I am asked the following--
Given real coefficients $a_{jk}, a_l$, define a general 2-variable linear differential operator by
$$L=a_{11}\partial_{xx} + 2a_{12}\partial_{xy} + a_{22}\partial_{yy} + a_{1}\partial_{x} + a_{2}\partial_{y} + a_0$$
Given the operator, define its discriminant by
$$D=a_{12}^2-a_{11}a_{22}$$
If $D=0$, prove that there exists a linear transformation $T:R^2\rightarrow R^2$,
$$(x,y)\mapsto T(x,y)=(\epsilon,\eta)$$
such that the operator becomes
$$L=\partial_{\epsilon\epsilon}+{lower \space order \space terms}$$
I am stuck, I don't know where to start... I know I am supposed to collapse the result into the first term in a matrix, but I'm not really sure how to get there. I think it involves breaking up the operators?
First note that
$$D=a_{12}^2-a_{11}a_{12}=0 \Rightarrow a_{12}=\sqrt{a_{11}a_{22}}$$
whence, expanding through the second-order derivatives,
$$L=a_{11}∂_{x}^2+2\sqrt{a_{11}a_{22}}∂_{x}(∂_{y}) +a_{22}∂_{y}^2+a_{1}∂_{x}+a_{2}∂_{y}+a_{0}$$ $$=(\sqrt{a_{11}}∂_{x}+\sqrt{a_{22}}∂_{y})^2+ a_{1}∂_{x}+a_{2}∂_{y}+a_{0}$$ $$=∂_{\epsilon}^2+ a_{1}∂_{x}+a_{2}∂_{y}+a_{0}=∂_{\epsilon \epsilon}+ a_{1}∂_{x}+a_{2}∂_{y}+a_{0}$$
where we have let $∂_{\epsilon} = \sqrt{a_{11}}∂_{x}+\sqrt{a_{22}}∂_{y}$. Hence the mapping is
$$T(x,y)=∂((\sqrt{a_{11}},\sqrt{a_{22}}),(0,0))$$