Solve the differential equation \(x\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + 2y = \sqrt {1 + {x^2}} \)

given that \(y = 1\) when \(x = \sqrt 3 \) .

Rewrite the equation in the form

\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{2}{x}y = \frac{{\sqrt {1 + {x^2}} }}{x}\)     M1A1

Integrating factor \( = {{\rm{e}}^{\int {\left( {\frac{2}{x}} \right)} {\rm{d}}x}}\)     M1

\( = {{\rm{e}}^{2\ln x}}\)     (A1)

\( = {x^2}\)     A1

The equation becomes

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

\(y{x^2} = \frac{1}{3}{(1 + {x^2})^{\frac{3}{2}}} + C\)     A1

\(3 = \frac{1}{3} \times 8 + C \to C = \frac{1}{3}\)    M1A1

\(\left[ {{\text{giving }}y{x^2} = \frac{1}{3}\left( {{{\left( {1 + {x^2}} \right)}^{\frac{3}{2}}} + 1} \right)} \right]\)

[9 marks]