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!
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
$$ u_x = u_m m_x + u_n n_x = -u_m+u_n\\ u_y = u_m m_y + u_n n_y = 2u_m+u_n\\ u_{xy} = (-u_m+u_n)_m m_y+ (-u_m+u_n)_n n_y = -2u_{mm}+2u_{nm}+u_{nn} $$
etc.
After that we obtain $2u_{xx}-u_{xy}-u_{yy} \equiv -9 u_{nm}$ and finally
$$ u_{nm}= -\frac n9 e^{nm} $$