# find general solution of PDE: $xu_{xy}+u_y=2xe^y$

$$xu_{xy}+u_y=2xe^y$$

I solved this equation like following:

Divided the equation by $$x$$: $$u_{xy}+ \frac {1}{x}u_y = 2e^y$$

Then integrated with respect to $$x$$: $$u_y + \ln(x)u_y = 2xe^y+f(y)$$

Then: $$u_y (1+\ln(x)) = 2xe^y +f(y)$$

Then integrated with respect to $$y$$: $$u = \frac{2x}{1+\ln(x)} e^y + \int f(y) + h(y)$$

However I am thinking that my 3rd step is incorrect. Can you help guys, if I am doing something wrong?

In fact the second step is incorrect: Since $$u$$ may depend on $$x$$, integrating $$f_x u_y$$ in general does not give $$f u_y$$.
Hint We can proceed as follows: Since $$u$$ only appears in the equation differentiated by $$y$$, we can produce a lower-order equation by writing $$v := u_y$$: $$x v_x + v = 2 x e^y .$$
Additional hint The left-hand side is just $$(x v)_x$$, so we can immediately integrate to obtain $$x v$$, thus $$v$$, and then integrate again to obtain $$u$$.