Rd Sharma XII Vol 1 2020 for Class 12 Science Maths Chapter 11 - Higher Order Derivatives

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Page No 11.16:

Question 1:

Find the second order derivatives of each of the following functions:

(i) x³ + tan x
(ii) sin (log x)
(iii) log (sin x)
(iv) e^x sin 5x
(v) e^6x cos 3x
(vi) x³ log x
(vii) tan⁻¹ x
(viii) x cos x
(ix) log (log x)

Answer:

(i) We have,

$y = x^{3} + \tan x Differentiating w . r . t . x, we get \frac{d y}{d x} = 3 x^{2} + \sec^{2} x Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = 6 x + 2 \sec^{2} x \tan x$

(ii) We have,

$y = \sin (\log x) Differentiating w . r . t . x, we get \frac{d y}{d x} = \cos (\log x) \times \frac{1}{x} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = - \sin (\log x) \frac{1}{x} \times \frac{1}{x} + \cos (\log x) \times \frac{- 1}{x^{2}} = \frac{- [\sin (\log x) + \cos (\log x)]}{x^{2}}$

(iii) We have,

$y = \log (\sin x) Differentiating w . r . t . x, we get \frac{d y}{d x} = \frac{1}{\sin x} \times \cos x = \cot x Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = - {cosec}^{2} x$

(iv) We have,

$y = e^{x} \sin (5 x) Differentiating w . r . t . x, we get \frac{d y}{d x} = e^{x} \sin 5 x + e^{x} \cos 5 x \times 5 Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = e^{x} \sin 5 x + e^{x} \cos 5 x \times 5 + 5 e^{x} (- \sin 5 x \times 5) + 5 e^{x} \cos 5 x = - 24 e^{x} \sin 5 x + 10 e^{x} \cos 5 x = 2 e^{x} (5 \cos 5 x - 12 \sin 5 x)$

(v) We have,
$y = e^{6 x} \cos 3 x Differentiating w . r . t . x, we get \frac{d y}{d x} = e^{6 x} \times 6 \times \cos 3 x + e^{6 x} (- \sin 3 x \times 3) = 6 e^{6 x} \cos 3 x - 3 e^{6 x} \sin 3 x Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = 6 e^{6 x} \cos 3 x \times 6 - 6 e^{6 x} \sin 3 x \times 3 - 3 \times 6 e^{6 x} \sin 3 x - 3 e^{6 x} \times 3 \cos 3 x = 27 e^{6 x} \cos 3 x - 36 e^{6 x} \sin 3 x = 9 e^{6 x} (3 \cos 3 x - 4 \sin 3 x)$

(vi) We have,
$y = x^{3} \log x Differentiating w . r . t . x, we get \frac{d y}{d x} = 3 x^{2} \log x + x^{3} \times \frac{1}{x} = 3 x^{2} \log x + x^{2} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = 6 x \log x + 3 x^{2} \times \frac{1}{x} + 2 x = 6 x \log x + 5 x$

(vii) We have,
$y = \tan^{- 1} x Differentiating w . r . t . x, we get \frac{d y}{d x} = \frac{1}{1 + x^{2}} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = - (2 x) \times \frac{1}{{(1 + x^{2})}^{2}} = \frac{- 2 x}{{(1 + x^{2})}^{2}}$

(viii) We have,
$y = x \cos x Differentiating w . r . t . x, we get \frac{d y}{d x} = \cos x - x \sin x Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = - \sin x - \sin x - x \cos x = - (2 \sin x + x \cos x)$

(ix) We have,

$y = \log (\log x) Differentiating w . r . t . x, we get \frac{d y}{d x} = \frac{1}{\log x} \times \frac{1}{x} = \frac{1}{x \log x} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{0 - (\log x + 1)}{{(x \log x)}^{2}} = - \frac{(1 + \log x)}{{(x \log x)}^{2}}$

Page No 11.16:

Question 2:

If y = e^−x cos x, show that $\frac{d^{2} y}{d x^{2}} = 2 e^{- x} \sin x$ .

Answer:

Here,

$y = e^{- x} \cos x Differentiating w . r . t . x, we get \frac{d y}{d x} = - e^{- x} \sin x - e^{- x} \cos x = - (e^{- x} \sin x + e^{- x} \cos x) Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = - (e^{- x} \cos x - e^{- x} \sin x - e^{- x} \sin x - e^{- x} \cos x) = 2 e^{- x} \sin x$

Hence proved.

Page No 11.16:

Question 3:

If y = x + tan x, show that $\cos^{2} x \frac{d^{2} y}{d x^{2}} - 2 y + 2 x = 0$ .

Answer:

Here,

$y = x + \tan x Differentiating w . r . t . x, we get \frac{d y}{d x} = 1 + \sec^{2} x Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = 2 \sec^{2} x \tan x Dividing both sides by \sec^{2} x, we get \cos^{2} x \frac{d^{2} y}{d x^{2}} = 2 \tan x \Rightarrow \cos^{2} x \frac{d^{2} y}{d x^{2}} = 2 (y - x) [∵ y = x + \tan x \Rightarrow \tan x = y - x] \Rightarrow \cos^{2} x \frac{d^{2} y}{d x^{2}} - 2 y + 2 x = 0$

Hence proved.

Page No 11.16:

Question 4:

If y = x³ log x, prove that $\frac{d^{4} y}{d x^{4}} = \frac{6}{x}$ .

Answer:

Here,
$y = x^{3} \log x Differentiating w . r . t . x, we get \frac{d y}{d x} = 3 x^{2} \log x + x^{3} \times \frac{1}{x} = 3 x^{2} \log x + x^{2} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = 6 x \log x + 3 x^{2} \times \frac{1}{x} + 2 x = 6 x \log x + 5 x Differentiating again w . r . t . x, we get \frac{d^{3} y}{d x^{3}} = 6 \log x + 6 x \times \frac{1}{x} + 5 = 6 \log x + 11 Differentiating again w . r . t . x, we get \frac{d^{4} y}{d x^{4}} = \frac{6}{x}$

Hence proved.

Page No 11.16:

Question 5:

If y = log (sin x), prove that $\frac{d^{3} y}{d x^{3}} = 2 \cos x {cosec}^{3} x$ .

Answer:

Here,
$y = \log (\sin x) Differentiating w . r . t . x, we get \frac{d y}{d x} = \frac{1}{\sin x} \times \cos x = \cot x Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = - {cosec}^{2} x Differentiating again w . r . t . x, we get \frac{d^{3} y}{d x^{3}} = - 2 cosec x \times (- cosec x \cot x) = 2 \cot x {cosec}^{2} x = 2 \cos x {cosec}^{3} x$

Hence proved.

Page No 11.16:

Question 6:

If y = 2 sin x + 3 cos x, show that $\frac{d^{2} y}{d x^{2}} + y = 0$ .

Answer:

Here,
$y = 2 \sin x + 3 \cos x Differentiating w . r . t . x, we get \frac{d y}{d x} = 2 \cos x - 3 \sin x Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = - 2 sinx - 3 cosx \Rightarrow \frac{d^{2} y}{d x^{2}} = - (2 \sin x + 3 \cos x) \Rightarrow \frac{d^{2} y}{d x^{2}} = - y \Rightarrow \frac{d^{2} y}{d x^{2}} + y = 0$

Hence proved.

Page No 11.16:

Question 7:

If $y = \frac{\log x}{x}$ , show that $\frac{d^{2} y}{d x^{2}} = \frac{2 \log x - 3}{x^{3}}$ .

Answer:

Here,

$y = \frac{\log x}{x} Differentiating w . r . t . x, we get \frac{d y}{d x} = \frac{1 - \log x}{x^{2}} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{- x - 2 x (1 - \log x)}{x^{4}} = \frac{- x - 2 x + 2 x \log x}{x^{4}} = \frac{- 3 + 2 \log x}{x^{3}} = \frac{2 \log x - 3}{x^{3}}$

Hence proved.

Page No 11.16:

Question 8:

If x = a sec θ, y = b tan θ, prove that $\frac{d^{2} y}{d x^{2}} = - \frac{b^{4}}{a^{2} y^{3}}$ .

Answer:

Here,
$x = a \sec θ and y = b \tan θ Differentiating w . r . t . θ, we get \frac{d x}{d θ} = a \sec θ \tan θ and \frac{d y}{d θ} = b \sec^{2} θ ∴ \frac{d y}{d x} = \frac{d y}{d θ} \times \frac{d θ}{d x} = \frac{b \sec^{2} θ}{a \sec θ \tan θ} = \frac{b cosec θ}{a} Differentiating w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{b}{a} \times - cosec θ \cot θ \times \frac{d θ}{d x} = - \frac{b}{a} \times cosec θ \cot θ \times \frac{1}{a \sec θ \tan θ} = - \frac{b}{a^{2}} \times \cot θ \times \frac{1}{\tan^{2} θ} = - \frac{b}{a^{2}} \times \frac{1}{\tan^{3} θ} = \frac{- b^{4}}{a^{2} y^{3}} [∵ y = b \tan θ]$

Hence proved.

Page No 11.16:

Question 9:

If x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), prove that $\frac{d^{2} x}{d θ^{2}} = a (\cos θ - θ \sin θ), \frac{d^{2} y}{d θ^{2}} = a (\sin θ + θ \cos θ) and \frac{d^{2} y}{d x^{2}} = \frac{\sec^{3} θ}{a θ} .$

Answer:

We have, $x = a (\cos θ + θ \sin θ)$
$\frac{d x}{d θ} = \frac{d [a (\cos θ + θ \sin θ)]}{d θ} \Rightarrow \frac{d x}{d θ} = - a \sin θ + a \sin θ + a θ \cos θ = a θ \cos θ . . . . . (i) \frac{d^{2} x}{d θ^{2}} = \frac{d}{d θ} (\frac{d x}{d θ}) = \frac{d (a θ \cos θ)}{d θ} \Rightarrow \frac{d^{2} x}{d θ^{2}} = a \cos θ - a θ \sin θ = a (\cos θ - θ \sin θ)$

$y = a (\sin θ - θ \cos θ) \frac{d y}{d θ} = \frac{d [a (\sin θ - θ \cos θ)]}{d θ} \Rightarrow \frac{d y}{d θ} = a \cos θ - a \cos θ + a θ \sin θ = a θ \sin θ . . . . . (ii) \frac{d^{2} y}{d θ^{2}} = \frac{d}{d θ} (\frac{d y}{d θ}) = \frac{d (a θ \sin θ)}{d θ} \Rightarrow \frac{d^{2} y}{d θ^{2}} = a \sin θ + a θ \cos θ = a (\sin θ + θ \cos θ)$
From (i) and (ii), we have
$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a θ \sin θ}{a θ \cos θ} = \tan θ \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d (\tan θ)}{d x} = \sec^{2} θ \frac{d θ}{d x} \Rightarrow \frac{d^{2} y}{d x^{2}} = (\sec^{2} θ) (\frac{1}{a θ \cos θ}) \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{\sec^{3} θ}{a θ}$
Hence proved.

Page No 11.16:

Question 10:

If y = e^x cos x, prove that $\frac{d^{2} y}{d x^{2}} = 2 e^{x} \cos (x + \frac{π}{2})$ .

Answer:

Here,
$y = e^{x} \cos x Differentiating w . r . t . x, we get \frac{d y}{d x} = e^{x} \cos x - e^{x} \sin x = e^{x} (\cos x - \sin x) Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = e^{x} (\cos x - \sin x) + e^{x} (- \sin x - \cos x) = e^{x} \cos x - e^{x} \sin x - e^{x} \sin x - e^{x} \cos x = - 2 e^{x} \sin x = 2 e^{x} \cos (x + \frac{π}{2})$

Hence proved.

Page No 11.16:

Question 11:

If x = a cos θ, y = b sin θ, show that $\frac{d^{2} y}{d x^{2}} = - \frac{b^{4}}{a^{2} y^{3}}$ .

Answer:

Here,

$x = a \cos θ and y = b \sin θ Differentiating w . r . t . θ, we get \frac{d x}{d θ} = - a \sin θ and \frac{d y}{d θ} = b \cos θ ∴ \frac{d y}{d x} = \frac{b \cos θ}{- a \sin θ} = \frac{- b}{a} \cot θ Differentiating w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = - \frac{b}{a} \times (- {cosec}^{2} θ) \frac{d θ}{d x} = \frac{b}{a} \times {cosec}^{2} θ \times \frac{1}{- a \sin θ} = - \frac{b}{a^{2}} \times \frac{1}{\sin^{3} θ} = - \frac{b}{a^{2}} \times \frac{b^{3}}{y^{3}} [∵ y = b \sin θ] = \frac{- b^{4}}{a^{2} y^{3}}$

Hence proved.

Page No 11.16:

Question 12:

If x = a (1 − cos³ θ), y = a sin³ θ, prove that $\frac{d^{2} y}{d x^{2}} = \frac{32}{27 a} at θ = \frac{π}{6}$ .

Answer:

Here,

$x = a (1 - \cos^{3} θ), y = a \sin^{3} θ Differentiating w . r . t . θ, we get \frac{d x}{d θ} = 3 a \cos^{2} θ \sin θ and \frac{d y}{d θ} = 3 a \sin^{2} θ \cos θ \Rightarrow \frac{d y}{d x} = \frac{3 a \sin^{2} θ \cos θ}{3 a \cos^{2} θ \sin θ} = \tan θ Differentiating w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \sec^{2} θ \frac{d θ}{d x} = \frac{\sec^{2} θ}{3 a \cos^{2} θ \sin θ} = \frac{\sec^{4} θ}{3 a \sin θ} ∴ \frac{d^{2} y}{d x^{2}} at θ = \frac{π}{6} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{{(\sec \frac{π}{6})}^{4}}{3 a \sin \frac{π}{6}} = \frac{32}{27 a}$

Page No 11.16:

Question 13:

If x = a (θ + sin θ), y = a (1 + cos θ), prove that $\frac{d^{2} y}{d x^{2}} = - \frac{a}{y^{2}}$ .

