If y = a x + x 2 - 1 n + b x - x 2 - 1 - n , then prove that x 2 - 1 d 2 y d x 2 + x d y d x - n 2 y = 0 .

Given, y = a x + x 2 - 1 n + b x - x 2 - 1 - n Differentiating both the sides of the equation with respect to x we get, d y d x = a n x + x 2 - 1 n - 1 1 + 1 × 2 x 2 x 2 - 1 + b - n x - x 2 - 1 - n - 1 1 - 2 x 2 x 2 - 1 d y d x = a n x + x 2 - 1 n x 2 - 1 + b n x - x 2 - 1 - n x 2 - 1 x 2 - 1 d y d x = a n x + x 2 - 1 n + b n x - x 2 - 1 - n x 2 - 1 d y d x = n y Squaring both the sides of the equation we get, x 2 - 1 d y d x 2 = n 2 y 2 Differentiating both the sides of the equation we get, 2 x 2 - 1 d y d x × d 2 y d x 2 + 2 x d y d x 2 = n 2 × 2 y d y d x Taking out the common factor 2 d y d x , we get: x 2 - 1 d 2 y d x 2 + x d y d x - n 2 y = 0

If x = sin θ , y = cos ρ θ , then prove that ( 1 - x 2 ) y 2 - x y 1 + ρ 2 y = 0 , where y 2 = d 2 y d x 2 and y 1 = d y d x .

Given, x = sin θ , y = cos ρ θ d x d θ = cos θ d y d θ = - ρ sin ρ θ = - ρ 1 - cos 2 ρ θ d y d x = d y d θ d x d θ = - ρ sin ρ θ cos θ d x d θ = - ρ sin ρ θ cos θ d x d θ = - ρ 1 - y 2 1 - x 2 Squaring on both the sides of the equation, we get: d y d x 2 = ρ 2 1 - y 2 1 - x 2 1 - x 2 d y d x 2 = ρ 2 1 - y 2 Differentiating on both the sides of the equation again, we get: 1 - x 2 2 d y d x d 2 y d x 2 - 2 x d y d x 2 = ρ 2 - 2 y d y d x Taking the common factor 2 d y d x from the equation, we get: 1 - x 2 d 2 y d x 2 - x d y d x + ρ 2 y = 0

If x 2 + y 2 + z 2 - 2 x y z = 1 , then show that d x 1 - x 2 + d y 1 - y 2 + d z 1 - z 2 = 0 .

Given, x 2 + y 2 + z 2 - 2 x y z = 1 Differentiating both the sides of the equation we get, 2 x d x + 2 y d y + 2 z d z - 2 x y d z + y z d x + z x d y = 0 x - y z d x + y - z x d y + z - x y d z = 0 d x y - z x z - x y + d y z - x y z - x y + d z x - y z y - z x = 0 . . . i (Dividing by x - y z y - z x z - x y ) Now, y - z x 2 = y 2 + z 2 x 2 - 2 x y z = z 2 x 2 + y 2 - 2 x y z = z 2 x 2 + 1 - x 2 - z 2 = 1 - x 2 1 - z 2 Similarly, z - x y 2 = 1 - x 2 1 - y 2 x - z y 2 = 1 - z 2 1 - y 2 by substitutinf these values in i , we will get d x 1 - x 2 + d y 1 - y 2 + d z 1 - z 2 = 0 .

If y is a twice differentiable function of x , then transform the expression 1 - x 2 d 2 y d x 2 - x d y d x + y by means of the transformation x = sin ⁡ t in terms of the independent variable t .

Given, 1 - x 2 d 2 y d x 2 - x d y d x + y ⇒ 1 - x 2 d d x d y d x - x d y d x + y ⇒ 1 - x 2 d d x d y d t d x d t - x d y d t d x d t + y . . . 1 Now, x = sin t d x d t = cos t Substituting the values in 1 , we get ⇒ 1 - sin 2 t d d x d y d t cos t - sin t d y d t cos t + y ⇒ cos 2 t × d d t d y d t × sec t d x d t - sin t cos t d y d t + y ⇒ cos 2 t × sec t d 2 y d t + d y d t × sec t tan t cos t - tan t d y d t + y ⇒ cos t sec t d 2 y d t 2 + tan t d y d t - tan d y d t + y ⇒ d 2 y d t 2 + y

If f x is a real function such that f 0 = 0 and f ' x = 1 1 - x 2 for - 1 < x < 1 , then show that f x + f a = f x 1 - a 2 + a 1 - x 2 without using integration.

