Question

If $y={\left({\cot }^{-1}x\right)}^2$, then show that ${\left(1+x^2\right)}^2\frac{d^{2}y}{d{x}^2}+2x\left(1+x^2\right)\frac{dy}{dx}=2.$

Answer 1

$y=({\cot }^{-1}x{)}^2$

$y_1=\frac{dy}{dx}=\frac{d}{dx}({\cot }^{-1}x{)}^2$

$y_1=\frac{-2{\cot }^{-1}x}{1+x^2}{y}_1=\frac{d}{dx}({\cot }^{-1}x{)}^2$

$(1+x^2)y_1=2{\cot }^{-1}x{y}_2=\frac{d^2({\cot }^{-1}x{)}^2}{d{x}^2}$

squaring both the sides

$(1+x^2{)}^2{y}_1^2=4({\cot }^{-1}x{)}^2$

differentiating w.r.t $x$

$2(1+x^2)2x=y_1^2+(1+x^2{)}^{2}2y,\forall y_2=4\times \frac{d}{dx}({\cot }^{-1}x{)}^2$

substituting $\left(i\right)$

$2(1+x^2)2x.y_1^2+(1+x^2{)}^{2}2y_1{y}_2=4.y_1$

$2y_1\left[2x\right(1+x^2)y_1+(1+x^2\left)y^2\right]=4y_1$

$\therefore (1+x^2{)}^2{y}_2+2x(1+x{)}^2{y}_1=2$

$(1+x^2{)}^2.\frac{d^{2}y}{d{x}^2}+2x(1+x^2)\frac{dy}{dx}=2$

Hence proved.