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I found this question in a Special Paper for Further Mathematics. I post my answer in full, but I wanted to verify that my Maclaurin series is correct. Thank you.

Given, $x = \cos \theta$, and $y = \cos 9\theta$, and given the differential equation, $(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} +81y = 0$

a). Prove that

$(1 - x^2) \frac{d^{n+2}y}{dx^{n+2}} - x(2n + 1) \frac{dy^{n+1}}{dx^{n+1}} - (n^2 - 81) \frac{d^ny}{dx^n} = 0$.

b). Denoting $\frac{d^ny}{dx^n}$ by $f^n$, deduce that, $f^{n + 2}(0) = (n^2 - 81)f^n(0)$

Use Maclaurin's theorem to obtain $y$ as a polynomial in $x$.

My workings:

a). Using Leibnitz's theorem to differentiate $n$-times the given differential equation,

Leibnitz's theorem: $\frac{d^n (uv)}{dx^n} = u_{n}v + {}^{n}C_{1}u_{n-1}v_{1} + {}^{n}C_{2}u_{n-2}v_{2} + {}^{n}C_{3}u_{n-3}v_{3} \dots + uv_{n}$

$\therefore$ differentiating n-times: $\frac{d^n}{dx^n} \left[ (1 - x^2) \frac{d^2y}{dx^2} \right] - \frac{d^n}{dx^n} \left[ x \frac{dy}{dx} \right] +81 \frac{dy^n}{dx^n} = 0 $

$ \therefore \left\{ \frac{d^n}{dx^n} \left(\frac{d^2y}{dx^2}(1 - x^2) \right) + n \frac{d^{n-1}}{dx^{n-1}} \left(\frac{d^2y}{dx^2} \right)(- 2x) + \frac{n(n - 1)}{2} \frac{d^{n-2}}{dx^{n-2}} \left(\frac{d^2y}{dx^2} \right)(- 2) \right \} - \left \{ \frac{d^n}{dx^n} \left (\frac{dy}{dx} \right )x + n \frac{d^{n - 1}}{dx^{n -1}} \left ( \frac{dy}{dx} \right) \times 1 \right \} + 81 \frac{d^ny}{dx} = 0 $

$ \Rightarrow \frac{d^{n + 2}y}{dx^{n + 2}} (1 - x^2) - 2nx \frac{d^{n + 1}y}{dx^{n + 1}} - n(n - 1) \frac{d^n y}{dx^n} - \left \{ \frac{d^{n + 1}y}{dx^{n + 1}} x + n \frac{d^n y}{dx^n} \right \} + 81 \frac{d^ny}{dx} = 0 $

Combining terms,

$ (1 - x^2) \frac{d^{n + 2}y}{dx^{n + 2}} - x(2n + 1) \frac{d^{n + 1}y}{dx^{n + 1}} - \left \{ n(n - 1) + n - 81 \right \} \frac{d^n y}{dx^n} = 0$

$ \therefore (1 - x^2) \frac{d^{n + 2}y}{dx^{n + 2}} - x(2n + 1) \frac{d^{n + 1}y}{dx^{n + 1}} - (n^2 - 81) \frac{d^n y}{dx^n} = 0$ ... Result 1, proved by Leibnitz's theorem!

(Induction also shows the same result holds true for all positive integers $n$).

b). When $x = 0$, Result 1 becomes, $(1 - x^2) \frac{d^{n + 2}y(0)}{dx^{n + 2}} - x(2n + 1) \frac{d^{n + 1}y(0)}{dx^{n + 1}} - (n^2 - 81) \frac{d^n y(0)}{dx^n} = 0$

$ \Rightarrow \frac{d^{n + 2}y(0)}{dx^{n + 2}} - (n^2 - 81) \frac{d^n y(0)}{dx^n} = 0$

With $ \frac{d^n y}{dx^n}$ denoted as $f^n$ $\Rightarrow f^{n + 2}(0) - (n^2 - 81)f^n(0) = 0$

$ \therefore$ we deduce that $ f^{n + 2}(0) = (n^2 - 81)f^n(0) $} ... Result 2

Maclaurin series coefficients are found using Result 2, for $ n \ge 0$

Maclaurin series: $f(x) = f(0) + xf^1(0) +\frac{x^2}{2!}f^2(0) + \frac{x^3}{3!}f^3(0) + \dots \frac{x^n}{n!}f^n(0) + \frac{x^{n + 1}}{(n + 1)!}f^{n + 1}(0) \dots $

Given that $x = \cos \theta$, and $y = \cos 9\theta$,

at $x = 0, \cos \theta = 0 \Rightarrow \theta = \pm \frac{\pi}{2}$

$ \therefore y(0) = f(0) = \cos (9 \frac{\pi}{2})$

$\therefore f(0) = 0$

Applying Result 2 with $ n = 0 $ gives, $ f^{2}(0) = (- 81)f(0) = 0 $

For, $ n = 2, 4, 6, \dots \Rightarrow f^{4}(0) = (4 - 81)f^{2}(0) = 0, f^{6}(0) = (16 - 81)f^{4}(0) = 0 \dots \Rightarrow f^{n + 2}(0) = 0$, for all EVEN $n$.

For, $ n = 1 $, differentiating $x$ and $y$ parametrically gives, $ f^1(x) = \frac{dy}{dx} = \frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}}$

$\therefore f^1(x) = \frac{9 \sin(9 \theta)}{\sin \theta}$

$ \Rightarrow f^1(0) = \frac{9 \sin (9 \frac{\pi}{2})}{\sin (\frac{\pi}{2})} = 9 $

Applying Result 2 with $ n = 1 \Rightarrow f^3(0) = (1 - 81)f^1(0) = - 80 \times 9 = - 720 $

For $ n = 3, \Rightarrow f^5(0) = (9 - 81)f^3(0) = (- 72)(- 80) \times 9 = 51840 $

For $ n = 5, \Rightarrow f^7(0) = (25 - 81)f^5(0) = (- 56)(- 72)(- 80) \times 9 = - 2903040 $

For $ n = 7, \Rightarrow f^9(0) = (49 - 81)f^7(0) = ( -32)(- 56)(- 72)(- 80) \times 9 = 92897280 $

For $ n = 9, \Rightarrow f^11(0) = (81 - 81)f^9(0) = 0 $

For $ n = 11, \Rightarrow f^13(0) = (121 - 81)f^11(0) = 0 $ ...

... for all ODD $n \ge 9$ the Maclaurin coefficients $= 0$

$ \therefore $ the Maclaurin series is, $ f(x) = - 720x + \frac{51840}{3!}x^3 - \frac{2903040}{5!}x^5 + \frac{92897280}{7!}x^7$

Therefore, $f(x) = - 720x + 8640x^3 - 24192x^5 + 18432x^7 $

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