We have, f ' x = 1 1 - x 2 ∴ f ' x 1 - a 2 + a 1 - x 2 = 1 1 - x 1 - a 2 + a 1 - x 2 2 = 1 1 - a 2 1 - x 2 - a x 2 = 1 1 - a 2 1 - x 2 - a x ⇒ 1 - a 2 1 - x 2 - a x 1 - x 2 f ' x 1 - a 2 + a 1 - x 2 = 1 1 - x 2 ⇒ d d x f x 1 - a 2 + a 1 - x 2 = d d x f x ∵ f ' x = 1 1 - x 2 ∴ f x 1 - a 2 + a 1 - x 2 = f x + C Putting x = 0 we get, f a = f 0 + C ⇒ C = f a ∵ f 0 = 0 ∴ f ( x 1 - a 2 + a 1 - x 2 ) = f x + f a

Prove that the expression y '' ' y ' - 3 2 y ' ' y 2 + 1 2 y ' y 2 remains unchanged, if y is replaced by 1 y 2 .

Let y 1 = 1 y 2 , then y 1 ' = - 2 y 3 y ' ⇒ y ' 1 y 1 = - 2 y ' y … i Differentiating both the sides w.r.t x , we get ⇒ y 1 '' y 1 - y 1 ' 2 y 1 2 = - 2 y '' y - y ' 2 y 2 … i i Differentiating w.r.t. x , we get ⇒ y 1 ' ' ' y 1 - 3 y 1 ' y 1 '' y 1 2 + 2 y 1 ' 3 y 1 3 = - 2 y ' ' ' y - 3 y ' y '' y 2 + 2 y ' 3 y 3 … i i i Dividing equations iii & ii by i , we get ⇒ y 1 ' ' y 1 ' - y 1 ' y 1 = y '' y ' - y ' y … i v ⇒ y 1 ' ' ' y 1 ' - 3 y 1 ' ' y 1 + 2 y 1 ' 2 y 1 2 = y '' ' y ' - 3 y ' ' y + 2 y ' 2 y 2 … v Subtracting 3 2 times the squares of equation i v from equation v , we get ⇒ y 1 ' ' ' y 1 ' - 3 2 y 1 '' y 1 2 + 1 2 y 1 ' y 1 2 = y '' ' y ' - 3 2 y ' ' y 2 + 1 2 y ' y 2 This shows that the given expression does not change, if y is replaced by 1 y 2 .

Show that the transformation x = cos ⁡ θ reduces the differential equation 1 - x 2 d 2 y d x 2 - x d y d x + y = 0 into d 2 y d θ 2 + y = 0 .

Given x = cos θ d x d θ = - sin θ d y d x = d y d θ × d θ d x d y d x = - d y d θ × 1 sin θ d y d x = - cosec θ d y d θ Differentiating both sides, we get d 2 y d x 2 = d d θ - cosec θ d y d θ . d θ d x = cosec θ cot θ d y d θ - cosec θ d 2 y d θ 2 × - cosec θ Now, 1 - x 2 d 2 y d x 2 - x d y d x + y = 0 1 - cos 2 θ - cosec 2 θ cot θ d y d θ + cos e c 2 θ d 2 y d θ 2 - cos θ - cosec θ d y d θ + y = 0 ⇒ - cot θ d y d θ + d 2 y d θ 2 + cot θ d y d θ + y = 0 ⇒ d 2 y d θ 2 + y = 0

Show that the transformation z = logtan x 2 reduces the differential equation d 2 y d x 2 + cosec x c o t x d y d x + 4 y cosec 2 x = 0 into d 2 y d z 2 + 4 y = 0 .

