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Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 21 maths textbook solution

Answers (1)

Answer:

        Proved

Hint:

You must know about the derivative of exponential function and tangent inverse x

Given:

y=e^{tan^{-1}x},\; \; Prove\: (1+x^{2})y_{2}+(2x-1)y_{1}=0

Solution:

Let\; \; y=e^{tan^{-1}x}

        \begin{aligned} &\frac{d y}{d x}=e^{\tan ^{-1} x} \times \frac{1}{\left(1+x^{2}\right)} \\ &\left(x^{2}+1\right) \frac{d y}{d x}=e^{\operatorname{tan}^{-1} x} \\ &\left(x^{2}+1\right) \frac{d^{2} y}{d x^{2}}+2 x \frac{d y}{d x}=e^{\tan ^{-1} x} \times \frac{1}{\left(1+x^{2}\right)} \end{aligned}

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

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Gurleen Kaur

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