I'm a little confused about GRE 8767 problem #40.
Let $y=f(x)$ be a solution of the differential equation $xdy+(y-xe^x)dx=0$ such that $y=0$ when $x=1$. What is the value of $f(2)$?
I see that this is an exact differential since $g(x,y)=xy+e^x-xe^x+C$ is such that $g_x=y-xe^x$ and $g_y=x$. How can I use this to find $f(2)$? I tried writing $ g(2,f(2))=2f(2)+e^2-2e^2+C=\cdots $ but I'm not sure what follows. Thanks. The final answer is $e^2/2$ by the way.