| Date | November 2016 | Marks available | 6 | Reference code | 16N.3ca.hl.TZ0.1 |
| Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
| Command term | Hence | Question number | 1 | Adapted from | N/A |
Question
Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} + \left( {\frac{{2x}}{{1 + {x^2}}}} \right)y = {x^2}\), given that \(y = 2\) when \(x = 0\).
Show that \(1 + {x^2}\) is an integrating factor for this differential equation.
Hence solve this differential equation. Give the answer in the form \(y = f(x)\).
Markscheme
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]
