Let $$u_{xx}-4u_{xy}+3u_{yy}=0.$$ Find the general solution given the solution $u(x,y)=f(\lambda x+y).$
My attempt was as follows: let $u(x,y)=e^{\lambda x+y}$. Then by computing $u_{xx},u_{xy}, \text{ and } u_{yy}$ we get $e^{\lambda x+y}(\lambda^2-4\lambda+3).$ This shows us that $\lambda =1$ or $\lambda =3$.
Is this the right track?
The idea is that the differential operator $\partial_{xx} - 4 \partial_{xy} + 3 \partial_{yy}$ decomposes as a product of two commuting operators of order $1$: $$\partial_{xx} - 4 \partial_{xy} + 3 \partial_{yy} = ( \partial_x - \partial_y)(\partial_x - 3 \partial_y)= (\partial_x - 3 \partial_y)( \partial_x - \partial_y)$$ Now for any functions $f$, $g$ in $1$ variable we have $$( \partial_x - \partial_y)( f(x+y) )= 0$$ and $$(\partial_x - 3 \partial_y) ( g (3x + y) ) =0$$ Therefore, any function $u$ of form $$u(x,y) = f(x+y) + g(3x + y)$$ is a solution of the equation. It is not hard to show that in this way we get all the solutions.