Question

If $y=(ta{n}^{-1}x{)}^2$ then show ${\left(x^2+1\right)}^2{y}_2+2x\left(x^2+1\right)y_1=2$

Answer 1

We have

$\begin{array}{c}y={\left({\tan }^{-1}x\right)}^2\\ Differentiatingw.r.t.x\\ \frac{dy}{dx}=\frac{d\left({\left({\tan }^{-1}x\right)}^2\right)}{dx}\\ \frac{dy}{dx}=2{\tan }^{-1}x.\frac{d\left({\tan }^{-1}x\right)}{dx}\\ \frac{dy}{dx}=2{\tan }^{-1}x.\frac{1}{1+x^2}\\ Hence,y_1=\frac{dy}{dx}=\frac{2{\tan }^{-1}x}{1+x^2}\\ Againdifferentiatingw.r.t.x\\ \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{2{\tan }^{-1}x}{1+x^2}\right)\\ \frac{d^{2}y}{d{x}^2}=-\frac{2d}{dx}\left(\frac{2{\tan }^{-1}x}{1+x^2}\right)\\ U\sin gquotientrule\\ As,{\left(\frac{u}{v}\right)}^{{}^{\prime }}=\frac{u^{{}^{\prime }}v-v^{{}^{\prime }}u}{v^2}\\ Whereu={\tan }^{-1}x\&+x^2\\ \frac{d^{2}y}{d{x}^2}=2\left[\frac{\frac{d\left(2{\tan }^{-1}x\right)}{dx}.\left(1+x^2\right)-\frac{d\left(1+x^2\right)}{dx}.{\tan }^{-1}x}{{\left(1+x^2\right)}^2}\right]\\ \frac{d^{2}y}{d{x}^2}=2\left[\frac{\frac{1}{1+x^2}.\left(1+x^2\right)-\left(\frac{d\left(1\right)}{dx}+\frac{d\left(x^2\right)}{dx}\right).{\tan }^{-1}x}{{\left(1+x^2\right)}^2}\right]\\ \frac{d^{2}y}{d{x}^2}=2\left[\frac{1-\left(0+2x\right){\tan }^{-1}x}{{\left(1+x^2\right)}^2}\right]\\ \frac{d^{2}y}{d{x}^2}=2\left[\frac{1-2x.{\tan }^{-1}x}{{\left(1+x^2\right)}^2}\right]\\ Thus,y_2=2\left[\frac{1-2x.{\tan }^{-1}x}{{\left(1+x^2\right)}^2}\right]\\ Weneedtoshow\\ {\left(x^2+1\right)}^2,y_2+2x\left(x^2+1\right)y_1=2\\ SolvingLHS\\ {\left(x^2+1\right)}^2{y}_2+2x\left(x^2+1\right)y_1\\ ={\left(x^2+1\right)}^2.2\left[\frac{1-2x.{\tan }^{-1}x}{{\left(1+x^2\right)}^2}\right]+2x\left(x^2+1\right).\left(\frac{2{\tan }^{-1}x}{1+x^2}\right)\\ =2\left(1-2x{\tan }^{-1}x\right)+4x{\tan }^{-1}x\\ =2-4x{\tan }^{-1}x+4x{\tan }^{-1}x\\ =2\\ Henceproved\end{array}$