12

12. [Maximum mark: 21]

Consider the differential equation $\frac{dy}{dx} - y \csc 2x = \sqrt{\tan x}$, where $0 < x < \frac{\pi}{2}$ and $y = \sqrt{\tan x}$ at $x = \frac{\pi}{4}$.

(a) Use Euler’s method with step length $\frac{\pi}{12}$ to find an approximate value of $y$ when $x = \frac{5\pi}{12}$. Give your answer correct to three significant figures. [3]

(b) Show that $\frac{d}{dx} \left( \frac{1}{2} \ln(\cot x) \right) = -\csc 2x$. [4]

(c) Show that $\sqrt{\cot x}$ is an integrating factor for this differential equation. [4]

(d) Hence, by solving the differential equation, show that $y = x\sqrt{\tan x}$. [5]

(e) Consider the curve $y = x\sqrt{\tan x}$ for $0 < x < \frac{\pi}{2}$ and the Euler’s method approximation calculated in part (a).
(i) Find the $y$‑coordinate at $x = \frac{5\pi}{12}$. Give your answer correct to three significant figures.

(ii) By considering the gradient of the curve, suggest a reason why Euler’s method does not give a good approximation for the $y$‑coordinate at $x = \frac{5\pi}{12}$.

(iii) State why this approximation is less than the $y$‑coordinate at $x = \frac{5\pi}{12}$. [3]

(f) By considering $\frac{dy}{dx} - y \csc 2x = \sqrt{\tan x}$, deduce that the curve $y = x\sqrt{\tan x}$ has a positive gradient for $0 < x < \frac{\pi}{2}$. [2]