Answer:

Here,
$x = a (θ + \sin θ) and y = a (1 + \cos θ) Differentiating w . r . t . θ, we get \frac{d x}{d θ} = a + a \cos θ and \frac{d y}{d θ} = - a \sin θ ∴ \frac{d y}{d x} = \frac{- a \sin θ}{a + a \cos θ} = \frac{- \sin θ}{1 + \cos θ} Differentiating w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = - \{\frac{(1 + \cos θ) \cos θ + \sin^{2} θ}{{(1 + cosθ)}^{2}}\} \frac{d θ}{d x} = \frac{- \cos θ - \cos^{2} θ - \sin^{2} θ}{{(1 + \cos θ)}^{2}} \times \frac{1}{a + a \cos θ} = \frac{- (1 + \cos θ)}{a {(1 + \cos θ)}^{3}} = \frac{- 1}{a {(1 + \cos θ)}^{2}} = \frac{- a}{y^{2}} [∵ y = a (1 + \cos θ)]$

Hence proved.

Page No 11.16:

Question 14:

If x = a (θ − sin θ), y = a (1 + cos θ) prove that, find $\frac{d^{2} y}{d x^{2}}$ .

Answer:

Here,

$x = a (θ - \sin θ) and y = a (1 + \cos θ) Differentiating w . r . t . θ, we get \frac{d x}{d θ} = a - a \cos θ, \frac{d y}{d θ} = - a \sin θ \Rightarrow \frac{d y}{d x} = \frac{- a \sin θ}{a - a \cos θ} = \frac{- \sin θ}{1 - \cos θ} Differentiating w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{- \cos θ + \cos^{2} θ + \sin^{2} θ}{{(1 - \cos θ)}^{2}} \times \frac{d θ}{d x} = \frac{- \cos θ + \cos^{2} θ + \sin^{2} θ}{{(1 - \cos θ)}^{2}} \times \frac{1}{a - a \cos θ} = \frac{(1 - \cos θ)}{a {(1 - cosθ)}^{3}} = \frac{1}{a {(1 - \cos θ)}^{2}}$

Page No 11.16:

Question 15:

If x = a(1 − cos θ), y = a(θ + sin θ), prove that $\frac{d^{2} y}{d x^{2}} = - \frac{1}{a} at θ = \frac{π}{2}$ .

Answer:

Here,
$x = a (1 - \cos θ) and y = a (θ + \sin θ) Differentiating w . r . t . x, we get \frac{d x}{d θ} = a \sin θ and \frac{d y}{d θ} = a + a \cos θ ∴ \frac{d y}{d x} = \frac{a + a \cos θ}{a \sin θ} = \frac{1 + \cos θ}{\sin θ} Differentiating w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \{\frac{- \sin^{2} θ - (1 + \cos θ) \cos θ}{\sin^{2} θ}\} \frac{d θ}{d x} = (- 1 - \frac{(1 + cosθ) cosθ}{\sin^{2} θ}) \frac{1}{a \sin θ} ∴ \frac{d^{2} y}{d x^{2}} at θ = \frac{π}{2} \Rightarrow {[\frac{d^{2} y}{d x^{2}}]}_{θ = \frac{π}{2}} = \frac{1}{a} (- 1 - \frac{0}{1}) = \frac{- 1}{a}$

Hence proved.

Page No 11.17:

Question 16:

If x = a (1 + cos θ), y = a(θ + sin θ), prove that $\frac{d^{2} y}{d x^{2}} = \frac{- 1}{a} at θ = \frac{π}{2}$ .

Answer:

Here,

$x = a (1 + \cos θ) and y = a (θ + \sin θ) Differentiating w . r . t . θ, we get \frac{d x}{d θ} = - a \sin θ and \frac{d y}{d θ} = a + a \cos θ ∴ \frac{d y}{d x} = \frac{a + a \cos θ}{- a \sin θ} = \frac{1 + \cos θ}{- \sin θ} Differentiating w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{d}{d θ} \{\frac{d y}{d x}\} \frac{d θ}{d x} \frac{d^{2} y}{d x^{2}} = - \{\frac{- \sin^{2} θ - \cos θ - \cos^{2} θ}{\sin^{2} θ}\} \frac{d θ}{d x} = \frac{1 + \cos θ}{\sin^{2} θ} \times \frac{- 1}{a \sin θ} = \frac{- (1 + \cos θ)}{a \sin^{3} θ} At θ = \frac{π}{2} : \frac{d^{2} y}{d x^{2}} = \frac{- (1 + \cos \frac{π}{2})}{a {(\sin \frac{π}{2})}^{3}} = \frac{- 1}{a}$

Page No 11.17:

Question 17:

If x = cos θ, y = sin³ θ, prove that $y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = 3 \sin^{2} θ (5 \cos^{2} θ - 1)$ .

Answer:

Here,
$x = \cos θ and y = \sin^{3} θ Differentiating w . r . t . θ, we get \frac{d x}{d θ} = - \sin θ and \frac{d y}{d θ} = 3 \sin^{2} θ \cos θ ∴ \frac{d y}{d x} = \frac{3 \sin^{2} θ \cos θ}{- \sin θ} = - 3 \sin θ \cos θ Differentiating w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = (- 3 \cos^{2} θ + 3 \sin^{2} θ) \frac{d θ}{d x} = \frac{(- 3 \cos^{2} θ + 3 \sin^{2} θ)}{- \sin θ} Now, LHS = y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = \sin^{3} θ \times \frac{(- 3 \cos^{2} θ + 3 \sin^{2} θ)}{- \sin θ} + {(- 3 \sin θ \cos θ)}^{2} = 3 \sin^{2} θ \cos^{2} θ - 3 \sin^{4} θ + 9 \sin^{2} θ \cos^{2} θ = 12 \sin^{2} θ \cos^{2} θ - 3 \sin^{4} θ = 3 \sin^{2} θ (4 \cos^{2} θ - \sin^{2} θ) = 3 \sin^{2} θ (5 \cos^{2} θ - 1) [∵ \cos^{2} + \sin^{2} θ = 1] = RHS$

Hence proved.

Page No 11.17:

Question 18:

If y = sin (sin x), prove that $\frac{d^{2} y}{d x^{2}} + \tan x \cdot \frac{d y}{d x} + y \cos^{2} x = 0$ .

Answer:

Here,
$y = \sin (\sin x) Differentiating w . r . t . x, we get \frac{d y}{d x} = \cos (\sin x) \cos x Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = - \sin (\sin x) \cos^{2} x - \cos (\sin x) \sin x \Rightarrow \frac{d^{2} y}{d x^{2}} = - \sin (\sin x) \cos^{2} x - \cos (\sin x) \cos x \tan x \Rightarrow \frac{d^{2} y}{d x^{2}} = - y \cos^{2} x - \tan x \frac{d y}{d x} \Rightarrow \frac{d^{2} y}{d x^{2}} + \tan x \frac{d y}{d x} + y \cos^{2} x = 0$

Hence proved.

Page No 11.17:

Question 19:

If x = sin t, y = sin pt, prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0$ .

Answer:

Here,
$x = \sin t and y = \sin p t Differentiating w . r . t . t, we get \frac{d x}{d t} = \cos t and \frac{d y}{d t} = p \cos p t \Rightarrow \frac{d y}{d x} = \frac{p \cos p t}{\cos t} Differentiating w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{- p^{2} \sin p t \cos t + p \cos p t \sin t}{\cos^{2} t} \times \frac{d t}{d x} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{- p^{2} \sin p t \cos t + p \cos p t \sin t}{\cos^{3} t} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{- p^{2} \sin p t \cos t}{\cos^{3} t} + \frac{p \cos p t \sin t}{\cos^{3} t} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{- p^{2} y}{\cos^{2} t} + \frac{x \frac{d y}{d x}}{\cos^{2} t} \Rightarrow \cos^{2} t \frac{d^{2} y}{d x^{2}} = - p^{2} y + x \frac{d y}{d x} \Rightarrow (1 - \sin^{2} t) \frac{d^{2} y}{d x^{2}} = - p^{2} y + x \frac{d y}{d x} \Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0$ .

Hence proved.

Page No 11.17:

Question 20:

If y = (sin⁻¹ x)², prove that (1 − x²) $\frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0$ .

Answer:

Here,

$y = {(\sin^{- 1} x)}^{2} Now, y_{1} = 2 \sin^{- 1} x \frac{1}{\sqrt{1 - x^{2}}} \Rightarrow y_{2} = \frac{2}{1 - x^{2}} + \frac{2 x \sin^{- 1} x}{{(1 - x^{2})}^{3 / 2}} \Rightarrow y_{2} = \frac{2}{1 - x^{2}} + \frac{2 x \sin^{- 1} x}{(1 - x^{2}) \sqrt{1 - x^{2}}} \Rightarrow y_{2} = \frac{2}{1 - x^{2}} + \frac{x y_{1}}{(1 - x^{2})} \Rightarrow y_{2} (1 - x^{2}) = 2 + x y_{1} \Rightarrow y_{2} (1 - x^{2}) - x y_{1} - 2 = 0$

Hence proved.

Page No 11.17:

Question 21:

If $y = e^{\tan^{- 1} x}$ , prove that (1 + x²)y₂ + (2x − 1)y₁ = 0.

Answer:

Here,
$y = e^{\tan^{- 1} x} Differentiating w . r . t . x, we get \frac{d y}{d x} = e^{\tan^{- 1} x} \times \frac{1}{1 + x^{2}} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = e^{\tan^{- 1} x} \frac{1}{{(1 + x^{2})}^{2}} + e^{\tan^{- 1} x} \frac{- 2 x}{{(1 + x^{2})}^{2}} \Rightarrow (1 + x^{2}) \frac{d^{2} y}{d x^{2}} = \frac{e^{\tan^{- 1} x}}{(1 + x^{2})} - \frac{2 x e^{\tan^{- 1} x}}{(1 + x^{2})} \Rightarrow (1 + x^{2}) \frac{d^{2} y}{d x^{2}} = \frac{d y}{d x} - 2 x \frac{d y}{d x} \Rightarrow (1 + x^{2}) \frac{d^{2} y}{d x^{2}} + (2 x - 1) \frac{d y}{d x} = 0$

Hence proved.

Page No 11.17:

Question 22:

If y = 3 cos (log x) + 4 sin (log x), prove that x²y₂ + xy₁ + y = 0.

Answer:

Here,
$y = 3 \cos (\log x) + 4 \sin (\log x) Differentiating w . r . t . x, we get y_{1} = - 3 \sin (\log x) \times \frac{1}{x} + 4 \cos (\log x) \times \frac{1}{x} = \frac{- 3 \sin (\log x) + 4 \cos (\log x)}{x} Differentiating again w . r . t . x, we get y_{2} = \frac{(\frac{- 3 \cos (\log x)}{x} - \frac{4 \sin (\log x)}{x}) \times x - \{- 3 \sin (\log x) + 4 \cos (\log x)\}}{x^{2}} \Rightarrow y_{2} = \frac{- 3 \cos (\log x) - 4 \sin (\log x) - \{- 3 \sin (\log x) + 4 \cos (\log x)\}}{x^{2}} \Rightarrow y_{2} = \frac{- 3 \cos (\log x) - 4 \sin (\log x) - \{- 3 \sin (\log x) + 4 \cos (\log x)\}}{x^{2}} \Rightarrow y_{2} = \frac{- 3 \cos (\log x) - 4 \sin (\log x)}{x^{2}} - \frac{\{- 3 \sin (\log x) + 4 \cos (\log x)\}}{x^{2}} \Rightarrow y_{2} = \frac{- \{3 \cos (\log x) + 4 \sin (\log x)\}}{x^{2}} - \frac{\{- 3 \sin (\log x) + 4 \cos (\log x)\}}{x^{2}} \Rightarrow y_{2} = \frac{- y}{x^{2}} - \frac{y_{1}}{x} \Rightarrow x^{2} y_{2} = - y - x y_{1} \Rightarrow x^{2} y_{2} + y + x y_{1} = 0$

Hence proved.

Page No 11.17:

Question 23:

If $y = e^{2 x} (a x + b)$ , show that $y_{2} - 4 y_{1} + 4 y = 0$ .

Answer:

Given,

$y = e^{2 x} (a x + b)$

To prove: $y_{2} - 4 y_{1} + 4 y = 0$

Proof:

We have,

$y = e^{2 x} (a x + b)$ ...(i)

$y_{1} = \frac{d y}{d x} = a e^{2 x} + 2 e^{2 x} (a x + b) . . . (ii) y_{2} = 2 a \times e^{2 x} + 4 e^{2 x} (a x + b) + 2 a e^{2 x} = 4 a e^{2 x} + 4 e^{2 x} (a x + b) . . . (iii) LHS = y_{2} - 4 y_{1} + 4 y = 4 a e^{2 x} + 4 e^{2 x} (a x + b) - 4 a e^{2 x} - 8 e^{2 x} (a x + b) + 4 e^{2 x} (a x + b) = 0 = RHS$

Page No 11.17:

Question 24:

If $x = \sin (\frac{1}{a} \log y)$ , show that (1 − x²)y₂ − xy₁ − a²y = 0.

