Question

If $y=({\tan }^{-1}x{)}^2$, then prove that $(1+x^2{)}^2{y}_2+2x(1+x^2)y_1=2.$

Answer 1

We have,
$y=(ta{n}^{-1}x{)}^2$
$\Rightarrow \frac{dy}{dx}=2\left(ta{n}^{-1}x{)}^{2-1}\frac{d}{dx}\right(ta{n}^{-1}x)$
$\Rightarrow \frac{dy}{dx}=\frac{2}{1+x^2}ta{n}^{-1}x$
$\Rightarrow (1+x^2)\frac{dy}{dx}=2ta{n}^{-1}x$
$\Rightarrow (1+x^2{)}^2{\left(\frac{dy}{dx}\right)}^2=4(ta{n}^{-1}x{)}^2$ [squaring both side]
$\Rightarrow (1+x^2{)}^2{\left(\frac{dy}{dx}\right)}^2=4y$
$\text{Differentiating both sides with respect to x, we get}$
$2(1+x^2)\times 2x{\left(\frac{dy}{dx}\right)}^2+2(1+x^2{)}^2\frac{dy}{dx}\frac{d^{2}y}{d{x}^2}=4\frac{dy}{dx}$
$\Rightarrow 2x(1+x^2)\frac{dy}{dx}+(1+x^2{)}^2\frac{d^{2}y}{d{x}^2}=2\Rightarrow (1+x^2{)}^2{y}_2+2x(1+x^2)y_1=2$