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x = sin t, y = sin pt (or cos pt): find (1 - x^2) y'' - x y'

The same question, asked 2 times: 25 Apr 2025 S2*, 11 Apr 2026 S2. Every wording below, oldest first.

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  1. Question 1

    25 Apr 2025 S2

    144
    If x=sin⁡tx= \sin t and y=sin⁡pty= \sin pt, then the value of (1−x2)d2y dx2−x dy dx+p2y=\left( 1 -x^{2} \right)\frac{d^{2}y}{\text{ }dx^{2}}-x\frac{\text{ }dy}{\text{ }dx}+p^{2}y=

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  2. Question 2

    11 Apr 2026 S2

    73
    If x=sin⁡(t),y=cos⁡(pt)x = \sin(t), y = \cos(pt), then (1−x2)d2ydx2−xdydx=(1 - x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} =

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