Find d 2 y d x 2 , when x = t 3 + 3 t + 1 , y = t 3 − 3 t + 1

We have, x = t 3 + 3 t + 1 , y = t 3 − 3 t + 1 Differentiating both sides w.r.t t , we get: d x d t = 3 t 2 + 3 and d y d t = 3 t 2 − 3 ∴ d y d x = d y d t d x d t = 3 t 2 − 3 3 t 2 + 3 = t 2 − 1 t 2 + 1 Again differentiating both sides w.r.t x , we get: d 2 y d x 2 = d d x t 2 − 1 t 2 + 1 = 2 t t 2 + 1 − t 2 − 1 × 2 t t 2 + 1 2 × d t d x = 2 t 3 + 2 t − 2 t 3 + 2 t t 2 + 1 2 × 1 3 t 2 + 1 = 4 t 3 t 2 + 1 3

If y = sin ⁡ m log ⁡ x + x 2 + 1 , then show that x 2 + 1 y 2 + x y 1 + m 2 y = 0 [where m is a constant.]

We have, y = sin ⁡ m log ⁡ x + x 2 + 1 . . . 1 Differentiating w.r.t x , we get: y 1 = cos ⁡ m log ⁡ x + x 2 + 1 × d d x m log ⁡ x + x 2 + 1 = cos ⁡ m log ⁡ x + x 2 + 1 × m x + x 2 + 1 × d d x x + x 2 + 1 = cos ⁡ m log ⁡ x + x 2 + 1 × m x + x 2 + 1 × 1 + x x 2 + 1 = m x 2 + 1 cos ⁡ m log ⁡ x + x 2 + 1 . . . ( 2 ) On differentiating again w.r.t x , we get: y 2 = − m x x 2 + 1 3 2 cos ⁡ m log ⁡ x + x 2 + 1 + m 1 + x 2 − sin ⁡ m log ⁡ x + x 2 + 1 × m x + x 2 + 1 × 1 + x x 2 + 1 ⇒ y 2 = − m x cos ⁡ m log ⁡ x + x 2 + 1 x 2 + 1 x 2 + 1 − m 2 x 2 + 1 sin ⁡ m log ⁡ x + x 2 + 1 ⇒ y 2 = − x y 1 x 2 + 1 − m 2 y x 2 + 1 [using 1 , 2 ] ⇒ x 2 + 1 y 2 + x y 1 + m 2 y = 0 Hence proved.

If y = e − a x cos ⁡ ( b x + c ) , then show that d 2 y d x 2 + 2 a d y d x + a 2 + b 2 y = 0

We have: y = e − a x cos ⁡ ( b x + c ) − ( 1 ) Differentiate w.r.t x , we get: d y d x = − a e − a x cos ⁡ ( b x + c ) − b e − a x sin ⁡ ( b x + c ) ⇒ d y d x = − a y − b e − a x sin ⁡ ( b x + c ) - 2 using 1 Differentiate w.r.t x , we get: d 2 y d x 2 = − a d y d x + a b e - a x sin b x + c - b 2 e - a x cos b x + c ⇒ d 2 y d x 2 = − a d y d x + a − a y − d y d x − b 2 y using 1 and 2 ⇒ d 2 y d x 2 = − 2 a d y d x − y a 2 + b 2 ⇒ d 2 y d x 2 + 2 a d y d x + a 2 + b 2 y = 0 . Hence proved

If x = A e − k t 2 cos ⁡ ( p t + ε ) and A , k , p , ε are constants, show that d 2 x d t 2 + k d x d t + n 2 x = 0 where n 2 = p 2 + 1 4 k 2

We have: x = A e − k t 2 cos ⁡ ( p t + ε ) . . . 1 Differentiate w.r.t t , we get: d x d t = − A k 2 e − k t 2 cos ⁡ ( p t + ε ) − p A e − k t 2 sin ⁡ ( p t + ε ) ⇒ d x d t = − k 2 x − p A e − k t 2 sin ⁡ ( p t + ϵ ) . . . 2 using 1 Differentiate w.r.t t , we get: d 2 x d t 2 = − k 2 d x d t + p k A 2 e − k t 2 sin ⁡ ( p t + ε ) − p 2 A e − k t 2 cos ⁡ ( p t + ε ) ⇒ d 2 x d t 2 = − k 2 d x d t + k 2 − k x 2 − d x d t − p 2 x using 1 and 2 ⇒ d 2 x d t 2 = − k d x d t − k 2 x 4 − p 2 x ⇒ d 2 x d t 2 + k d x d t + n 2 x = 0 where n 2 = p 2 + k 2 4 . Hence proved

If y = tan − 1 ⁡ 1 + x 2 − 1 x + cos − 1 ⁡ 1 − x 2 1 + x 2 , show that 1 + x 2 y 2 + 2 x y 1 = 0

