solving a second order nonlinear pde I would like to solve the following PDE,
$$f_{y}^{2} = 2 f f_{yy}$$
where $f= f(x,y)$ is a real function of two variables $x,y$.
My solution
: derivative of $f_{y}^{2}$ with respect to $y$ is itself, so $f_{y}^{2}= c(x)e^{y}$ and 
$$f_{y} = c(x) e^{\frac{y}{2}}$$
so, 
$f(x,y) = 2 c(x) e^{\frac{y}{2}} + d(x)$
but this does not satisfy in the equation $f_{y}^{2} = 2 f f_{yy}$ .
thanks for your help.
 A: This is really an ODE, it suffices to make two arbitrary integration constants in the general solution arbitrary functions of $x$. 
To solve this ODE, rewrite it as
$$\frac{f_{yy}}{f_y}=\frac12\frac{f_y}{f}\qquad \left(\ln f_y\right)_y=\frac12 \left(\ln f\right)_y,$$
which gives $\ln f_y=\frac12\ln f+ \mathrm{const}$ or, in other words, $f_y=2C_1\sqrt f$. The latter equation is separable, and its solution is given by
$$\sqrt f =C_1 y+C_2.$$
Hence the general solution is 
$$f(x,y)=\left[C_1(x)y+C_2(x)\right]^2.$$

Of course, knowing the answer, one can derive it much more quickly. Write $f=g^2$, then $f_y=2gg_y$, and $f_{yy}=2gg_{yy}+2g_y^2$, and the equation for $f$ transforms into $g_{yy}=0$.
A: $$
f_yf \left(\frac{f_y}{f} -2\frac{f_{yy}}{f_y}\right) = 0
$$
Thus we can obtain
$$
\dfrac{\partial}{\partial y}\ln f -2\dfrac{\partial }{\partial y}\ln f_y = 0
$$
Or
$$
\ln \left(\frac{f}{f_y^2}\right) = g(x)
$$
So you get
$$
f = g_2(x) f_y^2\implies f_y = \sqrt{\frac{f}{g_2(x)}}
$$
A: Hint: set $f(x,y)=g(x)h(y)$ and try and find polynomial solutions for $h(y)$.