Answer:

Here,
$x = \sin (\frac{1}{a} \log y) \Rightarrow \frac{1}{a} \log y = \sin^{- 1} x \Rightarrow y = e^{a \sin^{- 1} x} Differentiating w . r . t . x, we get y_{1} = e^{a \sin^{- 1} x} \times \frac{a}{\sqrt{1 - x^{2}}} \Rightarrow y_{1} = \frac{a y}{\sqrt{1 - x^{2}}} Differentiating again w . r . t . x, we get y_{2} = \frac{a y_{1} \sqrt{1 - x^{2}} + \frac{x a y}{\sqrt{1 - x^{2}}}}{(1 - x^{2})} \Rightarrow y_{2} = \frac{a y_{1} (1 - x^{2}) + x a y}{(1 - x^{2}) \sqrt{1 - x^{2}}} \Rightarrow y_{2} = \frac{a y_{1}}{\sqrt{1 - x^{2}}} + \frac{x a y}{(1 - x^{2}) \sqrt{1 - x^{2}}} \Rightarrow y_{2} = \frac{a^{2} y}{1 - x^{2}} + \frac{x y_{1}}{(1 - x^{2})} \Rightarrow (1 - x^{2}) y_{2} - x y_{1} - a^{2} y = 0$

Page No 11.17:

Question 25:

If log y = tan⁻¹ x, show that (1 + x²)y₂ + (2x − 1) y₁ = 0

Answer:

Here,
$\log y = \tan^{- 1} x Differentiating w . r . t . x, we get \frac{1}{y} \times y_{1} = \frac{1}{1 + x^{2}} \Rightarrow (1 + x^{2}) y_{1} = y \Rightarrow (1 + x^{2}) y_{2} + 2 x y_{1} = y_{1} \Rightarrow (1 + x^{2}) y_{2} + 2 x y_{1} - y_{1} = 0 \Rightarrow (1 + x^{2}) y_{2} + (2 x - 1) y_{1} = 0$

Hence proved.

Page No 11.17:

Question 26:

If y = tan⁻¹ x, show that $(1 + x^{2}) \frac{d^{2} y}{d x^{2}} + 2 x \frac{d y}{d x} = 0$ .

Answer:

Here,
$y = \tan^{- 1} x Differentiating w . r . t . x, we get \frac{d y}{d x} = \frac{1}{1 + x^{2}} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{- 2 x}{{(1 + x^{2})}^{2}} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{- 2 x}{1 + x^{2}} \times \frac{1}{1 + x^{2}} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{- 2 x \frac{d y}{d x}}{1 + x^{2}} \Rightarrow (1 + x^{2}) \frac{d^{2} y}{d x^{2}} = - 2 x \frac{d y}{d x} \Rightarrow (1 + x^{2}) \frac{d^{2} y}{d x^{2}} + 2 x \frac{d y}{d x} = 0$

Hence proved.

Page No 11.17:

Question 27:

If $y = {[\log (x + \sqrt{x^{2} + 1})]}^{2}$ , show that $(1 + x^{2}) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} = 2$ .

Answer:

Here,
$y = {[\log (x + \sqrt{x^{2} + 1})]}^{2} Differentiating w . r . t . x, we get \frac{d y}{d x} = \frac{2 \log (x + \sqrt{x^{2} + 1})}{(x + \sqrt{x^{2} + 1})} \times (1 + \frac{2 x}{2 \sqrt{x^{2} + 1}}) \Rightarrow \frac{d y}{d x} = \frac{2 \log (x + \sqrt{x^{2} + 1})}{(x + \sqrt{x^{2} + 1})} \times (\frac{\sqrt{x^{2} + 1} + x}{\sqrt{x^{2} + 1}}) \Rightarrow \frac{d y}{d x} = \frac{2 \log (x + \sqrt{x^{2} + 1})}{\sqrt{x^{2} + 1}} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{2 - \frac{2 x \log (x + \sqrt{x^{2} + 1})}{\sqrt{x^{2} + 1}}}{x^{2} + 1} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{2 - x \frac{d y}{d x}}{x^{2} + 1} \Rightarrow (x^{2} + 1) \frac{d^{2} y}{d x^{2}} = 2 - x \frac{d y}{d x} \Rightarrow (x^{2} + 1) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} = 2$

Page No 11.17:

Question 28:

If y = (tan⁻¹ x)², then prove that (1 + x²)² y₂ + 2x(1 + x²)y₁ = 2.

Answer:

Here,
$y = {(\tan^{- 1} x)}^{2} Differentiating w . r . t . x, we get y_{1} = \frac{2 \tan^{- 1} x}{1 + x^{2}} Differentiating again w . r . t . x, we get y_{2} = \frac{2 - 4 x \tan^{- 1} x}{{(1 + x^{2})}^{2}} \Rightarrow y_{2} = \frac{2}{{(1 + x^{2})}^{2}} - \frac{2 \tan^{- 1} x \times 2 x}{{(1 + x^{2})}^{2}} \Rightarrow y_{2} = \frac{2}{{(1 + x^{2})}^{2}} - \frac{2 x y_{1}}{(1 + x^{2})} \Rightarrow {(1 + x^{2})}^{2} y_{2} = 2 - 2 x (1 + x^{2}) y_{1} \Rightarrow {(1 + x^{2})}^{2} y_{2} + 2 x (1 + x^{2}) y_{1} = 2$

Hence proved.

Page No 11.17:

Question 29:

If y = cot x show that $\frac{d^{2} y}{d x^{2}} + 2 y \frac{d y}{d x} = 0$ .

Answer:

Here,
$y = \cot x Differentiating w . r . t . x, we get \frac{d y}{d x} = - {cosec}^{2} x Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = - 2 cosecx \times (- cosec x cotx) \Rightarrow \frac{d^{2} y}{d x^{2}} = 2 {cosec}^{2} x cotx \Rightarrow \frac{d^{2} y}{d x^{2}} = - 2 y \frac{d y}{d x} \Rightarrow \frac{d^{2} y}{d x^{2}} + 2 y \frac{d y}{d x} = 0$

Hence proved.

Page No 11.17:

Question 30:

Find $\frac{d^{2} y}{d x^{2}}$ , where $y = \log (\frac{x^{2}}{e^{2}})$ .

Answer:

Here,
$y = \log (\frac{x^{2}}{e^{2}}) Differentiating w . r . t . x, we get \frac{d y}{d x} = \frac{1}{\frac{x^{2}}{e^{2}}} \times \frac{2 x}{e^{2}} = \frac{2}{x} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{- 2}{x^{2}}$

Page No 11.17:

Question 31:

If y = ae^2x + be^−x, show that, $\frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 2 y = 0$ .

Answer:

Here,
$y = a e^{2 x} + b e^{- x} Differentiating w . r . t . x, we get \frac{d y}{d x} = 2 a e^{2 x} - b e^{- x} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = 4 a e^{2 x} + b e^{- x} \Rightarrow \frac{d^{2} y}{d x^{2}} = 2 a e^{2 x} - b e^{- x} + 2 (a e^{2 x} + b e^{- x}) \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{d y}{d x} + 2 y \Rightarrow \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 2 y = 0$

Hence proved.

Page No 11.17:

Question 32:

If y = e^x (sin x + cos x) prove that $\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0$ .

Answer:

Here,
$y = e^{x} (\sin x + \cos x) Differentiating w . r . t . x, we get \frac{d y}{d x} = e^{x} (\sin x + \cos x) + e^{x} (\cos x - \sin x) = 2 e^{x} \cos x Differentiating w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = 2 e^{x} \cos x - 2 e^{x} \sin x Now, LHS = \frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 2 e^{x} \cos x - 2 e^{x} \sin x - 4 e^{x} \cos x + 2 e^{x} (\sin x + \cos x) = 0 = RHS$

Hence proved.

Page No 11.17:

Question 33:

If y = cos⁻¹ x, find $\frac{d^{2} y}{d x^{2}}$ in terms of y alone.

Answer:

Here,
$y = \cos^{- 1} x Differentiating w . r . t . x, we get \frac{d y}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{- 2 x}{2 {\sqrt{1 - x^{2}}}^{3 / 2}} = \frac{- x}{{(1 - x^{2})}^{3 / 2}} Now, y = \cos^{- 1} x \Rightarrow x = \cos y \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{- \cos y}{{(1 - \cos^{2} y)}^{3 / 2}} = - \frac{\cos y}{{(\sin^{2} y)}^{3 / 2}} = - \cot y {cosec}^{2} y$

Page No 11.17:

Question 34:

If $y = e^{a \cos^{- 1}} x$ , prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - a^{2} y = 0$ .

Answer:

Here,
$y = e^{a \cos^{- 1} x} Differentiating w . r . t . x, we get \frac{d y}{d x} = - e^{a \cos^{- 1} x} \times \frac{a}{\sqrt{1 - x^{2}}} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = e^{a \cos^{- 1} x} \times \frac{a^{2}}{1 - x^{2}} + \frac{2 x a e^{a \cos^{- 1} x}}{2 {(1 - x^{2})}^{\frac{3}{2}}} \Rightarrow \frac{d^{2} y}{d x^{2}} = e^{a \cos^{- 1} x} \times \frac{a^{2}}{1 - x^{2}} + \frac{x a e^{a \cos^{- 1} x}}{(1 - x^{2}) \sqrt{1 - x^{2}}} \Rightarrow \frac{d^{2} y}{d x^{2}} = y \times \frac{a^{2}}{1 - x^{2}} - \frac{x \frac{d y}{d x}}{(1 - x^{2})} \Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} = a^{2} y - x \frac{d y}{d x} \Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} - a^{2} y = 0$

Hence proved.

Page No 11.17:

Question 35:

If y = 500 e^7x + 600 e^−7x, show that $\frac{d^{2} y}{d x^{2}} = 49 y$ .

Answer:

Here,

$y = 500 e^{7 x} + 600 e^{- 7 x} Differentiating w . r . t . x, we get \frac{d y}{d x} = 3500 e^{7 x} - 4200 e^{- 7 x} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = 24500 e^{7 x} + 29400 e^{- 7 x} = 49 (500 e^{7 x} + 600 e^{- 7 x}) = 49 y$

Page No 11.17:

Question 36:

If x = 2 cos t − cos 2t, y = 2 sin t − sin 2t, find $\frac{d^{2} y}{d x^{2}} at t = \frac{π}{2}$ .

Answer:

Here,
$x = 2 \cos t - \cos 2 t and y = 2 \sin t - \sin 2 t Differentiating w . r . t . t, we get \frac{d x}{d t} = - 2 \sin t + 2 \sin 2 t and \frac{d y}{d t} = 2 \cos t - 2 \cos 2 t ∴ \frac{d y}{d x} = \frac{2 \cos t - 2 \cos 2 t}{- 2 \sin t + 2 \sin 2 t} = \frac{\cos t - \cos 2 t}{- \sin t + \sin 2 t} Differentiating w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{(- \sin t + 2 \sin 2 t) (- \sin t + \sin 2 t) - (\cos t - \cos 2 t) (- \cos t + 2 \cos 2 t)}{{(- \sin t + \sin 2 t)}^{2}} \frac{d t}{d x} = \frac{(- \sin t + 2 \sin 2 t) (- \sin t + \sin 2 t) - (\cos t - \cos 2 t) (- \cos t + 2 \cos 2 t)}{{(- \sin t + \sin 2 t)}^{2} (- 2 \sin t + 2 \sin 2 t)} At t = \frac{π}{2} : \frac{d^{2} y}{d x^{2}} = \frac{(- 1 + 0) (- 1 + 0) - (0 + 1) (- 0 - 2)}{{(- 1 + 0)}^{2} (- 2 + 0)} = \frac{1 + 2}{- 2} = \frac{- 3}{2}$

Page No 11.17:

Question 37:

If x = 4z² + 5, y = 6z² + 7z + 3, find $\frac{d^{2} y}{d x^{2}}$ .

Answer:

Here,
$x = 4 z^{2} + 5 and y = 6 z^{2} + 7 z + 3 Differentiating w . r . t . z, we get \frac{d x}{d z} = 8 z and \frac{d y}{d z} = 12 z + 7 ∴ \frac{d y}{d x} = \frac{12 z + 7}{8 z} Differentiating w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{12 \times 8 z - 8 (12 z + 7)}{64 z^{2}} \times \frac{d z}{d x} = \frac{96 z - 96 z - 56}{512 z^{3}} = \frac{- 7}{64 z^{3}}$

Page No 11.17:

Question 38:

If y log (1 + cos x), prove that $\frac{d^{3} y}{d x^{3}} + \frac{d^{2} y}{d x^{2}} \cdot \frac{d y}{d x} = 0$

Answer:

Here,

$y = \log (1 + \cos x) Differentiating w . r . t . x, we get \frac{d y}{d x} = \frac{- \sin x}{1 + \cos x} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{- \cos x - \cos^{2} x - \sin^{2} x}{{(1 + \cos x)}^{2}} = \frac{- (\cos x + 1)}{(1 + \cos x)} = \frac{- 1}{1 + \cos x} Differentiating again w . r . t . x, we get \frac{d^{3} y}{d x^{3}} = \frac{- \sin x}{{(1 + \cos x)}^{2}} \Rightarrow \frac{d^{3} y}{d x^{3}} + \frac{\sin x}{{(1 + \cos x)}^{2}} = 0 \Rightarrow \frac{d^{3} y}{d x^{3}} + (\frac{- 1}{1 + \cos x}) (\frac{- \sin x}{1 + \cos x}) = 0 \Rightarrow \frac{d^{3} y}{d x^{3}} + \frac{d^{2} y}{d x^{2}} \times \frac{d y}{d x} = 0$

Page No 11.17:

Question 39:

If y = sin (log x), prove that $x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0$ .

Answer:

Here,
$y = \sin (\log x) Differentiating w . r . t . x, we get \frac{d y}{d x} = \frac{\cos (\log x)}{x} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{- \sin (\log x) - \cos (\log x)}{x^{2}} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{- \sin (\log x)}{x^{2}} - \frac{\cos (\log x)}{x^{2}} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{- y}{x^{2}} - \frac{1}{x} \times \frac{d y}{d x} \Rightarrow x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0$

Page No 11.18:

Question 40:

If y = 3 e^2x + 2 e^3x, prove that $\frac{d^{2} y}{d x^{2}} - 5 \frac{d y}{d x} + 6 y = 0$

Answer:

Here,
$y = 3 e^{2 x} + 2 e^{3 x} Differentiating w . r . t . x, we get \frac{d y}{d x} = 6 e^{2 x} + 6 e^{3 x} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = 12 e^{2 x} + 18 e^{3 x} \Rightarrow \frac{d^{2} y}{d x^{2}} = 5 (6 e^{2 x} + 6 e^{3 x}) - 6 (3 e^{2 x} + 2 e^{3 x}) \Rightarrow \frac{d^{2} y}{d x^{2}} = 5 (\frac{d y}{d x}) - 6 y \Rightarrow \frac{d^{2} y}{d x^{2}} - 5 (\frac{d y}{d x}) + 6 y = 0$

Page No 11.18:

Question 41:

If y = (cot⁻¹ x)², prove that y₂(x² + 1)² + 2x (x² + 1) y₁ = 2.

