21N.1.AHL.TZ0.8

pestleMathematicsAAHLPaper 121N· ahl-5-16-integration-by-substitution-parts-and-repeated-partssource ↗

Solve the differential equation $\frac{d y}{d x} = \frac{\ln 2 x}{x^{2}} - \frac{2 y}{x}, x > 0$ , given that $y = 4$ at $x = \frac{1}{2}$ .

Give your answer in the form $y = f (x)$ .

Markscheme / solution

$\frac{d y}{d x} + \frac{2 y}{x} = \frac{\ln 2 x}{x^{2}}$ (M1)

attempt to find integrating factor (M1)

$(e^{\int \frac{2}{x} d x} = e^{2 \ln x}) = x^{2}$ (A1)

$x^{2} \frac{d y}{d x} + 2 x y = \ln 2 x$

$\frac{d}{d x} (x^{2} y) = \ln 2 x$

$x^{2} y = \int \ln 2 x d x$

attempt to use integration by parts (M1)

$x^{2} y = x \ln 2 x - x (+ c)$ A1

$y = \frac{\ln 2 x}{x} - \frac{1}{x} + \frac{c}{x^{2}}$

substituting $x = \frac{1}{2}, y = 4$ into an integrated equation involving $c$ M1

$4 = 0 - 2 + 4 c$

$\Rightarrow c = \frac{3}{2}$

$y = \frac{\ln 2 x}{x} - \frac{1}{x} + \frac{3}{2 x^{2}}$ A1

[7 marks]

Examiners’ report

[N/A]