RD Sharma Class 12 Ex 12.1 Solutions Chapter 12 Higher Order Derivatives

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	12
Exercise	12.1
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 12.1 Solutions Chapter 12 Higher Order Derivatives

Find the second order derivative of following function

Question 1(i). x³+ tanx

Solution:

Let us considered
f(x) = x³+ tanx
On differentiating both sides w.r.t x,
f'(x) = 3x²+ sec²x
Again differentiating both sides w.r.t x,
f”(x) = 6x + 2(secx)(secx.tanx)
f”(x) = 6x + 2sec²x.tanx

Question 1(ii). sin(logx)

Solution:

Let us considered

y = sin(logx)

On differentiating both sides w.r.t x,

(dy/dx) = cos(logx) × (1/x)

(dy/dx) = cos(logx)/x

Again differentiating both sides w.r.t x,

d²y/dx²= d/dx[cos(logx)/x]

$=\frac{x\frac{d}{dx}cos(logx)-cos(logx)\frac{d}{dx}x}{x^2}$

= $\frac{-x[\frac{sin(logx)}{x}]-cos(logx).1}{x^2}$

= -[sin(logx) + cos(logx)]/x²

Question 1(iii). log(sinx)

Solution:

Let us considered
y = log(sinx)
On differentiating both sides w.r.t x,
(dy/dx) = (1/sinx) × (cosx)
(dy/dx) = cotx
Again differentiating both sides w.r.t x,
d²y/dx²= -cosec²x

Question 1(iv). e^xsin5x

Solution:

Let us considered
y = e^xsin5x
On differentiating both sides w.r.t x,
(dy/dx) = e^xsin5x + 5e^xcos5x
Again differentiating both sides w.r.t x,
d²y/dx²= e^xsin5x + 5e^xcos5x + 5(e^xcos5x – 5e^xsin5x)
d²y/dx²= -24e^xsin5x + 10e^xcos5x
d²y/dx² = 2e^x(5cos5x – 12sinx)

Question 1(v). e^6xcos3x

Solution:

Let us considered
y = e^6xcos3x
On differentiating both sides w.r.t x,
(dy/dx) = 6e^6xcos3x – 3e^6xsin3x
Again differentiating both sides w.r.t x,
d²y/dx²= 6(6e^6xcos3x – 3e^6xsin3x) – 3(6e^6xsin3x + 3e^6xcos3x)
d²y/dx²= 36e^6xcos3x – 18e^6xsin3x – 18e^6xsin3x – 9e^6xcos3x
d²y/dx²= 27e^6xcos3x – 36e^6xsin3x
d²y/dx²= 9e^6x(3cos3x – 4sin3x)

Question 1(vi). x³logx

Solution:

Let us considered
y = x³logx
On differentiating both sides w.r.t x,
(dy/dx) = logx.3x²+ x³(1/x)
(dy/dx) = logx.3x²+ x²
(dy/dx) = x²(1 + 3logx)
Again differentiating both sides w.r.t x,
d²y/dx²= (1 + 3logx).2x + x²(3/x)
d²y/dx²= 2x + 6xlogx + 3x
d²y/dx²= x(5 + 6logx)

Question 1(vii). tan^-1x

Solution:

Let us considered

y = tan^-1x

On differentiating both sides w.r.t x,

(dy/dx) = 1/(1 + x²)

Again differentiating both sides w.r.t x,

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

d²y/dx²= (-1)(1 + x²)^-2.2x

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

Question 1(viii). x.cosx

Solution:

Let us considered
y = x.cosx
On differentiating both sides w.r.t x,
(dy/dx) = cosx + x(-sinx)
(dy/dx) = cosx – xsinx
Again differentiating both sides w.r.t x,
d²y/dx²= -sinx – (sinx + xcosx)
d²y/dx²= -2sinx – xcosx
d²y/dx²= -(xcosx + 2sinx)

Question 1(ix). log(logx)

Solution:

Let us considered

y = log(logx)

On differentiating both sides w.r.t x,

(dy/dx) = (1/logx) × (1/x)

(dy/dx) = 1/xlogx

Again differentiating both sides w.r.t x,

$\frac{d^2y}{dx^2}=\frac{d}{dx}(xlogx)^{-1}$

d²y/dx²= (-1)(xlogx)^-2.[(d/dx)xlogx]

d²y/dx² = (-1)(xlogx)^-2[logx+x.(1/x)]

d²y/dx²= (-1)(xlogx)^-2.(logx+1)

$\frac{d^2y}{dx^2}=\frac{-(1+logx)}{(xlogx)^2}$

Question 2. If y = e^-x.cosx, show that d²y/dx²= 2e^-x.sinx.

