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प्रश्न
If \[y = \left[ \log \left( x + \sqrt{x^2 + 1} \right) \right]^2\] show that \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 2\] ?
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उत्तर
Here,
\[y = \left[ \log\left( x + \sqrt{x^2 + 1} \right) \right]^2 \]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{d y}{d x} = \frac{2\log\left( x + \sqrt{x^2 + 1} \right)}{\left( x + \sqrt{x^2 + 1} \right)} \times \left( 1 + \frac{2x}{2\sqrt{x^2 + 1}} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{2\log\left( x + \sqrt{x^2 + 1} \right)}{\left( x + \sqrt{x^2 + 1} \right)} \times \left( \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{2\log\left( x + \sqrt{x^2 + 1} \right)}{\sqrt{x^2 + 1}}\]
\[\text { Differentiating again w . r . t . x, we get }\]
\[\frac{d^2 y}{d x^2} = \frac{2 - \frac{2x\log\left( x + \sqrt{x^2 + 1} \right)}{\sqrt{x^2 + 1}}}{x^2 + 1}\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{2 - x\frac{dy}{dx}}{x^2 + 1}\]
\[ \Rightarrow \left( x^2 + 1 \right)\frac{d^2 y}{d x^2} = 2 - x\frac{dy}{dx}\]
\[ \Rightarrow \left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 2\]