Given z = logtan x 2 d z d x = 1 tan x 2 · sec 2 x 2 · 1 2 = 1 2 sin x 2 cos x 2 = 1 sin x = cosec x d y d x = d y d z · d z d x = cosec x d y d z Now, d 2 y d x 2 = d d x d y d x = d d x cosec x d y d z = - cosec x cot x d y d z + cosec x d 2 y d z 2 · d z d x = - cosec x cot x d y d z + cosec 2 x d 2 y d z 2 Given, d 2 y d x 2 + cot x cosec x d y d x + 4 y cosec 2 x = 0 - cosec x cot x d y d z + cosec 2 x d 2 y d z 2 + cot x cosec x d y d x + 4 y cosec 2 x = 0 cosec 2 x d 2 y d z 2 + 4 y = 0 ⇒ d 2 y d z 2 + 4 y = 0

E M B I B E

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Proficiency in 'Differentiation' Exercise 2

Amit M Agarwal Solutions for Chapter: Differentiation, Exercise 8: Proficiency in 'Differentiation' Exercise 2

Author:Amit M Agarwal

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Differentiation, Exercise 8: Proficiency in 'Differentiation' Exercise 2

Attempt the practice questions on Chapter 2: Differentiation, Exercise 8: Proficiency in 'Differentiation' Exercise 2 with hints and solutions to strengthen your understanding. Skills in Mathematics Differential Calculus for JEE Main & Advanced solutions are prepared by Experienced Embibe Experts.

Questions from Amit M Agarwal Solutions for Chapter: Differentiation, Exercise 8: Proficiency in 'Differentiation' Exercise 2 with Hints & Solutions

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Proficiency in 'Differentiation' Exercise 2>Q 3

If $y = a {(x + \sqrt{x^{2} - 1})}^{n} + b {(x - \sqrt{x^{2} - 1})}^{- n}$ , then prove that $(x^{2} - 1) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} - n^{2} y = 0$ .

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Proficiency in 'Differentiation' Exercise 2>Q 4

If $x = \sin θ, y = \cos ρ θ$ , then prove that $(x^{2} - 1) {(\frac{d y}{d x})}^{2} = n^{2} y^{2}$ where $y_{2} = \frac{d^{2} y}{d x^{2}}$ and $y_{1} = \frac{d y}{d x}$ .

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Proficiency in 'Differentiation' Exercise 2>Q 5

If $x^{2} + y^{2} + z^{2} - 2 x y z = 1,$ then show that $\frac{d x}{\sqrt{1 - x^{2}}} + \frac{d y}{\sqrt{1 - y^{2}}} + \frac{d z}{\sqrt{1 - z^{2}}} = 0$ .

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Proficiency in 'Differentiation' Exercise 2>Q 6

If $y$ is a twice differentiable function of $x$ , then transform the expression $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + y$ by means of the transformation $x = \sin t$ in terms of the independent variable $t$ .

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Proficiency in 'Differentiation' Exercise 2>Q 8

If $f (x)$ is a real function such that $f (0) = 0$ and $f^{'} (x) = \frac{1}{\sqrt{1 - x^{2}}}$ for $- 1 < x < 1$ , then show that $f (x) + f (a) = f (x \sqrt{1 - a^{2}} + a \sqrt{1 - x^{2}})$ without using integration.

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Proficiency in 'Differentiation' Exercise 2>Q 7

Prove that the expression $\frac{y^{'''}}{y^{'}} - \frac{3}{2} {(\frac{y^{''}}{y})}^{2} + \frac{1}{2} {(\frac{y^{'}}{y})}^{2}$ remains unchanged, if $y$ is replaced by $\frac{1}{y^{2}}$ .

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Proficiency in 'Differentiation' Exercise 2>Q 9

Show that the transformation $x = \cos θ$ reduces the differential equation $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + y = 0$ into $\frac{d^{2} y}{d θ^{2}} + y = 0$ .

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Proficiency in 'Differentiation' Exercise 2>Q 10

Show that the transformation $z = logtan \frac{x}{2}$ reduces the differential equation
$\frac{d^{2} y}{d x^{2}} + cosec x c o t x \frac{d y}{d x} + 4 y {cosec}^{2} x = 0$ into $\frac{d^{2} y}{d z^{2}} + 4 y = 0$ .

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