Need help solving what should probably be a very simple PDE Trying to teach myself PDEs, and I'm stumped on what should probably be a very simple exercise: 
Solve the equation $3u_{y}+u_{xy}=0$. And I am given the hint to let $v=u_{y}$ (it's a problem from Strauss' intro book).
Now, when you make the suggested substitution, $3u_{y}+u_{xy}=0$ becomes $3v+v_{x}=0$. The only problem is, I don't know how to solve this kind of PDE; the only types of PDE the book has really talked about at this point are ones of the form $au_{x}+bu_{y}=0$, where $a$ and $b$ are constants, and ones of the form $u_{x}+yu_{y}=0$.
What I've tried to do, therefore, is rewrite $v_{x}$ as $\frac{dv}{dx}$, subtract $3v$ from both sides, and turn it into a type of separable ODE type thing. Then, if I do that, and after substituting $v=u_{y}$ back in, I wind up getting that $u_{y}=\exp{(-3x)}\exp{(f(y))}$ Then, I suppose I'd have to integrate both sides with respect to $y$ to get the solution, but I'm worried it will take some messy integration by parts that never stops, and so I think that this method couldn't possibly be right. 
Could somebody tell me the RIGHT way to do this problem? Thanks!! :)
 A: We are given that
$$3u_y + u_{xy} = 0$$
As the hints suggest, we use the substitution: $v = u_y$.  Then, as you said before, the new ODE is
$$3v + v_x = 0$$
By the method of constant coefficients, setting $v = e^{rx}$, we have...
$$\begin{aligned}
re^{rx} + 3e^{rx} &= 0\\
r + 3 &= 0\\
r &= -3
\end{aligned}$$
So we have
$$v(x) = c_1e^{-3x}$$
where $c_1$ is any constant.  But since $v = u_y$,
$$u_y(x,y) = c_1e^{-3x}$$
which implies that a solution is
$$u(x,y) = c_1ye^{-3x}$$
A: Note.
$$
3u_y+u_{xy}=0 \quad\Longleftrightarrow\quad \frac{\partial}{\partial y}(3u+u_x)=0
\quad\Longleftrightarrow\quad 3u(x,y)+u_x(x,y)=f(x),
$$
for some function $f=f(x)$. Next
$$
\mathrm{e}^{3x}\big(3u(x,y)+u_x(x,y)\big)=\mathrm{e}^{3x}f(x)
\quad\Longleftrightarrow\quad \big(\mathrm{e}^{3x}u(x,y)\big)_x =\mathrm{e}^{3x}f(x),
$$
which is equivalent to
$$
\mathrm{e}^{3x}u(x,y) =u(0,y)+\int_0^x \mathrm{e}^{3t}f(t)\,dt,
$$
or
$$
u(x,y) =\mathrm{e}^{-3x}\,u(0,y)+\int_0^x \mathrm{e}^{-3(x-t)}f(t)\,dt,
$$
for a function $f$ which can be determined from the initial data.
