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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

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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

$$u(x,y) = -(x+2y)(x+3y) + (x+3y)^2 + a - a = \color{red} {3y^2 + xy} .$$