If x= \sin t,\: y= \sin pt,  prove that \left ( 1-x^{2} \right )\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}+p^{2}y= 0.

 

 

 

 
 
 
 
 

Answers (1)

x= \sin t,y= \sin pt\: \left ( given \right )
\therefore \frac{dx}{dt}= \cos t, \frac{dy}{dt}= p\cos pt
i.e \frac{dy}{dx}= \frac{dy}{dt} \cdot \frac{dt}{dx}= \frac{p\cos pt}{\cos t}\Rightarrow \frac{dy}{dx}= \frac{p\sqrt{1-\sin ^{2}pt}}{\sqrt{1-\sin ^{2}t}}
\frac{dy}{dx}= \frac{p\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}
\Rightarrow \sqrt{1-x^{2}}\left ( \frac{dy}{dx} \right )= p\sqrt{1-y^{2}}
\Rightarrow \left ( 1-x^{2} \right )\left ( \frac{dy}{dx} \right )^{2}= p^{2}\left ( 1-y^{2} \right )
\Rightarrow \left ( 1-x^{2} \right )\times 2\frac{dy}{dx}\times \frac{d^{2}y}{dx^{2}}-2x\left ( \frac{dy}{dx} \right )^{2}= -2p^{2}y\times \frac{dy}{dx}
\Rightarrow \left ( 1-x^{2} \right )\times \frac{d^{2}y}{dx^{2}}-x\left ( \frac{dy}{dx} \right )= -p^{2}y
\therefore \left ( 1-x^{2} \right )\times \frac{d^{2}y}{dx^{2}}-x\left ( \frac{dy}{dx} \right )+p^{2}y= 0

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