Solution:

Let us considered
y = e^-x.cosx
On differentiating both sides w.r.t x,
(dy/dx) = -e^-x.cosx – e^-x.sinx
Again differentiating both sides w.r.t x,
d²y/dx²= -(-e^-x.cosx – e^-x.sinx) – (-e^-x.sinx + e^-x.cosx)
d²y/dx²= e^-x.cosx – e^-x.cosx + e^-x.sinx + e^-x.sinx
d²y/dx²= 2e^-x.sinx
Hence Proved

Question 3. If y = x + tanx, Show that cos²x(d²y/dx²) – 2y + 2x = 0.

Solution:

Let us considered
y = x + tanx
On differentiating both sides w.r.t x,
(dy/dx) = 1 + sec²x
Again differentiating both sides w.r.t x,
d²y/dx²= 0 + (2secx)(secx.tanx)
d²y/dx²= 2sec²x.tanx
On multiplying both sides by cos²x
cos²x(d²y/dx²) = 2tanx
cos²x(d²y/dx²) = 2(y – x) [since, tanx = y – x]
cos²x(d²y/dx²) – 2y + 2x = 0
Hence Proved

Question 4. If y = x³logx, prove that (d⁴y/dx⁴) = (6/x).

Solution:

Let us considered
y = x³logx
On differentiating both sides w.r.t x,
(dy/dx) = logx.3x²+ x³(1/x)
(dy/dx) = logx.3x²+ x²
(dy/dx) = x²(1 + 3logx)
Again differentiating both sides w.r.t x,
d²y/dx²= (1 + 3logx).2x + x²(3/x)
d²y/dx2 = 2x + 6xlogx + 3x
d²y/dx²= 5x + 6xlogx
Again differentiating both sides w.r.t x,
d³y/dx³= 5 + 6[logx + (x/x)]
d³y/dx³= 11 + 6logx
Again differentiating both sides w.r.t x,
d⁴y/dx⁴= (6/x)
Hence Proved

Question 5. If y = log(sinx), prove that (d³y/dx³) = 2cosx.cosec³x.

Solution:

Let us considered
y = log(sinx)
On differentiating both sides w.r.t x,
(dy/dx) = (1/sinx) × (cosx)
(dy/dx) = cotx
Again differentiating both sides w.r.t x,
d²y/dx²= -cosec²x
Again differentiating both sides w.r.t x,
d³y/dx³= -2cosecx.(-cosesx.cotx)
d³y/dx³= 2cosec²x.cotx
d³y/dx³= 2cosec²x.(cosx/sinx)
d³y/dx³= cosx.cosec³x
Hence Proved

Question 6. If y = 2sinx + 3cosx, show that (d²y/dx²) + y = 0.

Solution:

Let us considered
y = 2sinx + 3cosx
On differentiating both sides w.r.t x,
(dy/dx) = 2cosx – 3sinx
Again differentiating both sides w.r.t x,
d²y/dx²= -2sinx – 3cosx
d²y/dx²= -(2sinx + 3cosx)
d²y/dx²= -y
d²y/dx²+ y = 0
Hence Proved

Question 7. If y = (logx/x), show that (d²y/dx²) = (2logx – 3)/x³

Solution:

Let us considered

y = (logx/x)

On differentiating both sides w.r.t x,

$\frac{dy}{dx}=\frac{x\frac{d}{dx}logx-logx\frac{dx}{dx}}{x^2}$

(dy/dx) = (1 – logx)/x²

Again differentiating both sides w.r.t x,

$\frac{d^2y}{dx^2}=\frac{x^2\frac{d}{dx}(1-logx)-(1-logx)\frac{dx}{dx}}{x^4}$

d²y/dx²= [-x – 2x(1 – logx)]/x⁴

d²y/dx²= (2xlogx – 3x)/x⁴

d²y/dx²= (2logx – 3)/x³

d2y/dx2 + y = 0

Hence Proved

Question 8. If x = a secθ, y = b tanθ, show that (d²y/dx²) = -b⁴/a²y³

Solution:

