Solve the following differential equation $$(x + 1)(x + 2)\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + y = x + 1$$ giving your answer in the form $y = f(x)$ .

Rewrite the equation in the form

$\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{y}{{(x + 1)(x + 2)}} = \frac{1}{{x + 2}}$     M1

Integrating factor $= \exp \left( {\int {\frac{{{\rm{d}}x}}{{(x + 1)(x + 2)}}} } \right)$     A1

$= \exp \left( {\int {\left( {\frac{1}{{x + 1}} - \frac{1}{{x + 2}}} \right){\rm{d}}x} } \right)$     M1A1

$= \exp \ln \left( {\frac{{x + 1}}{{x + 2}}} \right)$     A1

$= \frac{{x + 1}}{{x + 2}}$     A1

Multiplying by the integrating factor,

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

$= \frac{{x + 2}}{{{{(x + 2)}^2}}} - \frac{1}{{{{(x + 2)}^2}}}$     A1

Integrating,

$\left( {\frac{{x + 1}}{{x + 2}}} \right)y = \ln (x + 2) + \frac{1}{{x + 2}} + C$     A1A1

$y = \left( {\frac{{x + 2}}{{x + 1}}} \right)\left\{ {\ln (x + 2) + \frac{1}{{x + 2}} + C} \right\}$     A1

[11 marks]