Answer:

Here,
$y = {(\cot^{- 1} x)}^{2} Differentiating w . r . t . x, we get y_{1} = 2 c o t^{- 1} x \times \frac{- 1}{1 + x^{2}} = \frac{- 2 c o t^{- 1} x}{1 + x^{2}} Differentiating again w . r . t . x, we get y_{2} = \frac{2 + 4 x \cot^{- 1} x}{{(1 + x^{2})}^{2}} \Rightarrow y_{2} = \frac{2}{{(1 + x^{2})}^{2}} + \frac{2 x \times 2 \cot^{- 1} x}{(1 + x^{2}) (1 + x^{2})} \Rightarrow y_{2} = \frac{2}{{(1 + x^{2})}^{2}} - \frac{2 x y_{1}}{(1 + x^{2})} \Rightarrow {(1 + x^{2})}^{2} y_{2} = 2 - 2 x y_{1} (1 + x^{2}) \Rightarrow {(1 + x^{2})}^{2} y_{2} + 2 x y_{1} (1 + x^{2}) = 2$

Hence proved.

Page No 11.18:

Question 42:

If y = cosec⁻¹ x, x >1, then show that $x (x^{2} - 1) \frac{d^{2} y}{d x^{2}} + (2 x^{2} - 1) \frac{d y}{d x} = 0$ .

Answer:

Here,
$y = {cosec}^{- 1} x Differentiating w . r . t . x, we get \frac{d y}{d x} = \frac{- 1}{x \sqrt{x^{2} - 1}} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{\sqrt{x^{2} - 1} + \frac{x^{2}}{\sqrt{x^{2} - 1}}}{x^{2} (x^{2} - 1)} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{x^{2} - 1 + x^{2}}{x^{2} (x^{2} - 1) \sqrt{x^{2} - 1}} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{2 x^{2} - 1}{x^{2} (x^{2} - 1) \sqrt{x^{2} - 1}} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{2}{(x^{2} - 1) \sqrt{x^{2} - 1}} - \frac{1}{x^{2} (x^{2} - 1) \sqrt{x^{2} - 1}} \Rightarrow (x^{2} - 1) \frac{d^{2} y}{d x^{2}} = \frac{2}{\sqrt{x^{2} - 1}} - \frac{1}{x^{2} \sqrt{x^{2} - 1}} \Rightarrow (x^{2} - 1) \frac{d^{2} y}{d x^{2}} = - 2 x \frac{d y}{d x} + \frac{1}{x} \frac{d y}{d x} \Rightarrow x (x^{2} - 1) \frac{d^{2} y}{d x^{2}} = - (2 x^{2} - 1) \frac{d y}{d x} \Rightarrow x (x^{2} - 1) \frac{d^{2} y}{d x^{2}} + (2 x^{2} - 1) \frac{d y}{d x} = 0$

Hence proved.

Page No 11.18:

Question 43:

$If x = \cos t + \log \tan \frac{t}{2}, y = \sin t, then find the value of \frac{d^{2} y}{d t^{2}} and \frac{d^{2} y}{d x^{2}} at t = \frac{π}{4} .$

Answer:

$We have, x = \cos t + \log \tan \frac{t}{2} and y = \sin t On differentiating with respect to t, we get \frac{d x}{d t} = \frac{d}{d t} (\cos t + \log \tan \frac{t}{2}) = - \sin t + \frac{1}{\tan \frac{t}{2}} \times \sec^{2} \frac{t}{2} \times \frac{1}{2} = - \sin t + \frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}} = - \sin t + \frac{1}{\sin t} = \frac{- \sin^{2} t + 1}{\sin t} = \frac{- \sin^{2} t + 1}{\sin t} = \frac{\cos^{2} t}{\sin t} and \frac{d y}{d t} = \frac{d}{d t} (\sin t) = \cos t Now, \frac{d^{2} y}{d t^{2}} = \frac{d}{d t} (\frac{d y}{d t}) = \frac{d}{d t} (\cos t) = - \sin t {(\frac{d^{2} y}{d t^{2}})}_{t = \frac{π}{4}} = - \sin (\frac{π}{4}) = - \frac{1}{\sqrt{2}} . . . (1) Also, (\frac{d y}{d x}) = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\cos t}{\frac{\cos^{2} t}{\sin t}} = \frac{\sin t}{\cos t} = \tan t Now, \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\tan t) = \frac{d}{d t} (\tan t) \times \frac{d t}{d x} = \sec^{2} t \times \frac{\sin t}{\cos^{2} t} = \frac{\sin t}{\cos^{4} t} {(\frac{d^{2} y}{d x^{2}})}_{t = \frac{π}{4}} = \frac{\sin (\frac{π}{4})}{\cos^{4} (\frac{π}{4})} = 2 \sqrt{2} . . . (2) Hence, at t = \frac{π}{4}, \frac{d^{2} y}{d t^{2}} = - \frac{1}{\sqrt{2}} and \frac{d^{2} y}{d x^{2}} = 2 \sqrt{2} .$

Page No 11.18:

Question 44:

$If x = a \sin t and y = a (\cos t + \log \tan \frac{t}{2}), find \frac{d^{2} y}{d x^{2}} .$

Answer:

$We have, x = a \sin t and y = a (\cos t + \log \tan \frac{t}{2}) On differentiating with respect to t, we get \frac{d x}{d t} = \frac{d}{d t} (a \sin t) = a \cos t and \frac{d y}{d t} = \frac{d}{d t} [a (\cos t + \log \tan \frac{t}{2})] = a (- \sin t + \frac{1}{\tan \frac{t}{2}} \times \sec^{2} \frac{t}{2} \times \frac{1}{2}) = a (- \sin t + \frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}) = a (- \sin t + \frac{1}{\sin t}) = a (\frac{- \sin^{2} t + 1}{\sin t}) = a (\frac{\cos^{2} t}{\sin t}) Now, (\frac{d y}{d x}) = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{a (\frac{\cos^{2} t}{\sin t})}{a \cos t} = \cot (t) Therefore, \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\cot (t)) = \frac{d}{d t} (\cot (t)) \times \frac{d t}{d x} = - {cosec}^{2} t \times \frac{1}{a \cos t} = - (\frac{1}{a \sin^{2} t \cos t}) Hence, \frac{d^{2} y}{d x^{2}} = - (\frac{1}{a \sin^{2} t \cos t}) .$

Page No 11.18:

Question 45:

$If x = a (\cos t + t \sin t) and y = a (\sin t - t \cos t), then find the value of \frac{d^{2} y}{d x^{2}} at t = \frac{π}{4} .$

Answer:

$We have, x = a (\cos t + t \sin t) and y = a (\sin t - t \cos t) On differentiating with respect to t, we get \frac{d x}{d t} = \frac{d}{d t} [a (\cos t + t \sin t)] = - a \sin t + a \sin t + a t \cos t = a t \cos t and \frac{d y}{d t} = \frac{d}{d t} [a (\sin t - t \cos t)] = a \cos t - a \cos t + a t \sin t = a t \sin t Now, (\frac{d y}{d x}) = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{a t \sin t}{a t \cos t} = \tan t \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\tan t) = \frac{d}{d t} (\tan t) \times \frac{d t}{d x} = \sec^{2} t \times \frac{1}{a t \cos t} = \frac{1}{a t \cos^{3} t} {(\frac{d^{2} y}{d x^{2}})}_{t = \frac{π}{4}} = \frac{1}{a (\frac{π}{4}) \cos^{3} (\frac{π}{4})} = \frac{8 \sqrt{2}}{a π} Hence, at t = \frac{π}{4}, \frac{d^{2} y}{d x^{2}} = \frac{8 \sqrt{2}}{a π} .$

Page No 11.18:

Question 46:

$If x = a (\cos t + \log \tan \frac{t}{2}) and y = a (\sin t), evaluate \frac{d^{2} y}{d x^{2}} at t = \frac{π}{3} .$

Answer:

$We have, x = a (\cos t + \log \tan \frac{t}{2}) and y = a \sin t On differentiating with respect to t, we get \frac{d x}{d t} = \frac{d}{d t} [a (\cos t + \log \tan \frac{t}{2})] = a (- \sin t + \frac{1}{\tan \frac{t}{2}} \times \sec^{2} \frac{t}{2} \times \frac{1}{2}) = a (- \sin t + \frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}) = a (- \sin t + \frac{1}{\sin t}) = a (\frac{- \sin^{2} t + 1}{\sin t}) = a (\frac{\cos^{2} t}{\sin t}) and \frac{d y}{d t} = \frac{d}{d t} (a \sin t) = a \cos t Now, (\frac{d y}{d x}) = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{a \cos t}{a \frac{\cos^{2} t}{\sin t}} = \tan t Therefore, \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\tan (t)) = \frac{d}{d t} (\tan (t)) \times \frac{d t}{d x} = \sec^{2} t \times \frac{\sin t}{a \cos^{2} t} = (\frac{\sin t}{a \cos^{4} t}) {(\frac{d^{2} y}{d x^{2}})}_{t = \frac{π}{3}} = (\frac{\sin (\frac{π}{3})}{a \cos^{4} (\frac{π}{3})}) = \frac{\frac{\sqrt{3}}{2}}{a (\frac{1}{16})} = \frac{8 \sqrt{3}}{a} Hence, at t = \frac{π}{3}, \frac{d^{2} y}{d x^{2}} = \frac{8 \sqrt{3}}{a} .$

Page No 11.18:

Question 47:

$If x = a (\cos 2 t + 2 t \sin 2 t) and y = a (\sin 2 t - 2 t \cos 2 t), then find \frac{d^{2} y}{d x^{2}} .$

Answer:

$We have, x = a (\cos 2 t + 2 t \sin 2 t) and y = a (\sin 2 t - 2 t \cos 2 t) On differentiating with respect to t, we get \frac{d x}{d t} = \frac{d}{d t} [a (\cos 2 t + 2 t \sin 2 t)] = a (- 2 \sin 2 t + 2 \sin 2 t + 4 t \cos 2 t) = a (4 t \cos 2 t) and \frac{d y}{d t} = \frac{d}{d t} [a (\sin 2 t - 2 t \cos 2 t)] = a (2 \cos 2 t - 2 \cos 2 t + 4 t \sin 2 t) = a (4 t \sin 2 t) Now, (\frac{d y}{d x}) = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{a (4 t \sin 2 t)}{a (4 t \cos 2 t)} = \tan 2 t Therefore, \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\tan 2 t) = \frac{d}{d t} (\tan 2 t) \times \frac{d t}{d x} = 2 \sec^{2} 2 t \times \frac{1}{a (4 t \cos 2 t)} = \frac{1}{2 a t \cos^{3} 2 t} = \frac{1}{2 a t} \sec^{3} 2 t Hence, \frac{d^{2} y}{d x^{2}} = \frac{1}{2 a t} \sec^{3} 2 t .$

Page No 11.18:

Question 48:

If $x = 3 c o t - 2 \cos^{3} t, y = 3 \sin t - 2 \sin^{3} t,$ find $\frac{d^{2} y}{d x^{2}} .$

Answer:

We have,

$x = 3 \cos t - 2 \cos^{3} t \Rightarrow \frac{d x}{d t} = 3 (- \sin t) - 6 \cos^{2} t (- \sin t) = - 3 \sin t + 6 \sin t \cos^{2} t$

Also,

$y = 3 \sin t - 2 \sin^{3} t \Rightarrow \frac{d y}{d t} = 3 \cos t - 6 \sin^{2} t \cos t$

Now,

$\frac{d y}{d x} = \frac{(\frac{d y}{d t})}{(\frac{d x}{d t})} = \frac{3 \cos t - 6 \sin^{2} t \cos t}{- 3 \sin t + 6 \sin t \cos^{2} t} = \frac{3 \cos t (1 - 2 \sin^{2} t)}{3 \sin t (- 1 + 2 \cos^{2} t)} = \frac{\cot t (\cos 2 t)}{(\cos 2 t)} = \cot t$

So,

$So, \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\cot t) = - {cosec}^{2} t \frac{d t}{d x} = \frac{- {cosec}^{2} t}{(\frac{d x}{d t})} = \frac{- {cosec}^{2} t}{- 3 \sin t + 6 \sin t \cos^{2} t} = \frac{- {cosec}^{2} t}{- \sin t (1 - 2 \cos^{2} t)} = \frac{{cosec}^{3} t}{(- \cos 2 t)} = \frac{- {cosec}^{3} t}{\cos 2 t}$

Page No 11.18:

Question 49:

$If x = a \sin t - b \cos t, y = a \cos t + b \sin t, prove that \frac{d^{2} y}{d x^{2}} = - \frac{x^{2} + y^{2}}{y^{3}} .$

Answer:

