I would like to solve this PDE: $5u_x + u_{xy} = 0$. I start by substituting $v = u_x$ to obtain $5v + v_y = 0$ and proceed to solve for $v = C(x)e^{-5y}$. Since $v = u_x$, then I integrate w.r.t $x$ to $u$:
$$\int u_x dx =\int f'(x) e^{-5y}dx$$ where I let $C(x) = f'(x)$ just for notation purposes. Then I obtain the equation
$$u(x,y) = e^{-5y}f(x) + C(y)$$
How do I find what $C(y)$ and $f(x)$ is equal to?
Confirming your solution: Note that $u_{xy}=u_{yx}$ by smoothness assumptions, so $$ 5u_x=-u_{yx} $$ is our pde. Then integrate on both sides in $x$ to find $$ 5u(x,y)=-u_y+C(y) $$ which is now an ode we can solve using an integrating factor $$ 5u(x,y)=-u_y+C(y)\implies \frac{\partial}{\partial y}(e^{5y}u(x,y))=e^{5y}C(y) $$ and integrating in $y$ now introduces an arbitrary function of $x$, $$ e^{5y}u(x,y)=\int_0^y e^{5s}C(s)\mathrm ds+D(x) $$ renaming $e^{-5y}\int_0^y e^{5s}C(s)\mathrm ds\mathrm dy=A(y)$ we have $$ u(x,y)=A(y)+e^{-5y}D(x) $$ It's easy to check that any pair of twice differentiable functions $D$ and $A$ will satisfy the pde, and this is the point. Without more information, like boundary data, you cannot determine $A$ and $D$. Think in analogy to an ode where you need initial conditions to determine integration constants.