Solving $v-\eta v' = vv''+v'^2$ Question:
How to solve the differential equation
$$v - \eta \frac{dv}{d\eta} = v\frac{d^2v}{d\eta^2} + \bigg(\frac{dv}{d\eta}\bigg)^2$$

Attempt:
This equation looks quite disgusting to me. I tried writing it in the form
$$-\eta^2 \frac{d}{d\eta}\bigg(\frac{v}{\eta}\bigg) = \frac{d}{d\eta}\bigg( v\frac{dv}{d\eta}\bigg)$$
but that doesn't seem to help.
Honestly, I am at a loss at how to tackle this equation (not even sure if there is an easy solution for this).

Note:
This differential equation actually arose as part of my attempt to find a similarity solution for the PDE
$$\frac{\partial u}{\partial t} = \frac{\partial}{\partial x}\bigg(u\frac{\partial u}{\partial x}\bigg)$$
Letting $u(x,t) = t^a v(\eta)$ where $\eta = x/t^b$ and $a,b$ are constants to be specified later on, we get
\begin{align}
& \frac{\partial u}{\partial t} = \frac{\partial}{\partial x}\bigg(u\frac{\partial u}{\partial x}\bigg)\\
\implies & \frac{\partial u}{\partial t} = u\frac{\partial ^2u}{\partial x^2} + \bigg(\frac{\partial u}{\partial x}\bigg)^2 \\
\implies & at^{a - 1}v - bxt^{a-b-1}v' = t^{2a-2b}vv'' + t^{2a-2b} v'^2 \\
\implies & at^{a - 1}v - b\eta t^{a-1}v' = t^{2a-2b}vv'' + t^{2a-2b} v'^2
\end{align}
So we need $a=b=1$ in order for this to be a similarity solution. The equation thus becomes
$$v - \eta \frac{dv}{d\eta} = v\frac{d^2v}{d\eta^2} + \bigg(\frac{dv}{d\eta}\bigg)^2$$
which I am unable to solve.
 A: What makes you think that the equation $v - \eta \frac{dv}{d\eta} = v\frac{d^2v}{d\eta^2} + \bigg(\frac{dv}{d\eta}\bigg)^2$ can be solved in terms of a limited number of standard functions ?
In fact this arduous equation results from a guess about the existence of solutions of the PDE $$\frac{\partial u}{\partial t} = \frac{\partial}{\partial x}\bigg(u\frac{\partial u}{\partial x}\bigg)$$
on the particular form $u(x,t) = t^a v\left(x/t^b\right)$. 
More simply, why not looking for solutions of the form :
$$u(x,t)=f(x)g(t)$$
$$f(x)g'(t)=f(x)f''(x)(g(t))^2+(f'(x))^2(g(t))^2$$
$$\frac{g'(t)}{(g(t))^2}=f''(x)+\frac{f'(x))^2}{f(x)}=\lambda=\text{constant}$$
because a function of $t$ can be equal to a function of $x$ only if those fonctions are constant.
Solving $\frac{g'(t)}{(g(t))^2}=\lambda$ leads to 
$$g(t)=-\frac{1}{\lambda t+c_1}$$
$f''(x)+\frac{f'(x))^2}{f(x)}=\lambda$ is an autonomous ODE. The usual method to solve this kind of ODE leads to :
$$f(x)=\left(\lambda e^{-c_2(x-x_0)/2}+\frac{1}{c_2^2}e^{c_2(x-x_0)/2} \right)^2$$
This method of separation of variables leads to a family of solutions with four parameters $\lambda, c_0,c_1,c_2$ :
$$u(x,t)=-\frac{1}{\lambda t+c_1}\left(\lambda e^{-c_2(x-x_0)/2}+\frac{1}{c_2^2}e^{c_2(x-x_0)/2} \right)^2$$
Of course, other solutions of the PDE cannot derive from linear combination of those particular solutions since the PDE is not linear.
