The height $x$ metres, of a column of water in a fountain display satisfies the differential equation $$\frac{dx}{dr} = \frac{8\sin 2t}{3\sqrt{x}}$$, where $t$ is the time in seconds after the display begins - AQA - A-Level Maths Pure - Question 15 - 2017 - Paper 1

To solve the differential equation $\frac{dx}{dr} = \frac{8\sin 2t}{3\sqrt{x}}$, first we separate the variables:

$3\sqrt{x} \ dx = 8\sin 2t \ dr$.

Next, we can integrate both sides:

$\int 3\sqrt{x} \ dx = \int 8\sin 2t \ dr$.

The left side integrates to:

$\frac{3}{2} x^{\frac{3}{2}}$,

and the right side integrates to:

$-4\cos 2t + C$.

Setting the integrated equations equal gives:

$\frac{3}{2} x^{\frac{3}{2}} = -4\cos 2t + C$.

Given that initially, the column of water has zero height (at $t=0$, $x=0$), substituting these values helps us find $C$:

$\frac{3}{2}(0)^{\frac{3}{2}} = -4\cos(2\cdot0) + C \implies C = 4$.

Thus, we substitute $C$ back into the equation:

$\frac{3}{2} x^{\frac{3}{2}} = -4\cos 2t + 4$.

Now solving for $x$, we get:

$x^{\frac{3}{2}} = \frac{2}{3}(-4\cos 2t + 4)$,

and raising both sides to the power of $\frac{2}{3}$ leads to:

$$x = \left(\frac{2}{3}(4 - 4\cos 2t)\right)^{\frac{2}{3}}.$