Differential equation of the form $y^2y'' = a$ Can anyone help me in solving the differential equation of the form $$y^2y'' = a$$
where $y$ is a function of $x$ and $a$ is a constant.
I am new to solving differential equations and just out of curiosity I tried solving this, but I couldn't. It has been keeping me busy all the time and I wanted a solution to it. Can anyone please help me?
 A: Divide by $y^2$  and multiply both sides by $y′$  and integrate to give $\frac{y'^2}{2}=c1-\frac{a}{y}$. Square root and integrate again. $\int  \frac{dy}{\sqrt{c1-a/y}}= \sqrt{2}x +c2$. That gets nasty if $c1\ne0$ but Wolfram alpha can do it. http://www.wolframalpha.com/input/?i=integrate+dy+%2F+sqrt(c-a%2Fy). Which root you want should be clear from the boundary conditions or obvious from the task at hand.
A: Step 1.
$$
y^2y''=a \,\,\Longrightarrow\,\,y'y''=\frac{ay'}{y^2}
\,\,\Longrightarrow\,\, \left(\frac{1}{2}(y')^2\right)'=\left(-\frac{a}{y}\right)'
\,\,\Longrightarrow\,\, \frac{1}{2}(y')^2=-\frac{a}{y}+c \\
\,\,\Longrightarrow\,\, y'=\pm\left(2c-\frac{2a}{y}\right)^{1/2}
\,\,\Longrightarrow\,\, \left(\frac{y}{2cy-2a}\right)^{1/2}y'=\pm 1.
$$
A: Let  $x=at^2$, $y=2at$, then
\begin{align*}
  x' &= 2at \\
  y' &= 2a \\
  \frac{dy}{dx} &= \frac{y'}{x'} \\
  &= \frac{1}{t} \\
  \frac{d^2y}{dx^2}
  &= \frac{d}{dt} \left( \frac{dy}{dx} \right) \times \frac{dt}{dx} \\
  &= -\frac{1}{t^2} \times \frac{1}{2at} \\
  &= -\frac{1}{2at^3}
\end{align*}
substituting $t$ for $y$ will give the desired result.
