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Need solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 19

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You must know about derivative of sin pt


I\! f\; x=sin\; t,\; y=sin\; pt

\text { Prove that }\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+p^{2} y=0


Let\; x=sin\; t,\; y=sin\; pt

        \begin{aligned} &\frac{d x}{d t}=\cos t \text { and } \frac{d y}{d t}=p \cos p t \\ &\text { Now } \frac{d y}{d x}=\frac{p \cos p t}{\cos t} \\ &\frac{d y}{d x} \cos t=p \cos p t \end{aligned}

        \begin{aligned} &\left(\frac{d y}{d x}\right)^{2}\left(1-\sin ^{2} t\right)=p^{2}\left(1-\sin ^{2} p t\right) \\ &\left(\frac{d y}{d x}\right)^{2}\left(1-x^{2}\right)=p^{2}\left(1-y^{2}\right) \end{aligned}

Differentiating with x

        \begin{aligned} &\left(1-x^{2}\right) 2 \frac{d y}{d x} \times \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}(-2 x) \\ &=-p^{2} \times 2 y \frac{d y}{d x} \\ &2 \frac{d y}{d x}\left[\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}\right]=p^{2} y 2 \frac{d y}{d x} \\ &\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+p^{2} y=0 \end{aligned}

Hence proved

Posted by

Gurleen Kaur

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