(a)
Euler’s method: \( x_{n+1} = x_n + 0.1 \), \( y_{n+1} = y_n + 0.1 \cdot \frac{y_n^2 – 2x_n^2}{x_n^2} \)
Initial condition: \( x_0 = 1 \), \( y_0 = 3 \)
Step 1: \( x_1 = 1.1 \), \( \frac{dy}{dx} = \frac{3^2 – 2 \cdot 1^2}{1^2} = 7 \), \( y_1 = 3 + 0.1 \times 7 = 3.7 \)
Step 2: \( x_2 = 1.2 \), \( \frac{dy}{dx} = \frac{3.7^2 – 2 \cdot 1.1^2}{1.1^2} \approx 9.314 \), \( y_2 = 3.7 + 0.1 \times 9.314 \approx 4.6314 \)
Step 3: \( x_3 = 1.3 \), \( \frac{dy}{dx} = \frac{4.6314^2 – 2 \cdot 1.2^2}{1.2^2} \approx 12.8956 \), \( y_3 = 4.6314 + 0.1 \times 12.8956 \approx 5.92096 \)
Step 4: \( x_4 = 1.4 \), \( \frac{dy}{dx} = \frac{5.92096^2 – 2 \cdot 1.3^2}{1.3^2} \approx 18.7448 \), \( y_4 = 5.92096 + 0.1 \times 18.7448 \approx 7.79544 \)
Step 5: \( x_5 = 1.5 \), \( \frac{dy}{dx} = \frac{7.79544^2 – 2 \cdot 1.4^2}{1.4^2} \approx 29.0045 \), \( y_5 = 7.79544 + 0.1 \times 29.0045 \approx 10.69589 \approx 10.7 \)
Result: \( y \approx 10.7 \) [6]

(b)
Substitute \( y = vx \): \( \frac{dy}{dx} = v + x \frac{dv}{dx} \)
\( x^2 \left( v + x \frac{dv}{dx} \right) = (vx)^2 – 2x^2 \)
Divide by \( x^2 \): \( v + x \frac{dv}{dx} = v^2 – 2 \)
Simplify: \( x \frac{dv}{dx} = v^2 – v – 2 \)
Result: \( x \frac{dv}{dx} = v^2 – v – 2 \) [4]

(c)(i)
From (b): \( \frac{dv}{v^2 – v – 2} = \frac{dx}{x} \)
Factorize: \( v^2 – v – 2 = (v – 2)(v + 1) \)
Partial fractions: \( \frac{1}{(v – 2)(v + 1)} = \frac{1}{3} \left( \frac{1}{v – 2} – \frac{1}{v + 1} \right) \)
Integrate: \( \frac{1}{3} \left( \ln |v – 2| – \ln |v + 1| \right) = \ln |x| + c \)
Initial condition: \( x = 1 \), \( y = 3 \), \( v = 3 \implies \frac{1}{3} \ln \left| \frac{3 – 2}{3 + 1} \right| = \ln 1 + c \implies c = -\frac{1}{3} \ln 4 \)
Equation: \( \ln \left| \frac{v – 2}{v + 1} \right| = \ln \left( \frac{x^3}{4} \right) \)
Since \( v > 2 \), \( x > 0 \): \( \frac{v – 2}{v + 1} = \frac{x^3}{4} \)
Substitute \( v = \frac{y}{x} \): \( \frac{\frac{y}{x} – 2}{\frac{y}{x} + 1} = \frac{x^3}{4} \)
Solve: \( y – \frac{x^3 y}{4} = 2x + \frac{x^4}{4} \implies y = \frac{8x + x^4}{4 – x^3} \)
Verify: \( x = 1 \), \( y = \frac{8 \cdot 1 + 1^4}{4 – 1^3} = 3 \)
Result: \( y = \frac{8x + x^4}{4 – x^3} \) [6]

(c)(ii)
At \( x = 1.5 \): \( y = \frac{8 \cdot 1.5 + 1.5^4}{4 – 1.5^3} = \frac{12 + 5.0625}{4 – 3.375} = \frac{17.0625}{0.625} = 27.3 \)
Result: \( y = 27.3 \) [2]

(c)(iii)
The function has a vertical asymptote at \( x = \sqrt[3]{4} \approx 1.5874 \), causing the gradient to increase rapidly near \( x = 1.5 \), making Euler’s method inaccurate
Result: Rapid gradient change near asymptote [1]