Edit: @Etienne mentioned that I did a typo, writing $u_y' = -xye^{-y}$ instead of $u_y' = -xe^{-y}$. I've corrected that in the calculations and now it's closer to being correct! Though I still miss $e^{-y}$ multiplicated with $f_{uv}''$ to get the correct answer.
Problem: Transform the differential equation:
$$ x\frac{\partial ^2f}{\partial x^2} + \frac{\partial ^2f}{\partial x \,\partial y} + \frac{\partial f}{\partial x} = xe^{-2y} $$
by introducing: $$ \begin{cases} u = &xe^{-y} \\ v = &y \end{cases} $$
How I am solving it. Rewrite to: $$ xf_{xy}'' + f_{yx}'' + f_{x} = xe^{-2y} $$
solve $x$ and $y$ in $u$ and $v$: $$ \begin{cases} x = &\frac{u}{e^{-v}} \\ y = &v \end{cases} $$
And calculate: $$ \begin{cases} u_x' = & e^{-y} \\ u_y' = & -xe^{-y} \end{cases} $$ $$ \begin{cases} v_x' = & 0 \\ v_y' = & 1 \end{cases} $$
Find $f_x'$ by using the chainrule: $$ f_x' = f_u' \cdot u_x' + f_v' \cdot v_x' = f_u' \cdot e^{-y}+f_v'\cdot0=f_u'e^{-y} $$
set $g=f_u'$ and $h=f_v'$ and calculate $f_{xx}''$
$$ f_x' = f_u'e^{-y} \underbrace{= }_{g=f_u'} ge^{-y} $$ $$ f_{xx}'' = g_x' \cdot e^{-y} $$
Find $g_x'$ (using chainrule): $$ g_x' = g_u' \cdot u_x' + g_v' \cdot v_x' = g_u' \cdot e^{-y} + g_v' \cdot 0 = g_u' \cdot e^{-y} \underbrace{= }_{g=f_u'} (f_u')_u' \cdot e^{-y} = f_{uu}'' \cdot e^{-y} $$
insert $g_x'$ in $f_{xx}''$ gives: $$ f_{xx}'' \underbrace{=}_{g_x' = f_{uu}'' \cdot e^{-y}} (f_{uu}'' \cdot e^{-y}) \cdot e^{-y} = f_{uu}'' \cdot e^{-2y} \underbrace{=}_{y=v} f_{uu}'' \cdot e^{-2v} $$
Now I have $f_x'$ and $f_{xx}''$ and I only need $f_{yx}''$ to insert the differentials ito the original equation. But $f_{yx}''$ can be found using $(f_x')_y'$ since $f_{xy}'' = f_{yx}''$.
$$ f_{xy}'' = -ge^{-y} + g_y'e^{-y} $$
Find $g_y'$ (using chainrule): $$ g_y' = g_u' \cdot u_y' + g_v' \cdot v_y' = g_u' \cdot (-xe^{-y}) + g_v' \cdot 1 \underbrace{=}_{g=f_u'} (f_u')_u' \cdot (-xe^{-y}) + (f_u')_v' \cdot 1 = $$ $$ - f_{uu}'' \cdot xe^{-y} + f_{uv}'' = - f_{uu}'' \cdot \frac{u}{e^{-v}}e^{-v} + f_{uv}'' = - f_{uu}'' \cdot u + f_{uv}'' $$
replace $g$ and $g_y'$ and translate to $u$ and $v$ instead of $x$ and $y$
$$ f_{xy}'' = -f_u' e^{-v} + (- f_{uu}''u + f_{uv}'')e^{-v} = -f_u' e^{-v} - f_{uu}''ue^{-v} + f_{uv}''e^{-v} $$
Insert $f_{xx}''$, $f_{yx}''$ and $f_v'$ in the original equation:
$$ \frac{u}{e^{-v}} (f_{uu}'' e^{-2v}) + -f_u' e^{-v} - f_{uu}''ue^{-v} + f_{uv}''e^{-v} + f_u'e^{-v} = xe^{-2y} $$ $f_u' e^{-v}$ and $uf_{uu}''e^{-v}$ cancels and I get $$ f_{uv}'' = ue^{-v} = \frac{\partial ^2f}{\partial u \,\partial v} $$
Obviously I've missed something. The terms with $f_{uu}''$ is gonna cancel out each other, but I can't figure out whats wrong. I've recalculated this assignment 3 times and it must be something Im not doing correctly, since I probably wouldnt do the same error three times in a row..
The answer is supposed to be: $$ \frac{\partial ^2f}{\partial u \,\partial v} = u $$
So I'm missing a $v$ in the first term and $e^{-v}$ multiplied with $f_{uv}''$