if y=(tan^{-1}x)^{2}…(tan inverse x whole squared) , p.t. ((x^{2}+1)^{2} )y2 + 2x(x^{2}+1)y1=2

Answer : Given : y=(tan^{-1}x)^{2}

To Prove : ((x^{2} +1)^{2} )y2 + 2x(x^{2} +1)y1=2

now,

y=(tan^{-1}x)^{2}

diff both sides wrt to x

=> y1 = (2 tan^{-1}x ) / ( 1 + x^{2} ) { d( tan ^{-1} x) / dx = 1/ (1 + x^{2} ) }

=>( 1 + x2) y1 = 2 tan^{-1}x

diff again both sides wrt x

=> ( 1 + x^{2} ) y2 + 2x y1= 2 /( 1 + x^{2} )

=> ( 1 + x^{2} )^{2} y2 + 2x ( 1+x^{2} )y1= 2

hence proved

here y1 = dy / dx ( single differentiation) and

y2 = d^{2}y / dx^{2} ( double differentiation)