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This is related to this question. If anyone could help be spot the mistake it would really make my day!

I have redone the question. And found that the characteristics are indeed $x\pm 2\sqrt{-y}$ for the region $y<0$.

And I know  that $F(x,y)=f_1(x+2\sqrt{-y})+f_2(x-2\sqrt{-y})$ for arbitrary $f_1,f_2$ is indeed the solution to the equation $F_{xx}+yF_{yy}+{1\over 2}F_y=0$. (Verifiable by direct substitution into the equation.)

Now my problem is that I can't get the correct canonical form $F_{\alpha\beta}=0$ where $\alpha,\beta=x\pm 2\sqrt{-y}$ respectively. 

Here is what I have done:

$M_\pm={1\over c}(-b\pm\sqrt{b^2-ac})=\mp (-y)^{-1/2}$ So the characteristics are given by $\alpha=x+2\sqrt{-y}$ and $\beta=x-2\sqrt{-y}$.

We have $a(x,y)F_{xx}+2b(x,y)F_{xy}+c(x,y)F_{yy}+\cdots=H(x,y)$

and ${a\over c}\partial_{xx}+{2b\over c}\partial_{xy}+\partial_{yy}=(\partial_y-M_+\partial_x)(\partial_y-M_-\partial_x)+({\partial M_-\over \partial y}-M_+{\partial M_-\over \partial x})\partial_x$

where $(\partial_y-M_+\partial_x)(\partial_y-M_-\partial_x)=-\alpha_x\beta_x(M_+-M_-)^2\partial_{\alpha\beta}-\beta_x(M_--M_+)(\partial_\beta [\alpha_x(M_+-M_-)])\partial_\alpha$

For this particular question I calculated $(\partial_y-M_+\partial_x)(\partial_y-M_-\partial_x)=-4(-y)^{-1}\partial_{\alpha\beta}-2(-y)^{-3/2}\partial_\alpha$

$\implies(\partial_y-M_+\partial_x)(\partial_y-M_-\partial_x)+({\partial M_-\over \partial y}-M_+{\partial M_-\over \partial x})\partial_x=-4(-y)^{-1}\partial_{\alpha\beta}-{3\over 2}(-y)^{-3/2}\partial_\alpha+{1\over 2}(-y)^{-3/2}\partial_\beta$

$\implies 4F_{\alpha\beta}+{3\over 2}(-y)^{-1/2}F_\alpha-{1\over 2}(-y)^{-1/2}F_\beta-{1\over 2}(-y)^{-1/2}F_\alpha+{1\over 2}(-y)^{-1/2}F_\beta=0$

See my problem? I need the $F_\alpha, F_\beta$ terms to cancel. OK, the $F_\beta$ ones cancel, but not the $F_\alpha$ ones. 

Thanks in advance!

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Why not swap the roles of $x$ and $y$? $$\partial_\alpha = \frac12(\partial_x-(-y)^{1/2}\partial_y), \quad \partial_\beta=\frac12(\partial_x+(-y)^{1/2}\partial_y),$$ so $$4F_{\alpha\beta}= (\partial_x-(-y)^{1/2}\partial_y)(\partial_x+(-y)^{1/2}\partial_y)F= F_{xx}+yF_{yy}+(-y)^{1/2}(-y)^{-1/2}\frac12 F_y =0.$$

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