how to solve $y''=-\frac{1}{2y^2}?$ i've been learning special kind of second order non-linear ODE,of the form f(y,y') and f(t,y').I came across a question which I'm not able to reach to the solution.Can someone help me with the method...
 A: Maple's solution entails solving a transcendental equation.
To start.  Note that the variable $x$ does not appear explicitly.  (The equation is autonomous).  So we can make a substitution like
$u=y'$ to change it to a first-order equation for $u$ as a function of $y$...
$$
u=y'
\\
\frac{du}{dy} = \frac{u'}{y'} = \frac{y''}{y'} = \frac{-1}{2uy^2}
\\
2uy^2\frac{du}{dy} = -1
$$
Then solve as usual...
$$
2u\;du = -\frac{dy}{y^2}
\\
u^2 = \frac{1}{y}+C
$$
Now substitute $u=y'$ back in to get first-order (but non-inear) differential equation for $y$
$$
(y')^2=\frac{1}{y}+C
$$
Here is where we end up with a transcendental equation.
One case for each of two square roots, let's do this one
$$
y'= \sqrt{\frac{1}{y}+C}
\\
\frac{dy}{\sqrt{\frac{1}{y}+C}} = dx
\\
1/2\,{\frac {y}{\sqrt {y \left( Cy+1 \right) }{C}^{3/2}}\sqrt {{\frac 
{Cy+1}{y}}} \left( 2\,\sqrt {y \left( Cy+1 \right) }\sqrt {C}-\ln 
 \left( 1/2\,{\frac {2\,Cy+2\,\sqrt {y \left( Cy+1 \right) }\sqrt {C}+
1}{\sqrt {C}}} \right)  \right) }
= x+C_2
$$
(I let Maple do the integral.)  
And now merely solve for $y$...
