<s>[INST] Solve the PDE $xu_x-2yu_y+u=e^x,$ with the side condition $u(1,y)=y^2$ 1b. $xu_x-2yu_y+u=e^x,$ side condition $u(1,y)=y^2$
My attempt: This has been a super endurance and I hope I got the whole thing right. So anyway, here it goes ...oh and one more thing... can someone please show me how to solve the side condition step by step please. I know I'm kind of slow at the side condition part, but I really want to fully understand it...so far I can go up into have the $F(w)$ part isolated and then it has something to do with a dummy variable and afterwards substitute back, but where exactly? It would be greatly appreciated :).
$\frac{dy}{dx}=\frac{2y}{x}$
$\frac{-1dy}{2y}=\frac{1dx}{x}$
$\frac{-1}{2} \ln y = \ln x +C$
$\ln y^{\frac{-1}{2}} = \ln x +C$
$e^{\ln y^{\frac{-1}{2}}} = e^{\ln x +C}$
$y^{\frac{-1}{2}}=xe^C$
$\frac{y^{\frac{-1}{2}}}{x}=e^C$
$y^{\frac{-1}{2}}x^{-1}=e^C$
$y^{\frac{-1}{2}}x^{-1} = w \rightarrow x = \frac{w}{\sqrt{z}}$
$ z = y \rightarrow y =z$
$ W_x = -x^{-2}y^{\frac{-1}{2}}$
$W_y = \frac{-1}{2}x^{-1}y^{\frac{-3}{2}}$
$Z_y = 1$
$Z_x = 0$
$x[V_wW_x+V_zZ_x]-2y[V_wW_y+V_zZ_y]+v=e^{\frac{1}{w\sqrt{z}}}$
$-2yV_z+v=e^{\frac{1}{w\sqrt{z}}}$
$V_z+\frac{1}{-2y}v=\frac{1}{-2y}e^{\frac{1}{w\sqrt{z}}}$
$v(a) = e^{\int \frac{-1}{2y}} \rightarrow e^{-\frac{1}{2} \ln y} \rightarrow y^{\frac{-1}{2}}$
$y^{\frac{-1}{2}}V_z+y^{\frac{-1}{2}}\frac{1}{-2y}v=y^{\frac{-1}{2}}\frac{1}{-2y}e^{\frac{1}{w\sqrt{z}}}$
$y^{\frac{-1}{2}}V_z+y^{\frac{-3}{2}}\frac{1}{-2}v=y^{\frac{-3}{2}}\frac{1}{-2}e^{\frac{1}{w\sqrt{z}}}$
$y^{\frac{-1}{2}}V_z-y^{\frac{-3}{2}}\frac{1}{2}v=-y^{\frac{3}{2}}\frac{1}{2}e^{\frac{1}{w\sqrt{z}}}$
$y^{\frac{-1}{2}}v=\int \frac{-1}{2} y^{\frac{-3}{2}}e^{w^{-1}}e^{z^\frac{-1}{2}}$
$y^{\frac{-1}{2}}v=- y^{\frac{-3}{2}}e^{w^{-1}}e^{z^\frac{1}{2}} +F(w)$
$v=-y^{\frac{-1}{2}}e^{w^{-1}}e^{z^\frac{1}{2}}+F(w)y^{\frac{1}{2}}$
$u=-y^{\frac{-1}{2}}e^{{y^{(\frac{-1}{2}}x^{-1} )}^{-1}}e^{y^\frac{1}{2}}+F(y^{\frac{-1}{2}}x^{-1} )y^{\frac{1}{2}}$
Now for the side condition
$u(1,y)=y^2$
Recall that our solution is
$u=-y^{\frac{-1}{2}}e^{{y^{(\frac{-1}{2}}x^{-1} )}^{-1}}e^{y^\frac{1}{2}}+F(y^{\frac{-1}{2}}x^{-1} )y^{\frac{1}{2}}$
Under the side condition, $u(1,y)=y^2$, we have 
$-y^{\frac{-1}{2}}e^{{y^{(\frac{-1}{2}} )}^{-1}}e^{y^\frac{1}{2}}+F(y^{\frac{-1}{2}} )y^{\frac{1}{2}}=y^2$
$F(y^{\frac{-1}{2}})=y^{\frac{3}{2}}+e^{{y^{(\frac{-1}{2}} )}^{-1}}e^{y^\frac{1}{2}}$
So if I let $ y^{\frac{-1}{2}} =r$ Would it be $y=r^2$ because I am multiplying by $-2$ to get rid of the negative.
Edit: maybe not.. more like $y = r^{-2}$ 
 [/INST] Without looking at your attempt in details, a method of solving is shown below, so that you could compare with your calculus :

</s>