If y=( ^1 x)^2,show that (x^2+1)^2 y_2+ 2x(x^2+1)_y=2 - Vidyadip

Question

$\text{If y}=(\tan ^1 \text{x})^2,\text{show that }(\text{x}^2+1)^2 \ \text{y}_2+ 2\text{x}(\text{x}^2+1)_\text{y}=2$

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Answer

$\text{y}=(\tan^{-1}\text{x})^2\ \dots(1)$ $\therefore\ \frac{\text{dy}}{\text{dx}}=2 \tan ^{-1}\text{x}.\frac{1}{1+\text{x}^2}$ $\Rightarrow \ (1+ \text{x}^2) \frac{\text{dy}}{\text{dx}} =2 \tan^{-1} \text{x}$ $\Rightarrow\ (1+\text{x}^2)^2\bigg(\frac{\text{dy}}{\text{dx}}\bigg)^2=4 (\tan ^{-1}\text{x})^2$ $\Rightarrow\ (1+\text{x}^2)^2 \bigg(\frac{\text{dy}}{\text{dx}}\bigg)^2= 4\text{y}\ \ [\because\text{of } (1)]$ Differenitiating both sides w.r.t.x, we get,$(1 +\text{x}^2)^2.2 \frac{\text{dy}}{\text{dx}}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\bigg(\frac{\text{dy}}{\text{dx}}\bigg)^2.2(1+\text{x}^2).2 \text{x}=4 \frac{\text{dy}}{\text{dx}}$
Divide both sides by $2 \frac{\text{dy}}{\text{dx}},$ we get, $(1+\text{x}^2)^2.\frac{\text{d}^2\text{y}}{\text{dx}^2}+2 \text{x}(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}=2$ $\text{Or}\ \ (\text{x}{^2}+1)\text{y}_2+2 \text{x}(\text{x}^2+1)\text{y}_1=2$

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