16N.3.AHL.TZ0.HCA_1

pestleMathematicsAAHLPaper 316N· ahl-5-18-1st-order-des-euler-method-variables-separable-integrating-factor-homogeneous-de-using-sub-y=vxsource ↗

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.

[5]

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

[6]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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 $(1 + {x^2})y$

[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]

Examiners’ report

[N/A]