$We have, x = a \sin t - b \cos t, y = a \cos t + b \sin t On differentiating with respect to t, we get \frac{d x}{d t} = \frac{d}{d t} (a \sin t - b \cos t) = a \cos t + b \sin t and \frac{d y}{d t} = \frac{d}{d t} (a \cos t + b \sin t) = - a \sin t + b \cos t Now, (\frac{d y}{d x}) = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{- a \sin t + b \cos t}{a \cos t + b \sin t} Therefore, \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\frac{- a \sin t + b \cos t}{a \cos t + b \sin t}) = \frac{d}{d t} (\frac{- a \sin t + b \cos t}{a \cos t + b \sin t}) \times \frac{d t}{d x} = \frac{(a \cos t + b \sin t) \frac{d}{d t} (- a \sin t + b \cos t) - (- a \sin t + b \cos t) \frac{d}{d t} (a \cos t + b \sin t)}{{(a \cos t + b \sin t)}^{2}} \times \frac{1}{a \cos t + b \sin t} = \frac{(a \cos t + b \sin t) (- a \cos t - b \sin t) - (- a \sin t + b \cos t) (- a \sin t + b \cos t)}{{(a \cos t + b \sin t)}^{3}} = \frac{- {(a \cos t + b \sin t)}^{2} - {(- a \sin t + b \cos t)}^{2}}{{(a \cos t + b \sin t)}^{3}} = \frac{- {(a \cos t + b \sin t)}^{2} - {(a \sin t - b \cos t)}^{2}}{{(a \cos t + b \sin t)}^{3}} = \frac{- y^{2} - x^{2}}{y^{3}} Hence, \frac{d^{2} y}{d x^{2}} = - \frac{x^{2} + y^{2}}{y^{3}} .$

Page No 11.18:

Question 50:

$Find A and B so that y = A \sin 3 x + B \cos 3 x satisfies the equation \frac{d^{2} y}{d x^{2}} + 4 \frac{d y}{d x} + 3 y = 10 \cos 3 x .$

Answer:

$We have, y = A \sin 3 x + B \cos 3 x \frac{d y}{d x} = 3 A \cos 3 x - 3 B \sin 3 x \frac{d^{2} y}{d x^{2}} = - 9 A \sin 3 x - 9 B \cos 3 x Therefore, \frac{d^{2} y}{d x^{2}} + 4 \frac{d y}{d x} + 3 y = - 9 A \sin 3 x - 9 B \cos 3 x + 12 A \cos 3 x - 12 B \sin 3 x + 3 A \sin 3 x + 3 B \cos 3 x = (- 6 A - 12 B) \sin 3 x + (- 6 B + 12 A) \cos 3 x It is given that, \frac{d^{2} y}{d x^{2}} + 4 \frac{d y}{d x} + 3 y = 10 \cos 3 x Comparing the coefficients of \sin 3 x and \cos 3 x, we get - 6 A - 12 B = 0 and - 6 B + 12 A = 10 or A = \frac{2}{3} and B = - \frac{1}{3} Hence, A = \frac{2}{3} and B = - \frac{1}{3} .$

Page No 11.18:

Question 51:

$If y = A e^{- k t} \cos (p t + c), prove that \frac{d^{2} y}{d t^{2}} + 2 k \frac{d y}{d t} + n^{2} y = 0, where n^{2} = p^{2} + k^{2} .$

Answer:

$We have, y = A e^{- k t} \cos (p t + c) . . . (1) Differentiating y with respect to t, we get \frac{d y}{d t} = - k A e^{- k t} \cos (p t + c) - p A e^{- k t} \sin (p t + c) = - k y - p A e^{- k t} \sin (p t + c) [From (1)] \Rightarrow p A e^{- k t} \sin (p t + c) = - k y - \frac{d y}{d t} . . . (2) Differentiating \frac{d y}{d t} with respect to t, we get \frac{d^{2} y}{d t^{2}} = - k \frac{d y}{d t} + p k A e^{- k t} \sin (p t + c) - p^{2} A e^{- k t} \cos (p t + c) = - k \frac{d y}{d t} + k (- k y - \frac{d y}{d t}) - p^{2} y [From (1) and (2)] = - k \frac{d y}{d t} - k^{2} y - k \frac{d y}{d t} - p^{2} y = - 2 k \frac{d y}{d t} - (k^{2} + p^{2}) y \Rightarrow \frac{d^{2} y}{d t^{2}} + 2 k \frac{d y}{d t} + (k^{2} + p^{2}) y = 0 \Rightarrow \frac{d^{2} y}{d t^{2}} + 2 k \frac{d y}{d t} + n^{2} y = 0, where n^{2} = p^{2} + k^{2} . Hence, \frac{d^{2} y}{d t^{2}} + 2 k \frac{d y}{d t} + n^{2} y = 0, where n^{2} = p^{2} + k^{2} .$

Page No 11.18:

Question 52:

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

Disclaimer: There is a misprint in the question. It must be $x^{2} \frac{d^{2} y}{d x^{2}} + (1 - 2 n) x \frac{d y}{d x} + (1 + n^{2}) y = 0$ instead of $x^{2} \frac{d^{2} y}{d x^{2}} + (1 - 2 n) \frac{d y}{d x} + (1 + n^{2}) y = 0$ .

Answer:

$We have, y = x^{n} \{a \cos (\log x) + b \sin (\log x)\} . . . (1) Differentiating y with respect to x, we get \frac{d y}{d x} = n x^{n - 1} \{a \cos (\log x) + b \sin (\log x)\} + x^{n} \{- a \sin (\log x) \times \frac{1}{x} + b \cos (\log x) \times \frac{1}{x}\} = \frac{n}{x} x^{n} \{a \cos (\log x) + b \sin (\log x)\} + x^{n - 1} \{- a \sin (\log x) + b \cos (\log x)\} = \frac{n}{x} y + x^{n - 1} \{- a \sin (\log x) + b \cos (\log x)\} [From (1)] \Rightarrow x^{n - 1} \{- a \sin (\log x) + b \cos (\log x)\} = \frac{d y}{d x} - \frac{n}{x} y . . . (2) Differentiating \frac{d y}{d x} with respect to x, we get \frac{d^{2} y}{d x^{2}} = \frac{n}{x} \frac{d y}{d x} - \frac{n y}{x^{2}} + (n - 1) x^{n - 2} \{- a \sin (\log x) + b \cos (\log x)\} + x^{n - 1} \{- a \cos (\log x) \times \frac{1}{x} - b \sin (\log x) \times \frac{1}{x}\} = \frac{n}{x} \frac{d y}{d x} - \frac{n y}{x^{2}} + (n - 1) \frac{x^{n - 1}}{x} \{- a \sin (\log x) + b \cos (\log x)\} - \frac{x^{n}}{x^{2}} \{a \cos (\log x) + b \sin (\log x)\} = \frac{n}{x} \frac{d y}{d x} - \frac{n y}{x^{2}} + (\frac{n - 1}{x}) (\frac{d y}{d x} - \frac{n}{x} y) - \frac{y}{x^{2}} [From (1) and (2)] = \frac{n}{x} \frac{d y}{d x} - \frac{n y}{x^{2}} + (\frac{n - 1}{x}) \frac{d y}{d x} - \frac{n (n - 1) y}{x^{2}} - \frac{y}{x^{2}} = \frac{d y}{d x} (\frac{n + n - 1}{x}) - \frac{(n + n^{2} - n + 1) y}{x^{2}} = (\frac{2 n - 1}{x}) \frac{d y}{d x} - \frac{(n^{2} + 1) y}{x^{2}} \Rightarrow x^{2} \frac{d^{2} y}{d x^{2}} - x (2 n - 1) \frac{d y}{d x} + (n^{2} + 1) y = 0 Hence, x^{2} \frac{d^{2} y}{d x^{2}} + (1 - 2 n) x \frac{d y}{d x} + (1 + n^{2}) y = 0 .$

Page No 11.18:

Question 53:

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

Disclaimer: There is a misprint in the question, $(x^{2} + 1) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} - n^{2} y = 0$ must be written instead of $(x^{2} - 1) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} - n^{2} y = 0 .$

Answer:

$We have, y = a {\{x + \sqrt{x^{2} + 1}\}}^{n} + b {\{x - \sqrt{x^{2} + 1}\}}^{- n} . . . (1) Differentiating y with respect to x, we get \frac{d y}{d x} = a n {\{x + \sqrt{x^{2} + 1}\}}^{n - 1} (1 + \frac{1}{2 \sqrt{x^{2} + 1}} \times 2 x) - b n {\{x - \sqrt{x^{2} + 1}\}}^{- n - 1} (1 - \frac{1}{2 \sqrt{x^{2} + 1}} \times 2 x) = a n {\{x + \sqrt{x^{2} + 1}\}}^{n - 1} (1 + \frac{x}{\sqrt{x^{2} + 1}}) - b n {\{x - \sqrt{x^{2} + 1}\}}^{- n - 1} (1 - \frac{x}{\sqrt{x^{2} + 1}}) = a n {\{x + \sqrt{x^{2} + 1}\}}^{n - 1} (\frac{\sqrt{x^{2} + 1} + x}{\sqrt{x^{2} + 1}}) - b n {\{x - \sqrt{x^{2} + 1}\}}^{- n - 1} (\frac{\sqrt{x^{2} + 1} - x}{\sqrt{x^{2} + 1}}) = a n {\{x + \sqrt{x^{2} + 1}\}}^{n - 1} (\frac{x + \sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1}}) + b n {\{x - \sqrt{x^{2} + 1}\}}^{- n - 1} (\frac{x - \sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1}}) = \{a {\{x + \sqrt{x^{2} + 1}\}}^{n} (\frac{n}{\sqrt{x^{2} + 1}}) + b {\{x - \sqrt{x^{2} + 1}\}}^{- n}\} (\frac{n}{\sqrt{x^{2} + 1}}) = (\frac{n}{\sqrt{x^{2} + 1}}) y [From (1)] \Rightarrow \sqrt{x^{2} + 1} \frac{d y}{d x} = n y Squaring both sides, we get (x^{2} + 1) {(\frac{d y}{d x})}^{2} = n^{2} y^{2} . . . (2) Differentiating (2) with respect to x, we get (x^{2} + 1) 2 \frac{d y}{d x} \times \frac{d^{2} y}{d x^{2}} + 2 x {(\frac{d y}{d x})}^{2} = n^{2} (2 y \frac{d y}{d x}) \Rightarrow (x^{2} + 1) \frac{d^{2} y}{d x^{2}} + x (\frac{d y}{d x}) = n^{2} (y) \Rightarrow (x^{2} + 1) \frac{d^{2} y}{d x^{2}} + x (\frac{d y}{d x}) - n^{2} y = 0 Hence, (x^{2} + 1) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} - n^{2} y = 0 .$

Page No 11.22:

Question 1:

If x = a cos nt − b sin nt, then $\frac{d^{2} x}{d t^{2}}$ is

(a) n² x
(b) −n² x
(c) −nx
(d) nx

Answer:

(b) −n²x

^{Here,

$x = a \cos n t - b \sin n t Differentiating w . r . t . t, we get \frac{d x}{d t} = - a n \sin n t - b n \cos n t Differentiating again w . r . t . t, we get \frac{d^{2} x}{d t^{2}} = - a n^{2} \cos n t + b n^{2} \sin n t = - n^{2} (a \cos n t - b \sin n t) = - n^{2} x$}

Page No 11.22:

Question 2:

If x = at², y = 2 at, then $\frac{d^{2} y}{d x^{2}} =$

(a) $- \frac{1}{t^{2}}$
(b) $\frac{1}{2 a t^{3}}$
(c) $- \frac{1}{t^{3}}$
(d) $- \frac{1}{2 a t^{3}}$

Answer:

(d) $- \frac{1}{2 a t^{3}}$

Here,
$x = a t^{2} and y = 2 a t Differentiating w . r . t . t, we get \frac{d x}{d t} = 2 a t and \frac{d y}{d t} = 2 a ∴ \frac{d y}{d x} = \frac{2 a}{2 a t} = \frac{1}{t} Differentiating again w . r . t . t, we get \frac{d^{2} y}{d x^{2}} = \frac{- 1}{t^{2}} \frac{d t}{d x} = \frac{- 1}{2 a t^{3}}$

Page No 11.22:

Question 3:

If y = axⁿ⁺¹ + bx⁻ⁿ, then $x^{2} \frac{d^{2} y}{d x^{2}} =$

(a) n (n − 1)y
(b) n (n + 1)y
(c) ny
(d) n²y

Answer:

(b) n(n+1)y

Here,
$y = a x^{n + 1} + b x^{- n} \Rightarrow \frac{d y}{d x} = a (n + 1) x^{n} - b n x^{- n - 1} \Rightarrow \frac{d^{2} y}{d x^{2}} = a n (n + 1) x^{n - 1} + b n (n + 1) x^{- n - 2} ∴ x^{2} \frac{d^{2} y}{d x^{2}} = x^{2} \{a n (n + 1) x^{n - 1} + b n (n + 1) x^{- n - 2}\} = n (n + 1) (a x^{n + 1} + b x^{- n}) = n (n + 1) y$

Page No 11.22:

Question 4:

$\frac{d^{20}}{d x^{20}} (2 \cos x \cos 3 x) =$

(a) 2²⁰ (cos 2 x − 2²⁰ cos 4 x)
(b) 2²⁰ (cos 2 x + 2²⁰ cos 4 x)
(c) 2²⁰ (sin 2 x + 2²⁰ sin 4 x)
(d) 2²⁰(sin 2 x − 2²⁰ sin 4 x)

Answer:

(b) 2²⁰(cos2x + 2²⁰cos4x)

Here,
$y = 2 \cos x \cos 3 x = \cos (3 x - x) + \cos (3 x + x) = \cos 2 x + \cos 4 x \Rightarrow \frac{d y}{d x} = - 2 \sin 2 x - 4 \sin 4 x = - 2 (\sin 2 x + 2 \sin 4 x) \Rightarrow \frac{d^{2} y}{d x^{2}} = - 4 \cos 2 x - 16 \cos 4 x = - 2^{2} (\cos 2 x + 2^{2} \cos 4 x) \Rightarrow \frac{d^{3} y}{d x^{3}} = 2^{3} (\sin 2 x + 2^{3} \sin 4 x) \Rightarrow \frac{d^{4} y}{d x^{4}} = 2^{3} (2 \cos 2 x + 4 \times 2^{3} \cos 4 x) = 2^{4} (\cos 2 x + 2^{4} \cos 4 x) ∴ \frac{d^{20} (\cos 2 x + \cos 4 x)}{d x^{20}} = 2^{20} (\cos 2 x + 2^{20} \cos 4 x)$

