Transform the differential equation $$yu_x(x, y) – xu_y(x, y) = xyu(x, y)$$ by introducing new variables $s = x^2+y^2$ and $t = e^{-x^2/2}$.
Then determine the general solution to the equation.
I think I managed to transform the differential equation? First we find these partial derivatives of $u$
\begin{align}
u_x &= \frac{\partial u}{\partial s}2x – \frac{\partial u}{\partial t}tx \\
u_y &= \frac{\partial u}{\partial s}2y
\end{align}
then we plug them into our differential equation
$$
\frac{\partial u}{\partial s}2xy – \frac{\partial u}{\partial t}txy – \frac{\partial u}{\partial s}2xy = xyu(x, y) \implies -\frac{\partial u}{\partial t}txy = xyu(x, y).
$$
So our transformed differential equation is
$$-\frac{\partial u}{\partial t}txy = xyu(x, y).$$
I have no idea how I'm supposed to find the general solution. The answer for the general solution is $u(x, y) = F(x^2+y^2)\cdot e^{\frac{x^2}{2}}$. How did they arrive at this general solution, and did I do the transformation right?
Best Answer
You just need to finish. First, you can divide by $xy$. And you need to replace $u(x,y)$ with $u(s,t)$. What you get is $$-\frac{1}{u}\frac{\partial u}{\partial t}=\frac{1}{t}$$ We now search for a solution of the form $F(s)\cdot G(t)$. With some small manipulation, what you get is $$\frac{dG}{G}=-\frac{dt}{t}$$From here $\ln G=-\ln t +\ln C$, where $C$ is some constant (or function) independent of $t$. Taking the exponential, $G(t)=Ct^{-1}$. Now plugging it back into $u$, you can absorb $c$ into the function $F(s)$, and you get $$u(s,t)=F(s)t^{-1}$$ or if you write it in terms of $x$ and $y$: $$u(x,y)=F(x^2+y^2)e^{\frac{x^2}{2}}$$