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  • Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 24 maths

Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 24 maths

Answers (1)

Answer:

Proved

Hint:

You must know about the derivative of sin function and logarithm function

Given:

x=\sin \left(\frac{1}{a} \log y\right) \text { show that }\left(1-x^{2}\right) y_{2}-x y_{1}-a^{2} y=0

Solution:

        \begin{aligned} &x=\sin \left(\frac{1}{a} \log y\right)\\ &\log y=a \sin ^{-1} x\\ &y=e^{a \sin ^{-1} x}\; \; \; \; \; \; \; .....(1) \end{aligned}

        \begin{aligned} &\text { Let } t=a \sin ^{-1} x\\ &\frac{d t}{d x}=\frac{a}{\sqrt{1-x^{2}}} \quad\left[\frac{d}{d x} \sin ^{-1} x=\frac{1}{\sqrt{1-x^{2}}}\right]\\ &\text { and } y=e^{t}\\ &\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}\\ &\frac{d y}{d x}=e^{t} \frac{a}{\sqrt{1-x^{2}}}=\frac{a e^{a \sin ^{-1} x}}{\sqrt{1-x^{2}}}\; \; \; \; \; \; \; \; .......(2) \end{aligned}

        \begin{aligned} &\text { Again differentiating both sides }\\ &\frac{d^{2} y}{d x^{2}}=a e^{a \sin ^{-1} x} \frac{d}{d x}\left(\frac{1}{\sqrt{1-x^{2}}}\right)+\frac{a}{\sqrt{1-x^{2}}} \frac{d}{d x} e^{a \sin ^{-1} x} \end{aligned}

        \begin{aligned} &\text { Using chain rule and equation }\\ &\frac{d^{2} y}{d x^{2}}=\frac{-a e^{a \sin ^{-1} x}}{2\left(1-x^{2}\right) \sqrt{1-x^{2}}}(-2 x)+\frac{a^{2} e^{a \sin ^{-1} x}}{\left(1-x^{2}\right)}\\ &\frac{d^{2} y}{d x^{2}}=\frac{x a e^{a \sin ^{-1} x}}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}+\frac{a^{2} e^{a \sin ^{-1} x}}{\left(1-x^{2}\right)}\\ &\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=a^{2} e^{a \sin ^{-1} x}+\frac{x a e^{a \sin ^{-1} x}}{\sqrt{1-x^{2}}} \end{aligned}

        \begin{aligned} &\text { Using equation }(1) \text { and }(2)\\ &\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=a^{2} y+x \frac{d y}{d x}\\ &\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-a^{2} y-x \frac{d y}{d x}=0\\ &\left(1-x^{2}\right) y_{2}-x y_{1}-a^{2} y=0 \end{aligned}

Posted by

Gurleen Kaur

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