Solve $f^2(x)=\frac{k}{f''(x)}$ While studying physics I have many time stumbled upon forces that are directly affected by the position of a particle. Moreover given $f(x)$ how could we approach solving the equation:
$$f^2(x)=\frac{k}{f''(x)}$$
I am new to differential equations and I would really appreciate if someone could explain the thinking process and the steps needed to reach a solution. Thanks in advance.

Putting $y = f(x)$, we have
$$y^{\prime\prime} = \frac{k}{y^2}.$$ 
Now multiplying both sides by $2y^\prime$, we get
$$2y^\prime y^{\prime\prime} =  \frac{2ky^\prime}{y^2},$$
which can be rewritten as
$$\left( \left( y^\prime \right)^2 \right)^\prime  =  \frac{2ky^\prime}{y^2} = \frac{\mathrm{d}}{\mathrm{d} x} \left( - \frac{2k}{y} \right),$$
which upon integration of both sides gives
$$ \left( y^\prime \right)^2 = c -\frac{2k}{y},$$
where $c$ is an arbitrary constant of integration. So
$$ y^\prime = \left( c -\frac{2k}{y} \right)^{\frac{1}{2}},$$ and therefore 
$$ \frac{y^\prime  }{ \left( c -\frac{2k}{y} \right)^{\frac{1}{2}} } = 1,$$
which we can rewrite as 
$$ \frac{ y^\frac{1}{2} y^\prime }{\left( cy - 2k \right)^\frac{1}{2} } = 1, $$
or $$ \frac{1}{c^\frac{1}{2}} \left( \frac{y}{ y - \frac{2k}{c} } \right)^\frac{1}{2} y^\prime = 1.$$
Can you take it from here on?