We have,
x = a secθ and y = b tanθ
On differentiating both sides w.r.t θ,
(dx/dθ) = a secθ.tanθ, (dy/dθ) = b sec²θ
(dy/dx) = (dy/dθ) × (dθ/dx)
(dy/dx) = (b sec²θ)/(a secθ.tanθ)
(dy/dx) = (b/a).cosecθ
Again differentiating both sides w.r.t x,
(d²y/dx²) = (b/a).(-cosecθ.cotθ).(dθ/dx)
(d²y/dx²) = -(b/a).(cosecθ.cotθ).(1/a secθ.tanθ)
(d²y/dx²) = – (b/a²).(cotθ).(1/tan²θ)
d²y/dx²= -(b/a²).(1/tan³θ)
d²y/dx²= -(b/a²tan³θ).(b³/b³)
d²y/dx²= -(b⁴/a²y³)
Hence Proved

Question 9. If x = a(cost + tsint) and y = a(sint – tcost), prove that d²y/dx²= sec³t/ at 0 < t < π/2.

Solution:

We have,
x = a(cost + tsint)and y=a(sint – tcost)
On differentiating both sides w.r.t t,
(dx/dt) = a(-sint + sint + tcost), (dy/dt) = a(cost – cost + tsint)
(dx/dt) = atcost, (dy/dt) = atsint
(dy/dx) = (dy/dt) × (dt/dx)
(dy/dx) = atsint × [1/atcost]
(dy/dx) = tant
Again differentiating both sides w.r.t x,
(d²y/dx²) = sec²x.(dt/dx)
(d²y/dx²) = sec2x.[1/atcost]
(d²y/dx²) = sec³x/at
Hence Proved

Question 10. If y = e^xcosx, prove that d²y/dx²= 2e^xcos(x + π/2).

Solution:

We have,
y = e^xcosx
On differentiating both sides w.r.t x,
(dy/dx) = e^xcosx – e^xsinx
Again differentiating both sides w.r.t x,
d²y/dx²= (e^xcosx – e^xsinx) – (e^xsinx + e^xcosx)
d²y/dx²= e^xcosx – e^xcosx – e^xsinx – e^xsinx
d²y/dx²= -2e^xsinx
d²y/dx²= 2e^xcos(x + π/2)

Question 11. If x = a cosθ, y = b sinθ, show that (d²y/dx²) = -b⁴/a²y³

Solution:

We have,
x = a cosθ and y = b sinθ
On differentiating both sides w.r.t θ,
(dx/dθ) = -a sinθ, (dy/dθ) = b cosθ
(dy/dx) = (dy/dθ)×(dθ/dx)
(dy/dx) = (b cosθ)/(-a sinθ)
(dy/dx) = -(b/a).cotθ
Again differentiating both sides w.r.t x,
(d²y/dx²) = -(b/a).(-cosec²θ).(dθ/dx)
(d²y/dx²) = (b/a).(cosec²θ).(1/-a sinθ)
(d²y/dx²) = (b/a).(cosec²θ).(1/-a sinθ)
d²y/dx²= -(b/a²).(1/sin³θ)
d²y/dx²=-(b/a²sin³θ).(b³/b³)
d²y/dx2 = -(b⁴/a²y³) (since y = a sinθ)
Hence Proved

Question 12. If x = a(1 – cos³θ), y = a sin³θ, show that (d²y/dx²) = 32/27a, at θ = π/6.

Solution:

We have,

x = a(1 – cos³θ) and y = a sin³θ

On differentiating both sides w.r.t θ,

(dx/dθ) = a(3cos²θ.sinθ), (dy/dθ) = a 3sin²θcosθ

(dx/dθ) = 3acos²θ.sinθ, (dy/dθ) = 3asin²θ.cosθ

(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = (3asin²θ.cosθ) × (3acos²θ.sinθ)

(dy/dx) = tan²θ/tanθ

(dy/dx) = tanθ

Again differentiating both sides w.r.t x,

(d²y/dx²) = sec²θ(dθ/dx)

(d²y/dx²) = sec²θ.[1/3acos²θ.sinθ]

(d²y/dx²) = sec⁴θ/3asinθ

At θ = π/6

d²y/dx²= sec⁴(π/6)/3asin(π/6)

$\frac{d^2y}{dx^2}=\frac{(\frac{2}{\sqrt{3}})^4}{3a\frac{1}{2}}$

d²y/dx²= 32/27a

Hence Proved

Question 13. If x = a(θ + sinθ), y = a(1 + cosθ), prove that (d²y/dx²) = -(a/y²).

