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

Now, we integrate both sides: $\int 3\sqrt{x} \, dx = \int 8\sin 2t \, dt$ The left side becomes: $\frac{3}{\frac{3}{2}}x^{\frac{3}{2}} = 2x^{\frac{3}{2}}$ The right side integrates to: $-4\cos 2t + C$

To find the constant C, we substitute the initial condition where the column of water has zero height: When $t = 0$, $x = 0$: $2(0)^{\frac{3}{2}} = -4\cos(2\cdot0) + C$ This simplifies to: $0 = -4 + C \Rightarrow C = 4$

Substituting C back into our equation yields: $2x^{\frac{3}{2}} = -4\cos 2t + 4$ Rearranging this, we find: $x^{\frac{3}{2}} = 2 - 2\cos 2t$

Expressing in the required form: $x = (2 - 2\cos 2t)^{\frac{2}{3}}$