Partial differential equation $3u_y+u_{xy}=0$ I am only starting my PDE course and I have problems solving this easy equation.
$$3 \frac{\partial u}{\partial y} + \frac{\partial ^2 u}{\partial x \partial y} = 0$$
Here's what I've tried:
$$\frac{\partial ^2 u}{\partial x \partial y} = \frac{\partial ^2 u}{\partial y \partial x}$$
So we can substitute: $z = \frac{\partial u}{\partial y} $
Now we have $3z + \frac{\partial z}{\partial x} =0 $
We get $z(x,y) = C(y)e^{-3x}$.
Is this reasoning correct? Because now it seems impossible to compute $u(x,y)$.
What should I do?
 A: Hint
$$u(x,y) = \int C(y) e^{-3x} \mathrm dy = e^{-3x} \tilde C_2(y) + \tilde C_1(x)$$
Now look at the original PDE to infer something about the $\tilde C_i$s
$$u_y(x,y) = e^{-3x} \tilde C_2'(y)\\
u_x(x,y) = -3e^{-3x} \tilde C_2(y) + \tilde C_1'(x)\\
u_{yx}(x,y) = -3 e^{-3x} \tilde C_2'(y)\\
u_{xy}(x,y) = -3 e^{-3x} \tilde C_2'(y)$$
So you need $\tilde C_1 \in C^1(\mathbb R), \tilde C_2 \in C^1(\mathbb R)$ as the only requirements if no boundary conditions are imposed.

Example with given boundary conditions
If we impose the boundary conditions $u(x,0) = e^{-3x}, u_y(x,0) = 0\quad \forall x$ that translates to
$$e^{-3x} = \tilde C_2(0) e^{-3x} + \tilde C_1(x) \qquad \forall x\\
0 = e^{-3x} \tilde C_2'(0)$$
So we obtain $\tilde C_2'(0) = 0$ and $\tilde C_1(x) = e^{-3x}(1-\tilde C_2(0))$ so
$$u(x,y) = e^{-3x}\underbrace{(1-\tilde C_2(0) + \tilde C_2(y))}_{=\tilde C_3(y)}$$
Verify that $\tilde C_3' = \tilde C_2'$ to see that we may chose $\tilde C_1 \equiv 0$. Thus the solution space is precisely given by
$$u(x,y) = C(y) e^{-3x}$$
With $C\in C^1(\mathbb R), C(0) = 1, C'(0) = 0$. One possible choice is $C \equiv 1$ or $C \equiv \cos, C(y) = \frac1{1+y^2}$ would be a decaying solution.
