Find the most general solution of the equation $$6\frac{\partial^2u}{\partial x^2}-5\frac{\partial^2u}{\partial x\partial y}+\frac{\partial^2u}{\partial y^2}=1$$ by making the change of variables $$\xi=x+2y, \space \eta=x+3y.$$ Find the solution that satisfies $u=0$ and $\frac{\partial u}{\partial y}=x$ when $y=0$.
I eventually got to $$\frac{\partial^2u}{\partial\xi\partial\eta}=-1$$ integrating twice and substitutuingg $x$ and $y$ back in I got $$u(x,y)=-(x+2y)(x+3y)+f(x+3y)+g(x+2y)$$ where $f$ and $g$ are arbitrary functions.
I'm not sure how to solve now with the given initial conditions.
Any help is much appreciated, thank you
Assuming that your computations are correct (I haven't checked them), you have to impose the conditions $u(x,0) = 0$ and $\dfrac {\partial u} {\partial y} (x,0) = x$.
The first condition simply means that $f(x) + g(x) = x^2$. The second one means $3f'(x) + 2g'(x) = 6x$.
Differentiating the first condition gives $f'(x) + g'(x) = 2x$. This allows you to form a linear system:
$$\left\{ \begin{eqnarray} f'(x) + g'(x) = 2x \\ 3f'(x) + 2g'(x) = 6x \end{eqnarray} \right.$$
This has the (unique) solution $g'(x) = 0$ and $f'(x) = 2x$, whence $g(x) = a$ and $f(x) = x^2 + b$ with $a,b \in \Bbb R$ integration constants.
Plugging this back into $f(x) + g(x) = x^2$ produces $a+b = 0$, therefore $g(x) = a$ and $f(x) = x^2 - a$, so that
$$u(x,y) = -(x+2y)(x+3y) + (x+3y)^2 + a - a = \color{red} {3y^2 + xy} .$$