Mixed integration problem Well I am quite embarrassed asking this as I have done really advanced stuff in college but its been awhile and I have completely forgotten how to do it, since I don't do Math in university (in UK college is a compulsory extension of high school, sort of).
We have: $\dfrac{d^2y}{dx^2} = \dfrac{5}{y^2}$ Find: $dy/dx$
I think we need to rearrange the equation. It would be easy if on the right side we would've x, but it's y, hence the rearrangement is required.
I did something like this, but I doubt its correct:
Rearrange: $y^2 * d^2y = 5 * dx^2$
Integrate both sides: $\dfrac{y^3}{3} * dy = 5x * dx$
Rearrange back into required form: $\dfrac{dy}{dx} = \dfrac{3*5x}{y^3} + c$
 A: One standard way to solve a problem like this is to multiply both sides of the equation by $dy/dx$ and pull out a $d/dx$, which gives
\begin{align}
\frac{d^2 y}{dx^2}\cdot\frac{dy}{dx} &= \frac{5}{y^2}\cdot\frac{dy}{dx} \\
\implies \frac{d}{dx}\Big[\frac{1}{2}\Big(\frac{dy}{dx}\Big)^2\Big]&=\frac{d}{dx}\Big[-\frac{5}{y}\Big]
\end{align}
(Do you understand the trick here?)
It follows upon integrating with respect to $x$ that
\begin{align}
\Big(\frac{dy}{dx}\Big)^2 &= -\frac{10}{y} + C \\
\frac{dy}{dx} &= \sqrt{C-\frac{10}{y}}.
\end{align}
A: $$\frac{d^2y}{dx^2} = \frac{5}{y^2}$$
Now, $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$$
by definition.
Substitute, $t=\frac{dy}{dx}$
$$\frac{dt}{dx} = \frac{5}{y^2}$$
$$t\frac{dt}{dx} = \frac{5}{y^2}\frac{dy}{dx}$$
$$t\,dt = 5\frac{dy}{y^2}$$
On integrating, 
$$\frac{t^2}{2}+C = -\frac{5}{y}$$
$$\frac{dy}{dx} = \sqrt{C - \frac{10}{y}}$$
A: It won't be using a technique you learnt at school.
Multiply the equation by $dy/dx$:
$$\frac{d^2y}{dx^2} \frac{dy}{dx}= \frac{5}{y^2} \frac{dy}{dx}$$
Then notice the left hand side is half the derivative of $(dy/dx)^2$, so integrate with respect to x:
$$\frac{1}{2}(\frac{dy}{dx})^2 = \int \frac{5}{y^2} \frac{dy}{dx} dx = \int \frac{5}{y^2} dy = c - \frac{5}{y}$$
So 
$$\frac{dy}{dx} = \sqrt{c-\frac{10}{y}}$$
