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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?

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    $\begingroup$ You should not take a specific function $e^{\lambda x+y}$: in other words, just take $u(x,y)=f(\lambda x+y)$ as given. You should find that $u_{xx}$ etc are equally easy as what you did already, and you get the same values for $\lambda$. $\endgroup$ – David Sep 8 '14 at 1:47
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    $\begingroup$ Then take $t=\lambda x+y$ and you get an ODE after rewriting the x,y derivatives as t derivatives. $\endgroup$ – mathematician Sep 8 '14 at 1:50
  • $\begingroup$ So you mean, for example, $u_{xx}=f_{xx}(\lambda x+y)\lambda^2$ and so forth? Doesn't the $f_{xx}$ and sort forth complicate factoring? $\endgroup$ – emka Sep 8 '14 at 1:51
  • $\begingroup$ Do you know what they mean by the general solution in the question? $\endgroup$ – Mhenni Benghorbal Sep 8 '14 at 2:13
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    $\begingroup$ Instead of arbitrary constants, general solutions of PDE's may involve arbitrary functions. $\endgroup$ – Robert Israel Sep 8 '14 at 2:36
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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.

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