# Partial derivative

$$y\frac{\partial f}{\partial x}-x\frac{\partial f}{\partial y}=xyf(x, y)$$

$$u=x^2+y^2$$

$$v =e^{(-x^2/2)}$$

Find THE solution so that $$f(0, y)=y^2$$

I found that

$$v\frac{\partial f}{\partial v}+f(x,y)=0$$

But how should I proceed ? I don't know how to deal with $f$ being there.

$$y\frac{\partial f}{\partial x}-x\frac{\partial f}{\partial y}=xyf(x, y)$$ Try substitution $f=(x^2+y^2)g\to\frac{df}{dx}=2xg+(x^2+y^2)\frac{dg}{dx}, \frac{df}{dy}=2yg+(x^2+y^2)\frac{dg}{dy}$ $$2xyg+y(x^2+y^2)\frac{dg}{dx}-2xyg-x(x^2+y^2)\frac{dg}{dy}=xy(x^2+y^2)g\to\\ (x^2+y^2)\left(y\frac{dg}{dx}-x\frac{dg}{dy}\right)=(x^2+y^2)xyg$$ We returned to the equaton of the same form as initial (both sides multiplied by $x^2+y^2$). It means that if $g$ is a solution to the initial equation, then $f=(x^2+y^2)g$ is also a solution.

Now consider $f=g(x)$ and plug it to the initial equation: $$y\frac{dg}{dx}=xyg\to\frac{dg}{dx}=xg\to g=Ce^{\frac{x^2}{2}}$$ where $C-const$

Thus, we can get according to the written above $f=C(x^2+y^2)e^{\frac{x^2}{2}}$

The only thing left is to satisfy condition $f(0,y)=y^2$ $$f(0,y)=Cy^2e^0\to C=1$$ Finally, $$f(x,y)=(x^2+y^2)e^{\frac{x^2}{2}}$$