Find the solution to $$\frac{\partial}{\partial x}f(x,y) -x\frac{\partial}{\partial y}f(x,y) = y$$
that satisfies $f(x,0)=x^2+\frac{x^3}{3},$ by using the transformation
$$\left\{ \begin{array}{rcr} u & = & ax^2+y \\ v & = & x \\ \end{array} \right.$$ for an appropriate value of $a$.
So I have that $f'_x-xf'_y=f'_x+vf'_y=y=u-ax^2=u-av^2.$ Furthermore:
\begin{array}{lcl} u'_x & = & 2ax \\ u'_y & = & 1 \\ v'_x & = & 1 \\ v'_y & = & 0 \\ f'_x & = & f'_uu'_x+f'_vv'_x=2axf'_u+f'_v=2avf'_u+f'_v \\ f'_y & = & f'_uu'_y+f'_vv'_y=f'_u\cdot 1+f'_v\cdot 0=f'_u \end{array}
Plugging this into my PDE $f'_x-xf'_y=u-av^2$ I get
$$2avf'_u+f'_v-vf'_u = v(2af'_u-f'_u)+f'_v=u-av^2.$$
It's practical to choose $a=\frac{1}{2},$ then I get $$f'_v=u-\frac{v^2}{2} \ \implies f(u,v)=uv-\frac{v^3}{6} + g(u),$$ so
$$f(x,y)=\left(\frac{x^2}{2}+y\right)x-\frac{x^3}{6}+g\left(\frac{x^2}{2}+y\right).$$
Using the condition $f(x,0)=x^2+\frac{x^3}{3}$ I get
$$f(x,0)=\frac{x^3}{2}-\frac{x^3}{6}+g\left(\frac{x^2}{2}\right)= \Leftrightarrow g\left(\frac{x^2}{2}\right)=x^2.$$
Setting $t=\frac{x^2}{2}\Rightarrow x=\sqrt{2t}$ I get
$$g(t)=2t \implies g\left(\frac{x^2}{2}+y\right)= x^2+2y.$$
Finally:
$$f(x,y)=\left(\frac{x^2}{2}+y\right)x-\frac{x^3}{6}+x^2+2y.$$
But apparently this is wrong answer. Can anyone see where I made the error?
$$\begin{align} f(x,y)= & \left(\frac{x^2}{2}+y\right)x-\frac{x^3}{6}+g\left(\frac{x^2}{2}+y\right)\\ f(x,0)= &\left(\frac{x^2}{2}+0\right)x-\frac{x^3}{6}+g\left(\frac{x^2}{2}+0\right) \\ =& \frac{x^3}{2}-\frac{x^3}{6}+g\left(\frac{x^2}{2}\right) = \frac{x^3}{3}+g\left(\frac{x^2}{2}\right) \neq \frac{x^2}{2}-\frac{x^3}{6}+g\left(\frac{x^2}{2}\right) \\ \end{align}$$