How would I solve: $$ y^3 \eta_{xx} - y \eta_x =0 $$ where $\eta(x,y)$
I started by dividing by $y^3$ to get: $$ \eta_{xx} - \frac{1}{y^2}\eta_x = 0$$
I then thought to let $u = \eta_x$ and then rewrite my problem as: $$ u_x - \frac{1}{y^2} u = 0$$ but this doesn't seem correct. This method (i think) will only work if $u$ is a function in $x$ only and if the coefficient is also a function in $x$. Having the coefficient a function in $y$ and my function in $(x,y)$ is what is confusing me.
:(
Please help.
NOTE: this PDE arises from a system of PDEs found when solving a differential equation using symmetry methods. The preamble to it all is quite lengthy, essentially I get 4 PDEs which I use to determine $\eta(x,y)$ and $\ni(x,y)$.
The list of PDEs I get are as follows: $$ 2\eta - y\xi_x + 2y^3\eta_{xy} - y^3\xi_xx = 0 $$ $$ -2y\xi_y + y^3\eta_{yy} - 2y^3\xi_{xy} = 0 $$ $$ -y\eta_x + y^3\eta_{xx} = 0 $$ $$ y^3\xi_{yy} = 0$$
From the last equation I get: $\xi(x,y) = A(x)y + B(x)$
When I have solved for $\xi$ and $\eta$ they will then give me the Generator for the Lie Symmetries.
The notes where I found this example use MAPLE software to solve the PDEs, I need to be able to do so manually.
Observe that
$\eta_{xx} = y^{-2}\eta_x \tag 0$
may be written as
$(\ln \eta_x)_x = \dfrac{\eta_{xx}}{\eta_x} = y^{-2}. \tag{0.5}$
Only $x$-derivatives occur in the equation
$(\ln \eta_x)_x = y^{-2}; \tag 1$
therefore it may be treated as an ordinary differential equation, with $y$ playing the role of an independent parameter. As such, and based on its simple form, it easily submits to direct integration with respect to $x$:
$\ln \eta_x = y^{-2}x + C_1; \tag 2$
that is,
$\eta_x = e^{y^{-2}x + C_1} = e^{C_1}e^{y^{-2}x}; \tag 3$
then
$\eta(x, y) = e^{C_1}y^2 e^{y^{-2}x} + C_2, \tag 4$
which is easily checked; (4) yields
$\eta_x = e^{C_1}e^{y^{-2}x} = e^{y^{-2}x + C_1}; \tag 5$
$\ln \eta_x = y^{-2}x + C_1; \tag 6$
$(\ln \eta_x)_x = y^{-2}. \tag 7$
Here $C_1$ and $C_2$ are of course arbitrary constants of integration.