Question

Ify=(tan^-1 x)^2, show that(x^2 +1)^2y[2] + 2x(x^2 +1)y[1]=2

Answer 1

Given, $\bf{y=(tan^{-1}x)^2}$
we have to proof : $\bf{(x^2+1)^2y_2+2x(x^2+1)y_1=2}$

y = (tan^-1 x)²
differentiate y with respect to x,
dy/dx = y₁ = d{(tan^-1x)²}/dx
y₁ = 2(tan^-1x). 1/(1 + x²)
(x² + 1)y₁ = 2(tan^-1x)
differentiate once again,
d{(x² + 1).y₁ }/dx = 2. d(tan^-1x)/dx
(x² + 1). dy₁/dx + y₁ . d(x² + 1)/dx = 2. 1/(1 + x²)
(x² + 1).y₂ + y₁. 2x = 2/(1 + x²)
(x² + 1)²y₂ + 2x(x² +1)y₁ = 2 $\textbf{\underline{hence proved}}$

[ note :- y₁ denotes 1st order derivatives of function y and y₂ denotes 2nd order derivatives of function y ]