If y = 3 cos ( log x ) + 4 sin ( log x ) , prove that x 2 y 2 + x y 1 + y = 0

Given, y = 3 cos ( log x ) + 4 sin ( log x ) Differentiating with respect to x , we get y 1 = - 3 sin ( log x ) 1 x + 4 cos ( log x ) 1 x ⇒ y 1 = - 3 sin ( log x ) + 4 cos ( log x ) x Also, x y 1 = - 3 sin ( log x ) + 4 cos ( log x ) Again differentiating with respect to x , we get x y 2 + y 1 = - 3 cos ( log x ) × 1 x - 4 sin ( log x ) × 1 x ⇒ x 2 y 2 + x y 1 = - ( 3 cos ( log x ) + 4 sin ( log x ) ) ⇒ x 2 y 2 + x y 1 = - y ⇒ x 2 y 2 + x y 1 + y = 0 Hence proved.

If x = sin 1 a log y show that 1 - x 2 y 2 - x y 1 - a 2 y = 0 ?

Given, x = sin 1 a log y ⇒ 1 a log y = sin - 1 x ⇒ y = e a sin - 1 x Differentiating w . r . t . x , we get y 1 = e a sin - 1 x × a 1 - x 2 ⇒ y 1 = a y 1 - x 2 Differentiating again w . r . t . x , we get y 2 = a y 1 1 - x 2 + x a y 1 - x 2 1 - x 2 ⇒ y 2 = a y 1 1 - x 2 + x a y 1 - x 2 1 - x 2 ⇒ y 2 = a y 1 1 - x 2 + x a y 1 - x 2 1 - x 2 ⇒ y 2 = a 2 y 1 - x 2 + x y 1 1 - x 2 ⇒ 1 - x 2 y 2 - x y 1 - a 2 y = 0

If y = tan - 1 x , show that 1 + x 2 d 2 y d x 2 + 2 x d y d x = 0

Given, y = tan - 1 x Differentiating on both sides with respect to x , we get d y d x = 1 1 + x 2 ⇒ 1 + x 2 d y d x = 1 Again differentiating on both sides with respect to x , we get 1 + x 2 d 2 y d x 2 + 2 x d y d x = 0 (Using product rule) Hence proved.

If y = log x + x 2 + 1 2 show that 1 + x 2 d 2 y d x 2 + x d y d x = 2

Here, y = log x + x 2 + 1 2 Differentiating w.r.t x , we get d y d x = 2 log x + x 2 + 1 x + x 2 + 1 × d d x x + x 2 + 1 ⇒ d y d x = 2 log x + x 2 + 1 x + x 2 + 1 × 1 + 2 x 2 x 2 + 1 ⇒ d y d x = 2 log x + x 2 + 1 x + x 2 + 1 × x 2 + 1 + x x 2 + 1 ⇒ d y d x = 2 log x + x 2 + 1 x 2 + 1 Differentiating again w.r.t x , we get d 2 y d x 2 = 2 x 2 + 1 × 1 + 2 x 2 x 2 + 1 x + x 2 + 1 - 2 log x + x 2 + 1 × 2 x 2 x 2 + 1 x 2 + 1 2 ⇒ d 2 y d x 2 = 2 x 2 + 1 × x 2 + 1 + x x 2 + 1 x + x 2 + 1 - 2 log x + x 2 + 1 × x x 2 + 1 x 2 + 1 ⇒ d 2 y d x 2 = 2 x 2 + 1 × x 2 + 1 + x x 2 + 1 × x + x 2 + 1 - 2 log x + x 2 + 1 × x x 2 + 1 x 2 + 1 ⇒ d 2 y d x 2 = 2 - 2 x log x + x 2 + 1 x 2 + 1 x 2 + 1 ⇒ d 2 y d x 2 = 2 - x d y d x x 2 + 1 ∵ d y d x = 2 log x + x 2 + 1 x 2 + 1 ⇒ x 2 + 1 d 2 y d x 2 = 2 - x d y d x ⇒ x 2 + 1 d 2 y d x 2 + x d y d x = 2

If y = cot x show that d 2 y d x 2 + 2 y d y d x = 0

The given function is, y = cot x … … … 1 Differentiating both sides of equation 1 w.r.t. x , we get d d x y = d d y cot x d y d x = - cosec 2 x … … … 2 ∵ d d x cot x = - cosec 2 x Again differentiating both sides of equation 2 w.r.t. x , we get d d x d y d x = d d x - cosec 2 x d 2 y d x 2 = - 2 cosec x d d x cosec x [ Using chain rule ] d 2 y d x 2 = - 2 cosec x - cosec x · cot x ∵ d d x cosec x = - cosec x cot x d 2 y d x 2 = 2 cot x · cosec 2 x … … … 3 Using equation 1 and 2 , such that d 2 y d x 2 = - 2 y · d y d x [ ∵ y = cot x and d y d x = - cosec 2 x ] d 2 y d x 2 + 2 y d y d x = 0 Hence proved.