We have, y = tan − 1 ⁡ 1 + x 2 − 1 x + cos − 1 ⁡ 1 − x 2 1 + x 2 Put x = tan ⁡ θ ∴ y = tan − 1 ⁡ 1 + tan 2 ⁡ θ − 1 tan ⁡ θ + cos − 1 ⁡ 1 − tan 2 ⁡ θ 1 + tan 2 ⁡ θ ⇒ y = tan − 1 ⁡ sec ⁡ θ − 1 tan ⁡ θ + cos − 1 ⁡ ( cos ⁡ 2 θ ) ∵ cos ⁡ 2 A = 1 − tan 2 ⁡ A 1 + tan 2 ⁡ A ⇒ y = tan − 1 ⁡ 1 cos ⁡ θ − 1 sin ⁡ θ cos ⁡ θ + 2 θ ∵ cos − 1 ⁡ ( cos ⁡ A ) = A ⇒ y = tan − 1 ⁡ 1 − cos ⁡ θ sin ⁡ θ + 2 θ ⇒ y = tan − 1 ⁡ 2 sin 2 ⁡ θ 2 2 sin ⁡ θ 2 cos ⁡ θ 2 + 2 θ ∵ cos ⁡ 2 A = 1 − 2 sin 2 ⁡ A , sin ⁡ 2 A = 2 sin ⁡ A cos ⁡ A ⇒ y = tan − 1 ⁡ tan ⁡ θ 2 + 2 θ ⇒ y = θ 2 + 2 θ ∵ tan − 1 ⁡ ( tan ⁡ A ) = A ⇒ y = tan − 1 ⁡ x 2 + 2 tan − 1 ⁡ x [ ∵ x = tan ⁡ θ ] Differentiate w.r.t x , we get: y 1 = 1 2 1 + x 2 + 2 1 + x 2 = 5 2 1 + x 2 Differentiate again w.r.t x , we get: y 2 = 5 2 × − 2 x x 2 + 1 2 ⇒ y 2 = − 5 x x 2 + 1 2 ⇒ y 2 = 5 x 2 + 1 × ( − x ) x 2 + 1 ⇒ y 2 = − 2 y 1 x x 2 + 1 ∵ y 1 = 5 2 1 + x 2 ⇒ x 2 + 1 y 2 + 2 x y 1 = 0 Hence proved.

If x + y = e x − y , prove that y 2 = 4 e x − y e x − y + 1 3 .

We have: x + y = e x − y ⇒ log ⁡ ( x + y ) = x − y Differentiate w.r.t x , we get: 1 x + y 1 + d y d x = 1 − d y d x ⇒ 1 x + y + 1 x + y d y d x = 1 − d y d x ⇒ d y d x 1 x + y + 1 = 1 − 1 x + y ⇒ d y d x ( 1 + x + y ) = ( x + y − 1 ) ⇒ d y d x = x + y − 1 1 + x + y − ( 1 ) Differentiate w.r.t x , we get: d 2 y d x 2 = 1 + d y d x ( 1 + x + y ) − ( x + y − 1 ) 1 + d y d x ( 1 + x + y ) 2 ⇒ d 2 y d x 2 = 2 1 + d y d x ( 1 + x + y ) 2 ⇒ d 2 y d x 2 = 2 1 + x + y − 1 x + y + 1 ( 1 + x + y ) 2 using 1 ⇒ d 2 y d x 2 = 4 ( x + y ) ( 1 + x + y ) 3 ⇒ d 2 y d x 2 = 4 e x − y 1 + e x − y 3 or y 2 = 4 e x − y e x − y + 1 3 . Hence proved

If x = e t and d 2 y d t 2 + p 2 y = 0 , show that x 2 d 2 y d x 2 + x d y d x + p 2 y = 0 .

We have, x = e t Differentiate w.r.t t , we get: d x d t = e t and d 2 y d t 2 + p 2 y = 0 ⇒ d 2 y d t 2 = − p 2 y Integrating both sides, we get: ∫ d 2 y d t 2 d t = ∫ − p 2 y d t ⇒ d y d t = − p 2 y t ∴ d y d x = d y d t d x d t = − p 2 y t e t . . . 1 Differentiate w.r.t. x , we get: d 2 y d x 2 = − p 2 y e t + p 2 y t e t e t 2 × d t d x ⇒ d 2 y d x 2 = − p 2 y x + p 2 y t x x 2 × 1 x ∵ d x d t = e t , x = e t ⇒ d 2 y d x 2 = − p 2 y + p 2 y t x 2 ⇒ x 2 d 2 y d x 2 − p 2 y t + p 2 y = 0 ⇒ x 2 d 2 y d x 2 + x d y d x + p 2 y = 0 using 1 Hence proved

E M B I B E

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Calculus - A. Continuity and Differentiability>Chapter 9 - Second Order Differentiation>EXERCISE

Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Author:Parthasarathi Mukhopadhyay & Manabendra Nath Mukherjee

Parthasarathi Mukhopadhyay Mathematics Solutions for Exercise - Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Attempt the practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. RUDIMENTS of MATHEMATICS For Class 12 of +2 Level solutions are prepared by Experienced Embibe Experts.

Questions from Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE with Hints & Solutions

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 9 - Second Order Differentiation>EXERCISE>Q 1

Find $x$ , when $x = t^{3} + 3 t + 1, y = t^{3} - 3 t + 1$

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 9 - Second Order Differentiation>EXERCISE>Q 3

Eliminate the constants $C_{1} and C_{2}$ from the equation $y = C_{1} e^{2 x} + C_{2} x e^{2 x} + e^{x}$ .

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 9 - Second Order Differentiation>EXERCISE>Q 12

If $y = \sin [m \log (x + \sqrt{x^{2} + 1})]$ , then show that $(x^{2} + 1) y_{2} + x y_{1} + m^{2} y = 0$ [where $m$ is a constant.]

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 9 - Second Order Differentiation>EXERCISE>Q 15

If $y = e^{- a x} \cos (b x + c)$ , then show that $\frac{d^{2} y}{d x^{2}} + 2 a \frac{d y}{d x} + (a^{2} + b^{2}) y = 0$

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 9 - Second Order Differentiation>EXERCISE>Q 16

If $x = A e^{- \frac{k t}{2}} \cos (p t + ε)$ and $A, k, p, ε$ are constants, show that $\frac{d^{2} x}{d t^{2}} + k \frac{d x}{d t} + n^{2} x = 0 [where n^{2} = p^{2} + \frac{1}{4} k^{2}]$