Consider the inhomogeneous PDE $$2u_{xx}-u_{xy}-u_{yy}=(x+y)\exp(2y^2+xy-x^2).$$ Using the transformation $$m=-x+2y \ \ \ \ \ \text{and} \ \ \ \ \ n=x+y,$$ my goal is to transform this PDE into the form $$u_{mn}=aF(m,n).$$ I think this problem will use the chain rule a lot. From other problems I did before, I am pretty sure I have to start with the fact that $$\frac{\partial ^2u}{\partial x^2}=\frac{\partial }{\partial m}\left [ \frac{\partial u}{\partial x} \right ]\frac{\partial m}{\partial x}+\frac{\partial }{\partial n}\left [\frac{\partial u}{\partial x} \right ]\frac{\partial n}{\partial x}.$$However, I'm a little confused. I don't really know what to do with the function $$F(x,y) = (x+y)\exp(2y^2+xy-x^2).$$ Since I am very new to PDEs, I am sure that with time this stuff will become very easy. But as of now, I can't seem to pinpoint a specific method that will help me solve these problems. Any thoughts?

Thanks in advance!

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    $\begingroup$ $u_{mn}=-\frac 19 n e^{mn}$ $\endgroup$ – Cesareo Oct 14 '18 at 19:52
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    $\begingroup$ I made an error, sorry! $\endgroup$ – Calvin Khor Oct 14 '18 at 19:54
  • $\begingroup$ @Cesareo How did you come up with this? $\endgroup$ – MathIsLife12 Oct 14 '18 at 20:46
  • $\begingroup$ Just applying the chain rule. $\endgroup$ – Cesareo Oct 14 '18 at 21:39
  • $\begingroup$ @Cesareo Would you mind writing it down in the answer section so I can give you the upvote and correct answer check? $\endgroup$ – MathIsLife12 Oct 14 '18 at 22:05

As long as

$$ m = -x+2y\\ n = x+y $$

we have

$$ (x+y)e^{2y^2+x y - x^2} = ne^{nm} $$

and using the chain rule

$$ u_x = u_m m_x + u_n n_x = -u_m+u_n\\ u_y = u_m m_y + u_n n_y = 2u_m+u_n\\ u_{xy} = (-u_m+u_n)_m m_y+ (-u_m+u_n)_n n_y = -2u_{mm}+2u_{nm}+u_{nn} $$


After that we obtain $2u_{xx}-u_{xy}-u_{yy} \equiv -9 u_{nm}$ and finally

$$ u_{nm}= -\frac n9 e^{nm} $$


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