Solution:

We have,

x = a(θ + sinθ) and y = a(1 + cosθ)

On differentiating both sides w.r.t θ,

(dx/dθ) = a(1 + cosθ), (dy/dθ) = -asinθ

(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = [-asinθ] × [a(1 + cosθ)]

(dy/dx) = -sinθ/(1 + cosθ)

Again differentiating both sides w.r.t x,

$\frac{d^2y}{dx^2}=-[\frac{(1+cosθ)\frac{d}{dx}sinθ+sinθ\frac{d}{dx}(1+cosθ)}{(1+cosθ)^2}]\frac{dθ}{dx}$

$\frac{d^2y}{dx^2}=-[\frac{(1+cosθ)cosθ-sinθsinθ)}{(1+cosθ)^2}]\frac{1}{a(1+cosθ)}$

$\frac{d^2y}{dx^2}=[\frac{-cosθ-cos^2θ-sin^2θ}{(1+cosθ)^2}]\frac{1}{a(1+cosθ)}$

$\frac{d^2y}{dx^2}=\frac{-cosθ-1}{a(1+cosθ)^3}$

(d²y/dx²) = -(1 + cosθ)/a(1 + cosθ)³

(d²y/dx²) = -1/a(1 + cosθ)²

(d²y/dx²) = -[1/a(1 + cosθ)²](a/a)

d2y/dx2 = -a/y²

Hence Proved

Question 14. If x = a(θ – sinθ), y = a(1 + cosθ), find (d²y/dx²).

Solution:

We have,

x = a(θ – sinθ) and y = a(1 + cosθ)

On differentiating both sides w.r.t θ,

(dx/dθ) = a(1 – cosθ), (dy/dθ) = -asinθ

(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = [-asinθ] × [a(1 – cosθ)]

(dy/dx) = -sinθ/(1 – cosθ)

Again differentiating both sides w.r.t x,

$\frac{d^2y}{dx^2}=-[\frac{(1-cosθ)\frac{d}{dx}sinθ+sinθ\frac{d}{dx}(1-cosθ)}{(1-cosθ)^2}]\frac{dθ}{dx}$

$\frac{d^2y}{dx^2}=-[\frac{(1-cosθ)cosθ-sinθsinθ)}{(1-cosθ)^2}]\frac{1}{a(1-cosθ)}$

$\frac{d^2y}{dx^2}=[\frac{-cosθ+cos^2θ+sin^2θ}{(1-cosθ)^2}]\frac{1}{a(1-cosθ)}$

$\frac{d^2y}{dx^2}=\frac{1-cosθ}{a(1-cosθ)^3}$

(d²y/dx²) = 1/a(1 – cosθ)²

$\frac{d^2y}{dx^2}=\frac{1}{a(2sin^2\frac{θ}{2})^2}$

d²y/dx²= (1/4a)[cosec⁴(θ/2)]

Question 15. If x = a(1 – cosθ), y = a(θ + sinθ), prove that (d²y/dx²) = -1/a at θ = π/2.

Solution:

We have,

x = a(1 – cosθ) and y = a(θ + sinθ)

On differentiating both sides w.r.t θ,

(dx/dθ) = a(sinθ), (dy/dθ) = a(1 + cosθ)

(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = [a(1 + cosθ)] × [asinθ)]

(dy/dx) = (1 + cosθ)/sinθ

Again differentiating both sides w.r.t x,

$\frac{d^2y}{dx^2}=[\frac{sinθ\frac{d}{dx}(1+cosθ)-(1+cosθ)\frac{d}{dx}sinθ}{(sinθ)^2}]\frac{dθ}{dx}$

$\frac{d^2y}{dx^2}=[\frac{sinθ.sinθ-(1+cosθ)cosθ}{(sinθ)^2}]\frac{1}{asinθ}$

d²y/dx²= (-sin²θ – cosθ – cos²θ)/asin³θ

d²y/dx²= -(sin²θ + cosθ + cos²θ)/asin³θ

At θ = π/2,

d²y/dx²= -(1 + 0)/a

d²y/dx²= -(1/a)

Hence Proved

Question 16. If x = a(1 + cosθ), y = a(θ + sinθ), prove that (d²y/dx²) = -1/a at θ = π/2.

