I want to solve the problem : $$\dfrac{\partial u}{\partial t}=\left\vert \nabla u\right\vert ^{-2p}\left( \dfrac{\partial ^{2}u}{\partial \xi ^{2}}+\left( 1-p\right) \dfrac{\partial ^{2}u}{\partial \eta ^{2}}\right)$$ We search for a self-similarity solution, the general form of which is as follows $$u(x,y,t) = f \left( \xi \right), \text{ with } \xi = \dfrac{(x^{2}+y^{2})^{n}}{a (t)}$$ from which we obtain
$$\alpha \xi =\left( 1-p\right) \left( 2n\right) ^{-2p+2}\left( \left( \frac{1}{2n\left( 1-p\right) }+\frac{2n-1}{2n}\right) \left( \dfrac{df}{d\xi }\right) ^{-2p}+\xi \left( \dfrac{df}{d\xi }\right) ^{-2p-1}\dfrac{d^{2}f}{d\xi ^{2}}\right) $$ Now, I am very confused on how to solve the above equation and find the exact solution of $f(\xi)$, thereby finding the exact solution of $u(x,y,t)$. So my question is, does anyone know how to solve this ordinary differential equation?
$$\alpha \xi =\left( 1-p\right) \left( 2n\right) ^{-2p+2}\left( \left( \frac{1}{2n\left( 1-p\right) }+\frac{2n-1}{2n}\right) \left( \dfrac{df}{d\xi }\right) ^{-2p}+\xi \left( \dfrac{df}{d\xi }\right) ^{-2p-1}\dfrac{d^{2}f}{d\xi ^{2}}\right) $$ A bunch of constants that can be written as where I replaced $\xi = z$. $$C f_z^{2p+1} = \frac{a}{z} f_z + z f_{zz}.$$ Define $g = f_z$, then $$C g^{2p+1} = \frac{a}{z} g + z g_z$$ is in the form of Bernoulli's equation (https://en.wikipedia.org/wiki/Bernoulli_differential_equation)