First slide

Differentiability

Question

$If x = sint and y = sinpt then the value of (1 - x^{2}) \frac{d y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y =$

Moderate

A

0
B

1
C

2
D

-1

Solution

$\begin{array}{l} \frac{d x}{d t} = \cos t \\ \frac{d y}{d t} = p \cos p t \Rightarrow \frac{d y}{d x} = \frac{p \cos p t}{\cos t} = \frac{p \sqrt{1 - \sin^{2} p t}}{\sqrt{1 - \sin^{2} t}} \\ \frac{d y}{d x} = \frac{p \sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}} \\ {(\frac{d y}{d x})}^{2} (1 - x^{2}) = p^{2} (1 - y^{2}) \\ Diff \end{array}$
$\begin{array}{l} 2 \frac{d y}{d x} \frac{d^{2} y}{d x^{2}} (1 - x^{2}) + {(\frac{d y}{d x})}^{2} (- 2 x) = p^{2} (- 2 y) \frac{d y}{d x} \\ \frac{d^{2} y}{d x^{2}} (1 - x^{2}) - x \frac{d y}{d x} = - p^{2} y \\ (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0 \end{array}$

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