Show that \(1 + {x^2}\) is an integrating factor for this differential equation.

Hence solve this differential equation. Give the answer in the form \(y = f(x)\).

METHOD 1

attempting to find an integrating factor     (M1)

\(\int {\frac{{2x}}{{1 + {x^2}}}{\text{d}}x = \ln (1 + {x^2})} \)    (M1)A1

IF is \({{\text{e}}^{\ln (1 + {x^2})}}\)     (M1)A1

\( = 1 + {x^2}\)    AG

METHOD 2

multiply by the integrating factor

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

left hand side is equal to the derivative of \((1 + {x^2})y\)

A3

[5 marks]

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

\(\frac{{\text{d}}}{{{\text{d}}x}}\left[ {(1 + {x^2})y} \right] = {x^2} + {x^4}\)

\((1 + {x^2})y = \left( {\int {{x^2} + {x^4}{\text{d}}x = } } \right){\text{ }}\frac{{{x^3}}}{3} + \frac{{{x^5}}}{5}( + c)\)    A1A1

\(y = \frac{1}{{1 + {x^2}}}\left( {\frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} + c} \right)\)

\(x = 0,{\text{ }}y = 2 \Rightarrow c = 2\)    M1A1

\(y = \frac{1}{{1 + {x^2}}}\left( {\frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} + 2} \right)\)    A1

[6 marks]