2011 Paper 1 Question 14

$\sqrt3 \frac{dy}{dx} + y = 4 \sin x \qquad(*)$

(a)

$f(x)=\frac{1}{2}a_0 + \sum\limits_{r=1}^\infty a_r \cos \left(\frac{2\pi rx}{L}\right) + \sum\limits_{r=1}^\infty b_r \sin \left(\frac{2\pi rx}{L}\right)$

The period $L=2\pi$ so

$f(x)=\frac{1}{2}a_0 + \sum\limits_{r=1}^\infty a_r \cos \left( r x\right) + \sum\limits_{r=1}^\infty b_r \sin \left(r x \right)$

(b)

Assuming such an expansion exists for $y$

$\frac{dy}{dx} = \sum\limits_{r=1}^\infty a_r (-)\sin \left( r x\right)r + b_r \cos \left(r x \right) r$

Substitute in (*)

$\frac{1}{2}a_0 + \sum\limits_{r=1}^\infty \left( \sqrt3 r b_r + a_r \right) \cos{(rx)} + \left( b_r - \sqrt3 r a_r \right) \sin{(rx)} = 4 \sin{x}$

This yields the constraints on the coefficients, by comparing constants, coefficients of $\cos$ and $\sin$

$\begin{aligned}a_0 &= 0 \\ \sqrt3 r b_r + a_r &= 0 \\ b_r - \sqrt3 r a_r &= 4\delta_{1,r} \end{aligned}$

Here, $\delta_{1,r}$ is the Kronecker delta.

NST First Year Maths

Some unofficial solutions to Cambridge NST IA maths papers

2011 Paper 1 Question 14

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