Really easy PDE, confused about how to put in side conditions

[S.N.]

Hey, before you read this over I'll mention that I've checked the general solution and it works. So if you don't feel like following my steps to get the general solution just jump down to the end of my attempt, because the real problem for me is figuring out what to do with the side conditions.

Homework Statement

get the particular solution for this PDE, for side conditions u(x,1) = 0 ; u(0,y) = 0

yu_xy + 2u_x = x

Homework Equations

nothing really...

The Attempt at a Solution

first I integrate w.r.t x, which is suggested in the text:

yu_y + 2u = x²/2 + h(y) ; where y is some function of y

divide through by y...

u_y + 2u/y = x²/2y + h(y)/y

now take integrating factor y² (I think this next line follows obviously, if it is confusing please let me know and I'll show you how I got it):

y²u = (x²y²)/4 + ∫h(y)ydy + k(x)

at this point I have done the integral with respect to y on each side, basically. so some function k(x) shows up. Now I replace ∫h(y)ydy with just H(y), some new function of y because the whole expression ∫h(y)ydy will end up being only dependent on y so we just rename the thing:

y²u = (x²y²)/4 + H(y) + k(x)

and divide through by y²...

u = (x²)/4 + H(y)/(y²)+ k(x)/(y²)

so here's where I get confused: if I put u(x,1) into the general solution I get:
k(x) = -x²/4 - H(1)

and for u(0,y) I get:

H(y) = -k(0)

But I see no way of getting solutions for k(x) and H(y) on their own. Is there something I'm missing here? Some sort of "trick" or other thing I should be aware of? Thanks for any advice.

 

[S.N.]

aaaand just got it.