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प्रश्न
\[\text { If } y = \left( x + \sqrt{1 + x^2} \right)^n , \text { then show that }\]
\[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = n^2 y .\]
उत्तर
\[y = \left( x + \sqrt{1 + x^2} \right)^n \]
Differentiating both sides w.r.t. x, we get
\[\frac{dy}{dx} = n \left( x + \sqrt{1 + x^2} \right)^{n - 1} \times \frac{d}{dx}\left( x + \sqrt{1 + x^2} \right)\]
\[\Rightarrow \frac{dy}{dx} = n \left( x + \sqrt{1 + x^2} \right)^{n - 1} \times \left( 1 + \frac{2x}{2\sqrt{1 + x^2}} \right)\]
\[ \Rightarrow \frac{dy}{dx} = n \left( x + \sqrt{1 + x^2} \right)^{n - 1} \times \left( \frac{x + \sqrt{1 + x^2}}{\sqrt{1 + x^2}} \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{n \left( x + \sqrt{1 + x^2} \right)^n}{\sqrt{1 + x^2}}\]
\[\Rightarrow \frac{dy}{dx} = \frac{ny}{\sqrt{1 + x^2}}\]
\[ \Rightarrow \sqrt{1 + x^2}\frac{dy}{dx} = ny\]
Squaring both sides, we get
\[\left( 1 + x^2 \right) \left( \frac{dy}{dx} \right)^2 = n^2 y^2\]
Again differentiating both sides w.r.t. x, we get
\[\left( 1 + x^2 \right) \times \left( 2\frac{dy}{dx} \times \frac{d^2 y}{d x^2} \right) + \left( \frac{dy}{dx} \right)^2 \times 2x = 2 n^2 y\frac{dy}{dx}\]
\[ \Rightarrow 2\frac{dy}{dx}\left[ \left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} \right] = 2 n^2 y\frac{dy}{dx}\]
\[ \Rightarrow \left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = n^2 y\]
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