(a) Attempts to use Euler’s method:

\( x_{n+1} = x_n + \frac{\pi}{12} \), \( y_{n+1} = y_n + \frac{\pi}{12} \cdot \left( y_n \csc 2x_n + \sqrt{\tan x_n} \right) \), where \( \frac{dy}{dx} = y \csc 2x + \sqrt{\tan x} \) (M1).

Initial condition: \( x_0 = \frac{\pi}{4} \), \( y_0 = \frac{\pi}{4} \approx 0.785398 \).

Step 1: \( x_0 = \frac{\pi}{4} \), \( y_0 = 0.785398 \), \( 2x_0 = \frac{\pi}{2} \), \( \csc \frac{\pi}{2} = 1 \), \( \tan \frac{\pi}{4} = 1 \), \( \sqrt{\tan \frac{\pi}{4}} = 1 \).

\( \frac{dy}{dx} = 0.785398 \cdot 1 + 1 \approx 1.785398 \).

\( y_1 = 0.785398 + \frac{\pi}{12} \cdot 1.785398 \approx 0.785398 + 0.467414 \approx 1.252812 \) (A1).

Step 2: \( x_1 = \frac{\pi}{3} \), \( y_1 \approx 1.252812 \), \( 2x_1 = \frac{2\pi}{3} \), \( \csc \frac{2\pi}{3} = \frac{2}{\sqrt{3}} \approx 1.154701 \), \( \tan \frac{\pi}{3} = \sqrt{3} \approx 1.732051 \), \( \sqrt{\tan \frac{\pi}{3}} \approx 1.316074 \).

\( \frac{dy}{dx} = 1.252812 \cdot 1.154701 + 1.316074 \approx 1.446 + 1.316074 \approx 2.762074 \).

\( y_2 = 1.252812 + \frac{\pi}{12} \cdot 2.762074 \approx 1.252812 + 0.723262 \approx 1.976074 \) (A1).

At \( x = \frac{5\pi}{12} \), \( y \approx 1.976074 \approx 1.98 \) (to 3 significant figures) (A1). [3 marks]

(b) Show that \( \frac{d}{dx} \left( \frac{1}{2} \ln (\cot x) \right) = -\csc 2x \):

Let \( u = \cot x \), so \( y = \frac{1}{2} \ln u \), \( \frac{dy}{du} = \frac{1}{2u} \).

\( \frac{du}{dx} = -\csc^2 x \) (M1).

Chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2 \cot x} \cdot (-\csc^2 x) = -\frac{\csc^2 x}{2 \cot x} \).

\( \frac{\csc^2 x}{\cot x} = \frac{\frac{1}{\sin^2 x}}{\frac{\cos x}{\sin x}} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x} = 2 \csc 2x \).

\( \frac{dy}{dx} = -\frac{1}{2} \cdot 2 \csc 2x = -\csc 2x \) (A1) (AG). [2 marks]

OR

Let \( u = \frac{\cos x}{\sin x} \), so \( y = \frac{1}{2} \ln u \), \( \frac{dy}{du} = \frac{1}{2u} \).

\( \frac{du}{dx} = -\frac{1}{\sin^2 x} \), so \( \frac{dy}{dx} = \frac{1}{2 \cdot \frac{\cos x}{\sin x}} \cdot \left( -\frac{1}{\sin^2 x} \right) = -\frac{1}{2 \cos x \sin x} = -\csc 2x \) (A1).

(c) Show that \( \sqrt{\cot x} \) is an integrating factor:

METHOD 1:

Standard form: \( \frac{dy}{dx} – y \csc 2x = \sqrt{\tan x} \).

Integrating factor: \( I(x) = e^{\int -\csc 2x \, dx} \) (M1).

From (b): \( \int -\csc 2x \, dx = \frac{1}{2} \ln (\cot x) \).