Page No 11.22:

Question 5:

If x = t², y = t³, then $\frac{d^{2} y}{d x^{2}} =$

(a) 3/2
(b) 3/4t
(c) 3/2t
(d) 3t/2

Answer:

(b) 3/4t

Here,
$x = t^{2} and y = t^{3} \Rightarrow \frac{d x}{d t} = 2 t and \frac{d y}{d t} = 3 t^{2} ∴ \frac{d y}{d x} = \frac{3 t}{2} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{3}{2} \frac{d t}{d x} = \frac{3}{4 t}$

Page No 11.22:

Question 6:

If y = a + bx², a, b arbitrary constants, then

(a) $\frac{d^{2} y}{d x^{2}} = 2 x y$
(b) $x \frac{d^{2} y}{d x^{2}} = y_{1}$
(c) $x \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} + y = 0$
(d) $x \frac{d^{2} y}{d x^{2}} = 2 x y$

Answer:

(b) $x \frac{d^{2} y}{d x^{2}} = y_{1}$

Here,
$y = a + b x^{2} \Rightarrow y_{1} = 2 b x \Rightarrow y_{2} = 2 b Multiplying by x on both sides we get, x y_{2} = 2 b x = y_{1} \Rightarrow x \frac{d^{2} y}{d x^{2}} = y_{1}$

Page No 11.22:

Question 7:

If f(x) = (cos x + i sin x) (cos 2x + i sin 2x) (cos 3x + i sin 3x) ...... (cos nx + i sin nx) and f(1) = 1, then f'' (1) is equal to

(a) $\frac{n (n + 1)}{2}$
(b) ${\{\frac{n (n + 1)}{2}\}}^{2}$
(c) $- {\{\frac{n (n + 1)}{2}\}}^{2}$
(d) none of these

Answer:

(c) $- {\{\frac{n (n + 1)}{2}\}}^{2}$

Here,
$f (x) = (\cos x + i \sin x) (\cos 2 x + i \sin 2 x) . . . (\cos n x + i \sin n x) \Rightarrow f (x) = (\cos x + i \sin x) {(\cos x + i \sin x)}^{2} . . . {(\cos x + i \sin x)}^{n} \Rightarrow f (x) = {(\cos x + i \sin x)}^{1 + 2 + 3 . . . . . . . . . . . n} \Rightarrow f (x) = {(\cos x + i \sin x)}^{\frac{n (n + 1)}{2}} \Rightarrow f (x) = {(\cos x + i \sin x)}^{a} [where a = \frac{n (n + 1)}{2}] \Rightarrow f (x) = (\cos a x + i \sin a x) . . . (1) \Rightarrow f (1) = (\cos a + i \sin a) \Rightarrow 1 = (\cos a + i \sin a) . . . (2) [∵ f (1) = 1] Differentiating eqn . (1), we get, f' (x) = a (- \sin a x + i \cos a x) \Rightarrow f'' (x) = a^{2} (- \cos a x - i \sin a x) \Rightarrow f'' (x) = - a^{2} (\cos a x + i \sin a x) \Rightarrow f'' (x) = - {\{\frac{n (n + 1)}{2}\}}^{2} (\cos a x + i \sin a x) \Rightarrow f'' (1) = - {\{\frac{n (n + 1)}{2}\}}^{2} (\cos a + i \sin a) \Rightarrow f'' (1) = - {\{\frac{n (n + 1)}{2}\}}^{2} [Using (2)]$

Page No 11.22:

Question 8:

If y = a sin mx + b cos mx, then $\frac{d^{2} y}{d x^{2}}$ is equal to

(a) −m²y
(b) m²y
(c) −my
(d) my

Answer:

(a) −m²y

Here,

$y = a \sin m x + b \cos m x \Rightarrow \frac{d y}{d x} = a m \cos m x - b m \sin m x \Rightarrow \frac{d^{2} y}{d x^{2}} = - a m^{2} \sin m x - b m^{2} \cos m x = - m^{2} (a \sin m x + b \cos m x) = - m^{2} y$

Page No 11.22:

Question 9:

If $f (x) = \frac{\sin^{- 1} x}{\sqrt{1 - x^{2}}}$ , then (1 − x)² f '' (x) − xf(x) =

(a) 1
(b) −1
(c) 0
(d) none of these

Answer:

(a) 1

Here,
$f (x) = \frac{\sin^{- 1} x}{\sqrt{1 - x^{2}}} \Rightarrow \sqrt{1 - x^{2}} f (x) = \sin^{- 1} x Diffferentiating w . r . t . x, we get \sqrt{1 - x^{2}} f^{'} (x) - \frac{x f (x)}{\sqrt{1 - x^{2}}} = \frac{1}{\sqrt{1 - x^{2}}} \Rightarrow (1 - x^{2}) f^{'} (x) - x f (x) = 1$

DISCLAIMER : In the question instead of (1 − x)² f '' (x) − xf(x)
it should be (1 − x)² f ' (x) − xf(x)

Page No 11.22:

Question 10:

If $y = \tan^{- 1} \{\frac{\log_{e} (e / x^{2})}{\log_{e} (e x^{2})}\} + \tan^{- 1} (\frac{3 + 2 \log_{e} x}{1 - 6 \log_{e} x})$ , then $\frac{d^{2} y}{d x^{2}} =$

(a) 2
(b) 1
(c) 0
(d) −1

Answer:

(c) 0

$y = \tan^{- 1} \{\frac{\log_{e} (e / x^{2})}{\log_{e} (e x^{2})}\} + \tan^{- 1} (\frac{3 + 2 \log_{e} x}{1 - 6 \log_{e} x}) \Rightarrow y = \tan^{- 1} \{\frac{1 - 2 \log_{e} x}{1 + 2 \log_{e} x}\} + \tan^{- 1} (\frac{3 + 2 \log_{e} x}{1 - 6 \log_{e} x}) \Rightarrow y = \tan^{- 1} (\frac{\frac{1 - 2 \log_{e} x}{1 + 2 \log_{e} x} + \frac{3 + 2 \log_{e} x}{1 - 6 \log_{e} x}}{1 - (\frac{1 - 2 \log_{e} x}{1 + 2 \log_{e} x}) (\frac{3 + 2 \log_{e} x}{1 - 6 \log_{e} x})}) \Rightarrow y = \tan^{- 1} \{\frac{(1 - 2 \log_{e} x) (1 - 6 \log_{e} x) + (3 + 2 \log_{e} x) (1 + 2 \log_{e} x)}{(1 + 2 \log_{e} x) (1 - 6 \log_{e} x) - (1 - 2 \log_{e} x) (3 + 2 \log_{e} x)}\} \Rightarrow y = \tan^{- 1} \{\frac{1 - 8 \log_{e} x + 12 {(\log_{e} x)}^{2} + 3 + 8 \log_{e} x + 4 {(\log_{e} x)}^{2}}{1 - 4 \log_{e} x - 12 {(\log_{e} x)}^{2} - 3 + 4 \log_{e} x + 4 {(\log_{e} x)}^{2}}\} \Rightarrow y = \tan^{- 1} \{\frac{1 - 8 \log_{e} x + 12 {(\log_{e} x)}^{2} + 3 + 8 \log_{e} x + 4 {(\log_{e} x)}^{2}}{1 - 4 \log_{e} x - 12 {(\log_{e} x)}^{2} - 3 + 4 \log_{e} x + 4 {(\log_{e} x)}^{2}}\} \Rightarrow y = \tan^{- 1} \{\frac{4 + 16 {(\log_{e} x)}^{2}}{- 2 - 8 {(\log_{e} x)}^{2}}\} \Rightarrow y = \tan^{- 1} [\frac{4 \{1 + 4 {(\log_{e} x)}^{2}\}}{- 2 \{1 + 4 {(\log_{e} x)}^{2}\}}] \Rightarrow y = \tan^{- 1} [- 2] \Rightarrow \frac{d y}{d x} = 0 \Rightarrow \frac{d^{2} y}{d x^{2}} = 0$

Page No 11.22:

Question 11:

Let f(x) be a polynomial. Then, the second order derivative of f(e^x) is

(a) f'' (e^x) e^2x + f'(e^x) e^x
(b) f'' (e^x) e^x + f' (e^x)
(c) f'' (e^x) e^2x + f'' (e^x) e^x
(d) f'' (e^x)

Answer:

(a) f''(e^x)e^2x + f'(e^x)e^x

Since f(x) is a polynomial,

^{$∴ f^{'} (e^{x}) = f^{'} (e^{x}) e^{x} \Rightarrow f^{''} (e^{x}) = f^{''} (e^{x}) (e^{x})^{2} + f^{'} (e^{x}) e^{x} = f^{''} (e^{x}) e^{2 x} + f^{'} (e^{x}) e^{x}$}

Page No 11.23:

Question 12:

If y = a cos (log_e x) + b sin (log_e x), then x² y₂ + xy₁ =

(a) 0
(b) y
(c) −y
(d) none of these

Answer:

(c) −y

Here,
$y = a \cos (\log_{e} x) + b \sin (\log_{e} x) \Rightarrow y_{1} = - a \sin (\log_{e} x) \frac{1}{x} + b \cos (\log_{e} x) \frac{1}{x} \Rightarrow y_{2} = \frac{- a \sin (\log_{e} x) + b \cos (\log_{e} x)}{x} \Rightarrow y_{2} = \frac{- a \cos (\log_{e} x) - b \sin (\log_{e} x) - \{- a \sin (\log_{e} x) + b \cos (\log_{e} x)\}}{x^{2}} \Rightarrow x^{2} y_{2} = - \{a \cos (\log_{e} x) + b \sin (\log_{e} x)\} - \{- a \sin (\log_{e} x) + b \cos (\log_{e} x)\} \Rightarrow x^{2} y_{2} = - y - x y_{1} \Rightarrow x^{2} y_{2} + x y_{1} = - y$

Page No 11.23:

Question 13:

If x = 2 at, y = at², where a is a constant, then $\frac{d^{2} y}{d x^{2}} at x = \frac{1}{2}$ is

(a) 1/2a
(b) 1
(c) 2a
(d) none of these

Answer:

(a) 1/2a

Here,
$x = 2 a t and y = a t^{2} Differentiating w . r . t . t, we get \frac{d x}{d t} = 2 a and \frac{d y}{d t} = 2 a t ∴ \frac{d y}{d x} = \frac{2 a t}{2 a} = t Differentiating w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = 1 \times \frac{d t}{d x} = \frac{1}{2 a} Now, {[\frac{d^{2} y}{d x^{2}}]}_{x = \frac{1}{2}} = \frac{1}{2 a}$

Page No 11.23:

Question 14:

If x = f(t) and y = g(t), then $\frac{d^{2} y}{d x^{2}}$ is equal to

(a) $\frac{f' g'' - g' f''}{{(f')}^{3}}$
(b) $\frac{f' g'' - g' f''}{{(f')}^{2}}$
(c) $\frac{g''}{f''}$
(d) $\frac{f'' g' - g'' f'}{{(g')}^{3}}$

Answer:

(a) $\frac{f' g'' - g' f''}{{(f')}^{3}}$

Here,
x = f(t) and y = g(t)
$\Rightarrow \frac{d x}{d t} = f' (t) and \frac{d y}{d t} = g' (t) ∴ \frac{d y}{d x} = \frac{g' (t)}{f' (t)}$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{d}{d t} \{\frac{g^{'} (t)}{f^{'} (t)}\} \times \frac{d t}{d x} = \frac{f^{'} (t) g^{''} (t) - g^{'} (t) f^{''} (t)}{{[f^{'} (t)]}^{2}} \times \frac{1}{f' (t)} = \frac{f^{'} (t) g^{''} (t) - g^{'} (t) f^{''} (t)}{{[f^{'} (t)]}^{3}}$

Page No 11.23:

Question 15:

If y = sin (m sin⁻¹ x), then (1 − x²) y₂ − xy₁ is equal to

(a) m²y
(b) my
(c) −m²y
(d) none of these

Answer:

(c)−m²y

Here,
$y = \sin (m \sin^{- 1} x) \Rightarrow y_{1} = \cos (m \sin^{- 1} x) \frac{m}{\sqrt{1 - x^{2}}} \Rightarrow y_{2} = - \sin (m \sin^{- 1} x) \frac{m^{2}}{(1 - x^{2})} + \frac{m x \cos (m \sin^{- 1} x)}{{(1 - x^{2})}^{3 / 2}} \Rightarrow y_{2} = - \sin (m \sin^{- 1} x) \frac{m^{2}}{(1 - x^{2})} + \frac{x m \cos (m \sin^{- 1} x)}{(1 - x^{2}) \times \sqrt{1 - x^{2}}} \Rightarrow y_{2} = - \sin (m \sin^{- 1} x) \frac{m^{2}}{(1 - x^{2})} + \frac{x y_{1}}{(1 - x^{2})} \Rightarrow (1 - x^{2}) y_{2} = - y m^{2} + x y_{1} \Rightarrow (1 - x^{2}) y_{2} - x y_{1} = - m^{2} y$

Page No 11.23:

Question 16:

If y = (sin⁻¹ x)², then (1 − x²)y₂ is equal to

(a) xy₁ + 2
(b) xy₁ − 2
(c) −xy₁+2
(d) none of these

Answer:

(a) xy₁ + 2

Here,

$y = {(\sin^{- 1} x)}^{2} Now, y_{1} = 2 \sin^{- 1} x \frac{1}{\sqrt{1 - x^{2}}} \Rightarrow y_{2} = \frac{2}{1 - x^{2}} + \frac{2 x \sin^{- 1} x}{{(1 - x^{2})}^{3 / 2}} \Rightarrow y_{2} = \frac{2}{1 - x^{2}} + \frac{2 x \sin^{- 1} x}{(1 - x^{2}) \sqrt{1 - x^{2}}} \Rightarrow y_{2} = \frac{2}{1 - x^{2}} + \frac{x y_{1}}{(1 - x^{2})} \Rightarrow y_{2} (1 - x^{2}) = 2 + x y_{1}$

Page No 11.23:

Question 17:

If y = e^tan^x, then (cos² x)y₂ =

(a) (1 − sin 2x) y₁
(b) −(1 + sin 2x)y₁
(c) (1 + sin 2x)y₁
(d) none of these

Answer:

(c) (1 + sin 2x)y₁
_{Here,

$y = e^{\tan x} \Rightarrow y_{1} = e^{\tan x} \sec^{2} x \Rightarrow y_{2} = e^{\tan x} \sec^{4} x + e^{\tan x} \times 2 \sec x \sec x \tan x \Rightarrow y_{2} = \sec^{2} x (e^{\tan x} \sec^{2} x + e^{\tan x} \times 2 \tan x) \Rightarrow (\cos^{2} x) y_{2} = y_{1} + e^{\tan x} \times \frac{y_{1}}{\sec^{2} x} 2 \tan x \Rightarrow (\cos^{2} x) y_{2} = y_{1} + y_{1} \times 2 \sin x \cos x \Rightarrow (\cos^{2} x) y_{2} = y_{1} (1 + \sin 2 x)$}

Page No 11.23:

Question 18:

If $y = \frac{2}{\sqrt{a^{2} - b^{2}}} \tan^{- 1} (\frac{a - b}{a + b} \tan \frac{x}{2}), a > b > 0$ , then

(a) $y_{1} = \frac{- 1}{a + b \cos x}$
(b) $y_{2} = \frac{b \sin x}{{(a + b \cos x)}^{2}}$
(c) $y_{1} = \frac{1}{a - b \cos x}$
(d) $y_{2} = \frac{- b \sin x}{{(a - b \cos x)}^{2}}$

Answer:

Disclaimer: The question given in the book is wrong.

