(y)+C_1$, or in other words $\psi_1(x,y,u)=x+\ln(y)=C_1$
We also have $-\frac{dy}{x}=du$ and so $-\frac{y}{x}=u+C_2$ or in other words $\psi_2(x,y,u)=u+\frac{y}{x}=C_2$.
The general solution is then $\Phi(\psi_1(x,y,u),\psi_2(x,y,u))$ where $\Phi$ is any continuously differentiable function of $2$ variables.
You found wrongly about the general solution.
$xu_x-xyu_y-y=0$
$xu_x-xyu_y=y$
$u_x-yu_y=\dfrac{y}{x}$
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 $y=y_0e^{-t}=y_0e^{-x}$
$\dfrac{du}{dt}=\dfrac{y}{x}=\dfrac{y_0e^{-t}}{t}$ , we have $u(x,y)=y_0\int^t\dfrac{e^{-t}}{t}dt+f(y_0)=ye^x\int^x\dfrac{e^{-t}}{t}dt+f(ye^x)$
$u(s^2,s)=s^3$ :
$se^{s^2}\int^{s^2}\dfrac{e^{-t}}{t}dt+f(se^{s^2})=s^3$
$f(se^{s^2})=s^3-se^{s^2}\int^{s^2}\dfrac{e^{-t}}{t}dt$
$\therefore u(x,y)=ye^x\int^x\dfrac{e^{-t}}{t}dt+f(ye^x)$ , where $f(s)$ satisfy $f(se^{s^2})=s^3-se^{s^2}\int^{s^2}\dfrac{e^{-t}}{t}dt$