\( I(x) = e^{\frac{1}{2} \ln (\cot x)} = e^{\ln \sqrt{\cot x}} = \sqrt{\cot x} \) (A1). [2 marks]

METHOD 2:

Compute: \( \frac{d}{dx} (y \sqrt{\cot x}) = \sqrt{\cot x} \frac{dy}{dx} + y \frac{d}{dx} (\sqrt{\cot x}) \).

\( \frac{d}{dx} (\sqrt{\cot x}) = \frac{1}{2} (\cot x)^{-1/2} (-\csc^2 x) = -\frac{\csc^2 x}{2 \sqrt{\cot x}} \).

\( \frac{d}{dx} (y \sqrt{\cot x}) = \sqrt{\cot x} \frac{dy}{dx} – y \frac{\csc^2 x}{2 \sqrt{\cot x}} = \sqrt{\cot x} \left( \frac{dy}{dx} – y \frac{\csc^2 x}{2 \cot x} \right) \).

\( \frac{\csc^2 x}{\cot x} = 2 \csc 2x \), so \( \frac{d}{dx} (y \sqrt{\cot x}) = \sqrt{\cot x} \left( \frac{dy}{dx} – y \csc 2x \right) \) (A1).

Thus, \( \sqrt{\cot x} \) is an integrating factor (A1). [2 marks]

(d) Solve: \( \frac{dy}{dx} – y \csc 2x = \sqrt{\tan x} \).

Multiply by integrating factor \( \sqrt{\cot x} \):

\( \sqrt{\cot x} \frac{dy}{dx} – y \csc 2x \sqrt{\cot x} = \sqrt{\tan x} \sqrt{\cot x} \) (M1).

\( \sqrt{\tan x} \sqrt{\cot x} = \sqrt{\frac{\sin x}{\cos x} \cdot \frac{\cos x}{\sin x}} = 1 \).

Left-hand side: \( \frac{d}{dx} (y \sqrt{\cot x}) = 1 \).

Integrate: \( y \sqrt{\cot x} = \int 1 \, dx = x + C \) (A1).

\( y = (x + C) \sqrt{\tan x} \).

Apply initial condition \( x = \frac{\pi}{4} \), \( y = \frac{\pi}{4} \):

\( \tan \frac{\pi}{4} = 1 \), \( \sqrt{\tan \frac{\pi}{4}} = 1 \), \( \frac{\pi}{4} = \left( \frac{\pi}{4} + C \right) \cdot 1 \), so \( C = 0 \) (M1).

Thus, \( y = x \sqrt{\tan x} \) (A1) (AG). [4 marks]

(e) (i) At \( x = \frac{5\pi}{12} \), \( y = x \sqrt{\tan x} \):

\( \tan \frac{5\pi}{12} = \tan 75^\circ = 2 + \sqrt{3} \), \( \sqrt{\tan \frac{5\pi}{12}} \approx 1.931851653 \).

\( y = \frac{5\pi}{12} \cdot \sqrt{2 + \sqrt{3}} \approx 1.308996939 \cdot 1.931851653 \approx 2.528784 \).

\( y \approx 2.53 \) (to 3 significant figures) (A1). [1 mark]

(ii) The gradient changes substantially in the neighborhood of \( x = \frac{5\pi}{12} \) (A1).

Euler’s method assumes linear approximation, but rapid gradient increase leads to poor accuracy. [1 mark]

(iii) The curve is concave up (i.e., \( \frac{d^2 y}{dx^2} > 0 \) over the interval), so Euler’s method underestimates as tangent lines lie below the curve (A1). [1 mark]

(f) Given: \( \frac{dy}{dx} = y \csc 2x + \sqrt{\tan x} \).

Substitute \( y = x \sqrt{\tan x} \):

\( \csc 2x > 0 \), \( \sqrt{\tan x} > 0 \) for \( 0 < x < \frac{\pi}{2} \) (M1).

\( y = x \sqrt{\tan x} > 0 \), so \( y \csc 2x > 0 \).

Thus, \( \frac{dy}{dx} = y \csc 2x + \sqrt{\tan x} > 0 \) (A1).

The curve has a positive gradient for \( 0 < x < \frac{\pi}{2} \) (A1). [3 marks]