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प्रश्न
If x = sin t and y = sin pt, prove that \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] .
उत्तर
Given:
x = sint and y = sinpt
Differentiating both sides with respect to t, we get
\[\frac{d x}{d t} = \cos t and \frac{d y}{d t} = p\cos pt\]
\[ \Rightarrow \frac{d y}{d x} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{p\cos pt}{\cos t}\]
Differentiating both sides with respect to x, we get
\[\frac{d^2 y}{d x^2} = \frac{- p^2 \sin pt \cos t + p\cos pt\sin t}{\cos^2 t} \times \frac{dt}{dx}\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{- p^2 \sin pt \cos t + p\cos pt\sin t}{\cos^3 t}\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{- p^2 \sin pt \cos t}{\cos^3 t} + \frac{p\cos pt\sin t}{\cos^3 t}\]
\[\Rightarrow \frac{d^2 y}{d x^2} = \frac{- p^2 y}{\cos^2 t} + \frac{x\frac{d y}{d x}}{\cos^2 t}\]
\[ \Rightarrow \cos^2 t\frac{d^2 y}{d x^2} = - p^2 y + x\frac{d y}{d x}\]
\[ \Rightarrow \left( 1 - \sin^2 t \right)\frac{d^2 y}{d x^2} = - p^2 y + x\frac{d y}{d x}\]
\[ \Rightarrow \left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{d y}{d x} + p^2 y = 0\]
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