Page No 11.23:

Question 19:

If $y = \frac{a x + b}{x^{2} + c}$ , then (2xy₁ + y)y₃ =

(a) 3(xy₂ + y₁)y₂
(b) 3(xy₁ + y₂)y₂
(c) 3(xy₂ + y₁)y₁
(d) none of these

Answer:

(a) 3(xy₂ + y₁)y₂

_Here,

$y = \frac{a x + b}{x^{2} + c} \Rightarrow (x^{2} + c) y = a x + b Diffferentiating w . r . t . x, we get 2 x y + (x^{2} + c) \frac{d y}{d x} = a Diffferentiating w . r . t . x, we get 2 y + 2 x y_{1} + 2 x y_{1} + (x^{2} + c) y_{2} = 0 \Rightarrow 2 y + 4 x y_{1} + (x^{2} + c) y_{2} = 0 Diffferentiating again w . r . t . x, we get 2 y_{1} + 4 y_{1} + 4 x y_{2} + (x^{2} + c) y_{3} + 2 x y_{2} = 0 \Rightarrow 6 y_{1} + 6 x y_{2} + (x^{2} + c) y_{3} = 0 \Rightarrow 6 y_{1} + 6 x y_{2} + (\frac{- 2 y - 4 x y_{1}}{y_{2}}) y_{3} = 0 [∵ 2 y + 4 x y_{1} + (x^{2} + c) y_{2} = 0] \Rightarrow 6 y_{1} y_{2} + 6 x {(y_{2})}^{2} - 2 y - 4 x y_{1} y_{3} = 0 \Rightarrow 3 y_{1} y_{2} + 3 x {(y_{2})}^{2} - y - 2 x y_{1} y_{3} = 0 \Rightarrow (y_{1} + x y_{2}) 3 y_{2} = (2 x y_{1} + y) y_{3}$

Page No 11.23:

Question 20:

If $y = \log_{e} {(\frac{x}{a + b x})}^{x}$ , then x³ y₂ =

(a) (xy₁ − y)²
(b) (1 + y)²
(c) ${(\frac{y - x y_{1}}{y_{1}})}^{2}$
(d) none of these

Answer:

(a) (xy₁ − y)²

Here,
$y = \log_{e} {(\frac{x}{a + b x})}^{x} \Rightarrow y = x \log_{e} (\frac{x}{a + b x}) \Rightarrow y_{1} = \log_{e} (\frac{x}{a + b x}) + x \times \frac{a + b x}{x} (\frac{1}{a + b x} - \frac{b x}{{(a + b x)}^{2}}) \Rightarrow y_{1} = \log_{e} (\frac{x}{a + b x}) + (\frac{a}{a + b x}) . . . (1) \Rightarrow y_{1} = \frac{y}{x} + (\frac{a}{a + b x}) [∵ y = x \log_{e} (\frac{x}{a + b x})] \Rightarrow \frac{x y_{1} - y}{x} = \frac{a}{a + b x} . . . (2) Differentiating (1) we get, y_{2} = \frac{a + b x}{x} (\frac{a + b x - b x}{{(a + b x)}^{2}}) - \frac{b a}{{(a + b x)}^{2}} \Rightarrow y_{2} = \frac{a}{x (a + b x)} - \frac{b a}{{(a + b x)}^{2}} \Rightarrow y_{2} = \frac{a (a + b x) - a b x}{x {(a + b x)}^{2}} \Rightarrow y_{2} = \frac{a^{2}}{x {(a + b x)}^{2}} \Rightarrow y_{2} = \frac{{(x y_{1} - y)}^{2}}{x^{3}} [Using (2)] \Rightarrow x^{3} y_{2} = {(x y_{1} - y)}^{2}$

Page No 11.23:

Question 21:

If x = f(t) cos t − f' (t) sin t and y = f(t) sin t + f'(t) cos t, then ${(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2} =$

(a) f(t) − f''(t)
(b) {f(t) − f'' (t)}²
(c) {f(t) + f''(t)}²
(d) none of these

Answer:

(c){f(t) + f''(t)}²

Here,
$x = f (t) \cos t - f^{'} (t) \sin t and y = f (t) \sin t + f^{'} (t) \cos t \Rightarrow \frac{d x}{d t} = f^{'} (t) \cos t - f (t) \sin t - f^{''} (t) \sin t - f^{'} (t) \cos t and \frac{d y}{d t} = f^{'} (t) \sin t + f (t) \cos t + f^{''} (t) \cos t - f^{'} (t) \sin t \Rightarrow \frac{d x}{d t} = - f (t) \sin t - f^{''} (t) \sin t and \frac{d y}{d t} = f (t) \cos t + f^{''} (t) \cos t Thus, {(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2} = {\{- f (t) \sin t - f^{''} (t) \sin t\}}^{2} + {\{f (t) \cos t + f^{''} (t) \cos t\}}^{2} = {\{f (t) \sin t + f^{''} (t) \sin t\}}^{2} + {\{f (t) \cos t + f^{''} (t) \cos t\}}^{2} = \sin^{2} t {\{f (t) + f^{''} (t)\}}^{2} + \cos^{2} t {\{f (t) + f^{''} (t)\}}^{2} = {\{f (t) + f^{''} (t)\}}^{2} (\sin^{2} t + \cos^{2} t) = {\{f (t) + f^{''} (t)\}}^{2}$

Page No 11.23:

Question 22:

If $y^{\frac{1}{n}} + y^{- \frac{1}{n}} = 2 x, then find (x^{2} - 1) y_{2} + x y_{1} =$

Answer:

$(c) n^{2} y y^{\frac{1}{n}} + y^{- \frac{1}{n}} = 2 x Differentiating the above equation with respect to x (\frac{1}{n} y^{\frac{1}{n} - 1} - \frac{1}{n} y^{- \frac{1}{n} - 1}) y_{1} = 2 \frac{1}{n y} (y^{\frac{1}{n}} - y^{- \frac{1}{n}}) y_{1} = 2 (y^{\frac{1}{n}} - y^{- \frac{1}{n}}) y_{1} = 2 n y . . . . . (1) (y^{\frac{1}{n}} - y^{- \frac{1}{n}}) y_{2} + y_{1} (\frac{1}{n} y^{\frac{1}{n} - 1} + \frac{1}{n} y^{- \frac{1}{n} - 1}) y_{1} = 2 n y_{1} n y (y^{\frac{1}{n}} - y^{- \frac{1}{n}}) y_{2} + {y_{1}}^{2} (y^{\frac{1}{n}} + y^{- \frac{1}{n}}) = 2 n^{2} y y_{1} Dividing the above equation by y_{1} \frac{n y}{y_{1}} (y^{\frac{1}{n}} - y^{- \frac{1}{n}}) y_{2} + y_{1} (y^{\frac{1}{n}} + y^{- \frac{1}{n}}) = 2 n^{2} y Putting y_{1} from equation (1) \frac{{(y^{\frac{1}{n}} - y^{- \frac{1}{n}})}^{2}}{2} y_{2} + y_{1} (y^{\frac{1}{n}} + y^{- \frac{1}{n}}) = 2 n^{2} y . . . . . (2) Now, {(y^{\frac{1}{n}} - y^{- \frac{1}{n}})}^{2} = {(y^{\frac{1}{n}} + y^{- \frac{1}{n}})}^{2} - 4 {(y^{\frac{1}{n}} - y^{- \frac{1}{n}})}^{2} = 4 x^{2} - 4 . . . . . (3) Putting the value of (3) in (2) \frac{4 (x^{2} - 1) y_{2}}{2} + 2 x y_{1} = 2 n^{2} y (x^{2} - 1) y_{2} + x y_{1} = n^{2} y$

Page No 11.23:

Question 23:

If $\frac{d}{d x} [x^{n} - a_{1} x^{n - 1} + a_{2} x^{n - 2} + . . . + {(- 1)}^{n} a_{n}] e^{x} = x^{n} e^{x}$ , then the value of a_r, 0 < r ≤ n, is equal to

(a) $\frac{n!}{r!}$
(b) $\frac{(n - r)!}{r!}$
(c) $\frac{n!}{(n - r)!}$
(d) none of these

Answer:

(c) $\frac{n!}{(n - r)!}$

According to the given equation,
$\frac{d}{d x} [x^{n} - a_{1} x^{n - 1} + a_{2} x^{n - 2} + . . . + {(- 1)}^{n} a_{n}] e^{x} = x^{n} e^{x} \Rightarrow \frac{d}{d x} [x^{n} - a_{1} x^{n - 1} + a_{2} x^{n - 2} + . . . + {(- 1)}^{n} a_{n}] e^{x} = \frac{d}{d x} [x^{n} - n x^{n - 1} + n (n - 1) x^{n - 2} + . . . + {(- 1)}^{n} a_{n}] e^{x} Comparing the coefficients of the above equation we get, a_{1} = n a_{2} = n (n - 1) Similarly, a_{r} = n (n - 1) (n - 2) (n - 3) . . . (n - r + 1) \Rightarrow a_{r} = \frac{n!}{(n - r)!}$

Page No 11.23:

Question 24:

If y = xⁿ⁻¹ log x then x² y₂ + (3 − 2n) xy₁ is equal to

(a) −(n − 1)² y
(b) (n − 1)²y
(c) −n²y
(d) n²y

Answer:

(a) −(n − 1)² y

Here,

$y = x^{n - 1} \log x \Rightarrow y_{1} = (n - 1) x^{n - 2} \log x + \frac{x^{n - 1}}{x} \Rightarrow y_{1} = \frac{(n - 1) x^{n - 1} \log x + x^{n - 1}}{x} \Rightarrow x y_{1} = (n - 1) y + x^{n - 1} \Rightarrow x y_{2} + y_{1} = (n - 1) y_{1} + (n - 1) x^{n - 2} \Rightarrow x y_{2} + y_{1} = (n - 1) y_{1} + \frac{(n - 1) x^{n - 1}}{x} \Rightarrow x^{2} y_{2} + x y_{1} = x (n - 1) y_{1} + (n - 1) x^{n - 1} \Rightarrow x^{2} y_{2} + x y_{1} = x (n - 1) y_{1} + (n - 1) \{x y_{1} - (n - 1) y\} \Rightarrow x^{2} y_{2} + x y_{1} = x (n - 1) y_{1} + (n - 1) x y_{1} - {(n - 1)}^{2} y \Rightarrow x^{2} y_{2} + x y_{1} = 2 x (n - 1) y_{1} - {(n - 1)}^{2} y \Rightarrow x^{2} y_{2} + x y_{1} - 2 x (n - 1) y_{1} = - {(n - 1)}^{2} y \Rightarrow x^{2} y_{2} + x y_{1} (1 - 2 n + 2) = - {(n - 1)}^{2} y \Rightarrow x^{2} y_{2} + (3 - 2 n) x y_{1} = - {(n - 1)}^{2} y$

Page No 11.23:

Question 25:

If xy − log_e y = 1 satisfies the equation $x (y y_{2} + y_{1}^{2}) - y_{2} + λ y y_{1} = 0$ , then λ =

(a) −3
(b) 1
(c) 3
(d) none of these

Answer:

(c) 3

Here,

$x y - \log_{e} y = 1 \Rightarrow x y_{1} + y - \frac{y_{1}}{y} = 0 \Rightarrow x y y_{1} + y^{2} - y_{1} = 0 \Rightarrow y y_{1} + x y_{1} y_{1} + x y y_{2} + 2 y y_{1} - y_{2} = 0 \Rightarrow x ({y_{1}}^{2} + y y_{2}) - y_{2} + 3 y y_{1} = 0 ∴ λ = 3$

Page No 11.24:

Question 26:

If y² = ax² + bx + c, then $y^{3} \frac{d^{2} y}{d x^{2}}$ is

(a) a constant
(b) a function of x only
(c) a function of y only
(d) a function of x and y

Answer:

(a) a constant

Here,

$y^{2} = a x^{2} + b x + c Now, 2 y \frac{d y}{d x} = 2 a x + b \Rightarrow 2 y \frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{2} = 2 a \Rightarrow y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = a \Rightarrow y \frac{d^{2} y}{d x^{2}} + {(\frac{2 a x + b}{2 y})}^{2} = a [∵ 2 y \frac{d y}{d x} = 2 a x + b] \Rightarrow 4 y^{3} \frac{d^{2} y}{d x^{2}} + {(2 a x + b)}^{2} = 4 a y^{2} \Rightarrow y^{3} \frac{d^{2} y}{d x^{2}} = \frac{4 a y^{2} - {(2 a x + b)}^{2}}{4} \Rightarrow y^{3} \frac{d^{2} y}{d x^{2}} = \frac{4 a (a x^{2} + b x + c) - {(2 a x + b)}^{2}}{4} [∵ y^{2} = a x^{2} + b x + c] \Rightarrow y^{3} \frac{d^{2} y}{d x^{2}} = \frac{4 a^{2} x^{2} + 4 a b x + 4 a c - 4 a^{2} x^{2} - b^{2} - 4 a x b}{4} \Rightarrow y^{3} \frac{d^{2} y}{d x^{2}} = \frac{4 a c - b^{2}}{4} = a constant$

Page No 11.24:

Question 1:

If y = t¹⁰ + 1 and x = t⁸ + 1, then $\frac{d^{2} y}{d x^{2}}$ = ___________________.

