(a)     Find an integrating factor for this differential equation.

(b)     Solve the differential equation given that \(y = 1\) when \(x = 1\) , giving your answer in the forms \(y = f(x)\) .

(a)     Rewrite the equation in the form

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} - \frac{2}{x}y = \frac{{{x^2}}}{{{x^2} + 1}}\)     M1A1

Integrating factor \( = {{\text{e}}^{\int { - \frac{2}{x}{\text{d}}x} }}\)     M1

\( = {{\text{e}}^{ - 2\ln x}}\)     A1

\( = \frac{1}{{{x^2}}}\)     A1

Note: Accept \(\frac{1}{{{x^3}}}\) as applied to the original equation.

[5 marks]

(b)     Multiplying the equation,

\(\frac{1}{{{x^2}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} - \frac{2}{{{x^3}}}y = \frac{1}{{{x^2} + 1}}\)     (M1)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{y}{{{x^2}}}} \right) = \frac{1}{{{x^2} + 1}}\)     (M1)(A1)

\(\frac{y}{{{x^2}}} = \int {\frac{{{\text{d}}x}}{{{x^2} + 1}}} \)     M1

\( = \arctan x + C\)     A1

Substitute \(x = 1,{\text{ }}y = 1\) .     M1

\(1 = \frac{\pi }{4} + C \Rightarrow C = 1 - \frac{\pi }{4}\)     A1

\(y = {x^2}\left( {\arctan x + 1 - \frac{\pi }{4}} \right)\)     A1

[8 marks]

Total [13 marks]

The response to this question was often disappointing. Many candidates were unable to find the integrating factor successfully.