Solving a PDE with mixed derivatives Let $u_{xy}+u_{y}=e^{x}.$
To solve this, I attempted to use the following substitution.  Let $V=u_x$. I tend to hit a roadblock as this is not a homogenous equation.  Could someone get me started on how to proceed?
 A: Put $u=\exp(-x)w$, then your equation becomes $w_{xy}=\exp(2x)$ with the general solution $w(x,y)=f(x)+g(y)+y\exp(2x)/2$, where $f$ and $g$ are arbitrary smooth functions.
A: Put $v = u_y$, so you get $$v_x + v = e^x$$
every solution to this is of the form $v(x,y)=h(x,y) + p(x,y)$, where $h$ satisfies the homogeneous equation $$h_x + h = 0$$ and $p$ is any solution to the original equation. $p(x,y) = \frac12 e^{x}$ is a solution to the original equation (from guessing; multiples of $e^x$ are a good guess as these differentiate to multiples of $e^x$). 
The general solution to $h_x + h = 0$ is $h(x,y) = e^{-x}a(y)$ for functions $a:\mathbb R \rightarrow \mathbb R$; this follows from just using an integrating factor in $x$; multiplying by $e^x$ turns it into $$h_x e^x + h e^x=0$$, i.e. $$\frac{\partial }{\partial x}\left(h e^x \right) = 0$$ which shows $he^{x}$ depends on $y$ only. 
So we now know $$u_y(x,y) = \frac12e^x + e^{-x}a(y),$$ for some $a:\mathbb R \rightarrow \mathbb R$. Choosing $A:\mathbb R \rightarrow \mathbb R$ such that $A'=a$ ($A$ can be anything differentiable by choosing $a$ correctly). Hence the final answer is that the general solution is $$u(x,y) = \frac y2 e^x + e^{-x}A(y)+B(x),$$ where $A,B:\mathbb R \rightarrow \mathbb R$.
