If x = t and y = pt, prove that (1 - x^ 2 ) d^ 2 y dx^ 2 - x dy d… - Vidyadip

Question

If $x = \sin \text{t} $ and $\text{y} = \sin \text{pt,}$ prove that $(1 - x^{2}) \frac{\text{d}^{2} \text{y}}{\text{dx}^{2}} - x \frac{\text{dy}}{\text{dx}} + \text{P}^{2}\text{y} = 0.$

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Answer

$\text{x} = \sin \text{t} \Rightarrow \frac{\text{dx}}{\text{dt}} = \cos \text{t}$
$\text{} = \sin \text{pt} \Rightarrow \frac{\text{dy}}{\text{dt}} = \text{p} \cos \text{pt}$
$\frac{\text{dy}}{\text{dx}} = \frac{\text{p} \cos \text{pt}}{\cos \text{t}}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = \frac{\cos \text{t} (-\text{p}^{2} \sin \text{pt)} - \text{p} \cos \text{pt} (-\sin \text{t})}{\cos^{2}\text{t}}. \frac{\text{dt}}{\text{dx}}$
$= \frac{\text{-p}^{2} \sin \text{pt} \cos \text{t} + \text{p} \cos \text{pt} \sin \text{t}}{\cos^{3} \text{t}}$
$\text{Now} (1 - \text{x}^{2}) \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} - \text{x} \frac{\text{dy}}{\text{dx}} + \text{p}^{2} \text{y} = 0 \Bigg[ \text{Substituting values of y,}\frac{\text{dy}}{\text{dx}} \& \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Bigg]$

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