Answer:

Given, y = t¹⁰ + 1 and x = t⁸ + 1.

$y = t^{10} + 1$

Differentiating both sides with respect to t, we get

$\frac{d y}{d t} = 10 t^{9}$

$x = t^{8} + 1$

Differentiating both sides with respect to t, we get

$\frac{d x}{d t} = 8 t^{7}$

$∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} \Rightarrow \frac{d y}{d x} = \frac{10 t^{9}}{8 t^{7}} \Rightarrow \frac{d y}{d x} = \frac{5 t^{2}}{4}$

Differentiating both sides with respect to x, we get

$\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\frac{5 t^{2}}{4})$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{5}{4} \times 2 t \frac{d t}{d x}$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{5}{2} t \times \frac{1}{8 t^{7}} (\frac{d x}{d t} = 8 t^{7} \Rightarrow \frac{d t}{d x} = \frac{1}{8 t^{7}})$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{5}{16 t^{6}}$

If y = t¹⁰ + 1 and x = t⁸ + 1, then $\frac{d^{2} y}{d x^{2}}$ = $\frac{5}{16 t^{6}}$ .

Page No 11.24:

Question 2:

If x = a sin θ and y = b cos θ, then $\frac{d^{2} y}{d x^{2}}$ = ______________________.

Answer:

Given, $x = a \sin θ$ and $y = b \cos θ$ .

$x = a \sin θ$

Differentiating both sides with respect to θ, we get

$\frac{d x}{d θ} = a \cos θ$

$y = b \cos θ$

Differentiating both sides with respect to θ, we get

$\frac{d y}{d θ} = - b \sin θ$

$∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$

$\Rightarrow \frac{d y}{d x} = \frac{- b \sin θ}{a \cos θ}$

$\Rightarrow \frac{d y}{d x} = - \frac{b \tan θ}{a}$

Differentiating both sides with respect to x, we get

$\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (- \frac{b \tan θ}{a})$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - \frac{b}{a} \sec^{2} θ \frac{d θ}{d x}$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - \frac{b}{a} \sec^{2} θ \times \frac{1}{a \cos θ} (\frac{d x}{d θ} = a \cos θ \Rightarrow \frac{d θ}{d x} = \frac{1}{a \cos θ})$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - \frac{b}{a^{2}} \sec^{3} θ$

If x = a sin θ and y = b cos θ, then $\frac{d^{2} y}{d x^{2}}$ = $- \frac{b}{a^{2}} \sec^{3} θ$ .

Page No 11.24:

Question 3:

If y = x + e^x, then $\frac{d^{2} y}{d x^{2}}$ = _____________________.

Answer:

$y = x + e^{x}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{d}{d x} (x + e^{x})$

$\Rightarrow \frac{d y}{d x} = 1 + e^{x}$

Again differentiating both sides with respect to x, we get

$\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (1 + e^{x})$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = 0 + e^{x} = e^{x}$

$∴ \frac{d^{2} y}{d x^{2}} = e^{x}$

If y = x + e^x, then $\frac{d^{2} y}{d x^{2}}$ = $e^{x}$ .

Page No 11.24:

Question 4:

If $y = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x 4}{4!}_________________, then \frac{d^{2} y}{d x^{2}} =__________________.$

Answer:

$y = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x^{4}}{4!} - . . .$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{d}{d x} (1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x^{4}}{4!} - . . .)$

$\Rightarrow \frac{d y}{d x} = \frac{d}{d x} (1) - \frac{d}{d x} (x) + \frac{d}{d x} (\frac{x^{2}}{2!}) - \frac{d}{d x} (\frac{x^{3}}{3!}) + \frac{d}{d x} (\frac{x^{4}}{4!}) - . . .$

$\Rightarrow \frac{d y}{d x} = 0 - 1 + \frac{2 x}{2!} - \frac{3 x^{2}}{3!} + \frac{4 x^{3}}{4!} - . . .$

$\Rightarrow \frac{d y}{d x} = - 1 + \frac{x}{1!} - \frac{x^{2}}{2!} + \frac{x^{3}}{3!} - . . .$

Again differentiating both sides with respect to x, we get

$\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (- 1 + \frac{x}{1!} - \frac{x^{2}}{2!} + \frac{x^{3}}{3!} - . . .)$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (- 1) + \frac{d}{d x} (\frac{x}{1!}) - \frac{d}{d x} (\frac{x^{2}}{2!}) + \frac{d}{d x} (\frac{x^{3}}{3!}) - . . .$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = 0 + 1 - \frac{2 x}{2!} + \frac{3 x^{2}}{3!} - . . .$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + . . .$

If $y = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x^{4}}{4!} -_________________, then \frac{d^{2} y}{d x^{2}} = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + . . . .$

Page No 11.24:

Question 5:

If y = x + e^x , then $\frac{d^{2} x}{d y^{2}}$ = ______________.

Answer:

$y = x + e^{x}$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{d}{d x} (x + e^{x})$

$\Rightarrow \frac{d y}{d x} = 1 + e^{x}$

$\Rightarrow \frac{d x}{d y} = \frac{1}{1 + e^{x}} (\frac{d x}{d y} = \frac{1}{\frac{d y}{d x}})$

Differentiating both sides with respect to y, we get

$\frac{d}{d y} (\frac{d x}{d y}) = \frac{d}{d y} (\frac{1}{1 + e^{x}})$

$\Rightarrow \frac{d^{2} x}{d y^{2}} = \frac{0 - 1 \times \frac{d}{d y} (1 + e^{x})}{{(1 + e^{x})}^{2}}$

$\Rightarrow \frac{d^{2} x}{d y^{2}} = - \frac{0 + e^{x} \frac{d x}{d y}}{{(1 + e^{x})}^{2}}$

$\Rightarrow \frac{d^{2} x}{d y^{2}} = - \frac{e^{x} (\frac{1}{1 + e^{x}})}{{(1 + e^{x})}^{2}} (\frac{d x}{d y} = \frac{1}{1 + e^{x}})$

$\Rightarrow \frac{d^{2} x}{d y^{2}} = - \frac{e^{x}}{{(1 + e^{x})}^{3}}$

If y = x + e^x , then $\frac{d^{2} x}{d y^{2}}$ = $- \frac{e^{x}}{{(1 + e^{x})}^{3}}$ .

Page No 11.24:

Question 1:

If y = a xⁿ^{+ 1} + bx⁻ⁿ and $x^{2} \frac{d^{2} y}{d x^{2}} = λ y$ , then write the value of λ.

Answer:

Here,
$y = a x^{n + 1} + b x^{- n} and x^{2} \frac{d^{2} y}{d x^{2}} = λ y Now, \frac{d y}{d x} = a (n + 1) x^{n} - b n x^{- n - 1} and \frac{d^{2} y}{d x^{2}} = a n (n + 1) x^{n - 1} - b n (- n - 1) x^{- n - 2} Now, x^{2} \frac{d^{2} y}{d x^{2}} = λ y [Given] \Rightarrow x^{2} [a n (n + 1) x^{n - 1} + b n (n + 1) x^{- n - 2}] = λ (a x^{n + 1} + b x^{- n}) \Rightarrow a n (n + 1) x^{n + 1} + b n (n + 1) x^{- n} = λ (a x^{n + 1} + b x^{- n}) \Rightarrow n (n + 1) (a x^{n + 1} + b x^{- n}) = λ (a x^{n + 1} + b x^{- n}) \Rightarrow λ = n (n + 1)$

Page No 11.24:

Question 2:

If x = a cos nt − b sin nt and $\frac{d^{2} x}{d t} = λ x$ , then find the value of λ.

Answer:

Here,

$x = a \cos n t - b \sin n t Now, \frac{d x}{d t} = - a n \sin n t - b n \cos n t \frac{d^{2} x}{d t^{2}} = - a n^{2} \cos n t + b n^{2} \sin n t Also, \frac{d^{2} x}{d t^{2}} = λ x [Given] \Rightarrow - a n^{2} \cos n t + b n^{2} \sin n t = λ (a \cos n t - b \sin n t) \Rightarrow - n^{2} (a \cos n t - b \sin n t) = λ (a \cos n t - b \sin n t) \Rightarrow λ = - n^{2}$

Page No 11.24:

Question 3:

If x = t² and y = t³, find $\frac{d^{2} y}{d x^{2}}$ .

Answer:

Here,
$x = t^{2} and y = t^{3} \Rightarrow \frac{d x}{d t} = 2 t and \frac{d y}{d t} = 3 t^{2} ∴ \frac{d y}{d x} = \frac{3 t}{2} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{3}{2} \frac{d t}{d x} = \frac{3}{4 t}$

Page No 11.24:

Question 4:

If x = 2at, y = at², where a is a constant, then find $\frac{d^{2} y}{d x^{2}} at x = \frac{1}{2}$ .

Answer:

Here,
$x = 2 a t and y = a t^{2} Differentiating w . r . t . t, we get \frac{d x}{d t} = 2 a and \frac{d y}{d t} = 2 a t ∴ \frac{d y}{d x} = \frac{2 a t}{2 a} = t Differentiating again w . r . t . t, we get \frac{d^{2} y}{d x^{2}} = 1 \times \frac{d t}{d x} = \frac{1}{2 a} Now, {[\frac{d^{2} y}{d x^{2}}]}_{x = \frac{1}{2}} = \frac{1}{2 a}$

Page No 11.24:

Question 5:

If x = f(t) and y = g(t), then write the value of $\frac{d^{2} y}{d x^{2}}$ .

Answer:

Here.
x = f(t) and y = g(t)
$\Rightarrow \frac{d x}{d t} = f' (t) and \frac{d y}{d t} = g' (t) ∴ \frac{d y}{d x} = \frac{g' (t)}{f' (t)}$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{d}{d t} \{\frac{g^{'} (t)}{f^{'} (t)}\} \times \frac{d t}{d x} = \frac{f^{'} (t) g^{''} (t) - g^{'} (t) f^{''} (t)}{{[f^{'} (t)]}^{2}} \times \frac{1}{f' (t)} = \frac{f^{'} (t) g^{''} (t) - g^{'} (t) f^{''} (t)}{{[f^{'} (t)]}^{3}}$

Page No 11.24:

Question 6:

If $y = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x^{4}}{4!}$ .....to ∞, then write $\frac{d^{2} y}{d x^{2}}$ in terms of y.

Answer:

Here,
$y = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + . . . \infty Thus, \Rightarrow \frac{d y}{d x} = - 1 + \frac{2 x}{2!} - \frac{3 x^{2}}{3!} + \frac{4 x^{3}}{4!} . . . \infty = - 1 + x - \frac{x^{2}}{2!} + \frac{x^{3}}{3!} - . . . \infty \Rightarrow \frac{d^{2} y}{d x^{2}} = 1 - \frac{2 x}{2!} + \frac{3 x^{2}}{3!} - \frac{4 x^{3}}{4!} + . . . \infty = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + . . . \infty = y$

Page No 11.24:

Question 7:

If y = x + e^x, find $\frac{d^{2} x}{d y^{2}}$ .

Answer:

Here,
$y = x + e^{x} \Rightarrow \frac{d y}{d x} = 1 + e^{x} \Rightarrow \frac{d x}{d y} = \frac{1}{1 + e^{x}} \Rightarrow \frac{d^{2} x}{d y^{2}} = \frac{- e^{x}}{{(1 + e^{x})}^{2}} \frac{d x}{d y} = \frac{- e^{x}}{{(1 + e^{x})}^{3}}$

Page No 11.25:

Question 8:

If y = |x − x²|, then find $\frac{d^{2} y}{d x^{2}}$ .

Answer:

Here,
$y = |x - x^{2}| = \{\begin{cases} x - x^{2} if 0 < x < 1 \\ - x + x^{2} if x > 1, x < 0 \end{cases} \Rightarrow \frac{d y}{d x} = \{\begin{cases} 1 - 2 x if 0 < x < 1 \\ - 1 + 2 x if x > 1, x < 0 \end{cases} \Rightarrow \frac{d^{2} y}{d x^{2}} = \{\begin{cases} - 2 if 0 < x < 1 \\ 2 if x > 1, x < 0 \end{cases}$

Page No 11.25:

Question 9:

If $y = |\log_{e} x|$ , find $\frac{d^{2} y}{d x^{2}}$ .

Answer:

Here,

$y = |\log_{e} x| = \{\begin{cases} - \log_{e} x if 0 < x < 1 \\ \log_{e} x if x > 1 \end{cases} Differentiating w . r . t . x, we get \frac{d y}{d x} = \{\begin{cases} \frac{- 1}{x} if 0 < x < 1 \\ \frac{1}{x} if x > 1 \end{cases} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \{\begin{cases} \frac{1}{x^{2}} if 0 < x < 1 \\ \frac{- 1}{x^{2}} if x > 1 \end{cases}$

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