How to solve this PDE: $u_{xy}+\frac{1}{x}u_y-y=0$? Question: A function $u(x, y)$ obeys the PDE:
$$\frac{\partial^2 u}{\partial x \partial y}+\frac{1}{x} \frac{\partial u}{\partial y}=y.$$
Find the general solution for $u(x, t)$.
Find the solution obeying the Cauchy data $u=0$ and $u_x=0$ on the line $y=2x$.
Generally, when we solve a second order linear PDE, first we need to find its characteristics equations. Then by integrating the characteristic equations, we define two new characteristics variables. Then we use these new variables to reduce the PDE to canonical form. Then we could easily find the general solution to that PDE.
However, in this question, it is already in the normal form, and there is no other condition given. I have no clue how to solve this PDE. Can anyone help please.
Thank you for your attention, I am looking forward to your reply.
 A: It is essentially an ODE by making the substitution $v = u_{y}$. Indeed with that substitution
$$v_{x}+\frac{1}{x}v =y$$
can you take it from here?
A: I am going to answer this question for completeness, because this is such a fun question!
Integrate with respect to y:
$$
u_x + \frac{1}{x}u = \frac{1}{2}y^2 + f(x)
$$
Multiply by an integrating factor $x$:
$$
\frac{d}{dx}[ux] = \frac{1}{2}xy^2 + xf(x)
$$
Integrate with respect to $x$:
$$
ux = \frac{1}{4}x^2y^2 + f(x) + g(y)
$$
We thus have the general solution:
$$
u = \frac{1}{4}xy^2 + f(x) +\frac{g(y)}{x}
$$
Subject to the boundary conditions $u=0$ and $u_x=0$ over $y=2x$ we have that:
$$
0 = x^2 + f'(x) - \frac{g(2x)}{x^2}
$$
$$
0 = x^3 + f(x) + \frac{g(2x)}{x}
$$
From which we find that:
$$
f' + \frac{1}{x}f = -2x^2
$$
Which has the solution:
$$
f(x) = -\frac{1}{2}x^3 + C
$$
Now we find that:
$$
g(2x) = -\frac{1}{2}x^4 -Cx
$$
Hence:
$$
g(y) = -\frac{1}{32}y^4 -\frac{C}{2}y
$$
Thus the particular solution is given by
$$
u = -\frac{1}{32}\frac{y^4}{x} - \frac{1}{2}x^3 +\frac{1}{4}xy^2+C\left(1-\frac{y}{2x}\right)
$$
