Question

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

Answer 1

Let

$y=(ta{n}^{-1}x{)}^2$

Differentiate w,r.t. $x$.

$\frac{dy}{dx}=2.ta{n}^{-1}x\frac{1}{(1+x^2)}$

$y_1(1+x^2)=2ta{n}^{-1}x$

Again differentiating w.r.t $x$.

$y_2(1+x^2)+y_1\left(2x\right)=\frac{2}{(1+x^2)}$

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