How would you solve this differential equation: $\frac{d^2y}{dx^2} = \frac{100}{y}$? The equation is:
$$\frac{d^2y}{dx^2} = \frac{100}{y}$$
Also if $y = f(x)$, 
$f'(0) = 0$ and $f(0) = 10$
How would you solve this for $y$?
 A: $$y''=\frac {100} y$$
is autonomous (ie $x$ is not present in the equation). The standard procedure therefore is to substitute $u(y)=y'$ with
$$ y'' = \frac{du}{dx} = \frac{du}{dy}\frac{dy}{dx} = u'u $$
solve
$$ u'u = 100/y $$
with $u'u = \frac{1}{2}(u^2)'$ we get
$$ u(y)^2 = 2\left(c_1 + 100\ln(y)\right). $$
This equation "can be integrated"
$$ y' = \pm\sqrt{2\left(c_1 + 100\ln(y)\right)} $$
$$ \Rightarrow \int\frac{1}{\sqrt{2\left(c_1 + 100\ln(y)\right)}}dy = \mp\int 1\,dx $$
This is the point where it gets ugly...
The integral can be evaluated to
$$ \sqrt{\frac{\pi}{200}} e^{-c_1/100}\operatorname{erfi}\left(\sqrt{\frac{c_1}{100}+\ln(y)}\right) + c_2 = \mp x $$
where $\operatorname{erfi}$ is the imaginary error function. Solve for $y$ and you are done...
A: Multiply both sides by $y'$ and integrate. You will get here:
$$\frac{dy}{dx}=\pm\sqrt{2k_1+200\log (y)}$$
$$\int^x\frac{dy}{\sqrt{2k_1+200\log (y)}}=x+k_2$$
That is not an elementary integral, you can't get a solution as elementary functions. Wolfram uses the error function to give a solution that has no integrals in it.
A: This answer is similar to that of the user example, but I though this answer might give you so more intuition about how to solve similar problems and when certain things are equal to each other.
In physics when you have a dynamic system where the force/acceleration on an object only depends on its position, then the potential energy of that object will also only be a function of its position and energy will be conserved. In this case the force $F$ can be expressed as follows
$$
y'' = F = -\nabla U(y) = \frac{100}{y},
$$
such that the potential $U(y)$ can be found with
$$
U(y) = -\int{\frac{100}{y}dy} = -100 \log{y} + c_1.
$$
In physics the conservation of energy means that the sum of the kinetic energy and potential energy are constant
$$
\frac{y'^2}{2} - 100 \log{y} + c_1 = c_2,
$$
$$
y' = \frac{dy}{dx} = \sqrt{200 \log{y} + c_3}.
$$
By applying separation of variables the solution can be found with
$$
\int\frac{dy}{\sqrt{200 \log{y} + c_3}} = x + c_4,
$$
which has the implicit solution
$$
\sqrt{\frac{\pi}{200}} e^{-\frac{c_3}{200}} \text{erfi}\left(\frac{c_3}{200} + \log(y)\right) = x + c_4,
$$
where $\text{erfi}(x)$ is the imaginary error function ($\text{erfi}(x) = -i\, \text{erf}(i\, x)$). Rewriting this to an expression for $y(x)$ yields
$$
y(x) = \exp\left(\text{erfi}^{-1}\left(\sqrt{\frac{200}{\pi}} e^{\frac{c_3}{200}} (x + c_4)\right) - \frac{c_3}{200}\right).
$$
A: Wolfram alpha gives this result which shows it is not easy to find. 
