How to solve $v f^{'}_v + f = 0$, $f(0, y) = y^2$

Question

So I was presented with this equation in my textbook, currently studying multi variable calculus:

$$yf^{'}_x(x,y) - xf^{'}_y(x,y) = f(x,y)$$

Using the substitution:

$$x^2 + y^2 = u$$ $$e^{{-x^2}{/2}} = v$$

I get the equation (which is correct):

$$v f^{'}_v + f = 0$$

My next task is to find the solution, $f(x,y)$ that satisfies $f(0, y) = y^2$. The correct answer should be $f(x,y) = (x^2 + y^2)e^{{-x^2}{/2}}$.

But here I am stuck, I have solved similar tasks but I don't know how to handle the $f$. For example $v f^{'}_v = x$ or similar would be easy to solve.

Answer 1

Hint: the left-hand side equals $(v f)'_v$ (derivative of a product).