Solution:

We have,

x = a(1 + cosθ) and y = a(θ + sinθ)

On differentiating both sides w.r.t θ,

(dx/dθ) = a(-sinθ), (dy/dθ) = a(1 + cosθ)

(dy/dx) = (dy/dθ) × (dθ/dx)

(dy/dx) = [a(1 + cosθ)] × [-asinθ)]

(dy/dx) = -(1 + cosθ)/sinθ

Again differentiating both sides w.r.t x,

$\frac{d^2y}{dx^2}=-[\frac{sinθ\frac{d}{dx}(1+cosθ)-(1+cosθ)\frac{d}{dx}sinθ}{(sinθ)^2}]\frac{dθ}{dx}$

$\frac{d^2y}{dx^2}=-[\frac{sinθ.sinθ-(1+cosθ)cosθ}{(sinθ)^2}]\frac{1}{-asinθ}$

d²y/dx²= (-sin²θ – cosθ – cos²θ)/asin³θ

d²y/dx²= -(sin²θ + cosθ + cos²θ)/asin³θ

At θ = π/2,

d²y/dx²= -(1 + 0)/a

d²y/dx²= -(1/a)

Hence Proved

Question 17. If x = cosθ, y = sin³θ, prove that y(d²y/dx²) + (dy/dx)²= 3sin²θ(5cos²θ – 1).

Solution:

We have,
x = cosθ and y = sin³θ
On differentiating both sides w.r.t θ,
(dx/dθ) = -sinθ, (dy/dθ) = 3sin²θ.cosθ
(dy/dx) = (dy/dθ) × (dθ/dx)
(dy/dx) = [3sin²θ.cosθ] × [-sinθ]
(dy/dx) = -3sinθ.cosθ
Again differentiating both sides w.r.t x,
d²y/dx²= -3[sinθ(-sinθ) + cosθ.cosθ](dθ/dx)
d²y/dx²= (3sin²θ – 3cos²θ)/-sinθ
d²y/dx²= -(3sin²θ – 3cos²θ)/sinθ
L.H.S,
y(d²y/dx²) + (dy/dx)²= -sin³θ[(3sin²θ – 3cos²θ)/sinθ] + (-3sinθ.cosθ)²
= 3sin²θ.cos²θ – 3sin⁴θ + 9sin²θ.cos²θ
= 12sin²θ.cos²θ – 3sin⁴θ
= 3sin²θ(4cos²θ – sin²θ)
= 3sin²θ(4cos²θ – sin²θ – cos²θ + cos²θ)
= 3sin²θ[5cos²θ – (sin²θ + cos²θ)]
= 3sin²θ(5cos²θ – 1)
= R.H.S
L.H.S = R.H.S
Hence Proved

Question 18. If y = sin(sinx), prove that (d²y/dx²) + tanx.(dy/dx) + ycos²x = 0

Solution:

We have,
y = sin(sinx)
On differentiating both sides w.r.t x,
(dy/dx) = cos(sinx).cosx
Again differentiating both sides w.r.t x,
d²y/dx²= -sin(sinx).cosx.cosx – cos(sinx).sinx
d²y/dx²= -sin(sinx).cos²x – cos(sinx).sinx
d²y/dx²= -sin(sinx).cos²x – cos(sinx).cosx.tanx
d²y/dx²= -ycos²x – (dy/dx)tanx
d²y/dx²+ ycos²x + (dy/dx)tanx = 0
Hence Proved

Question 19. If x = sin t, y = sin pt, prove that (1 – x²)(d²y/dx²) – x.(dy/dx) + p²y = 0

Solution:

We have,

x = sin t, and y = sin pt

On differentiating both sides w.r.t t,

(dx/dt) = cos t, (dy/dt) = pcos pt

(dy/dx) = (dy/dt) × (dt/dx)

(dy/dx) = pcos pt×[1/cos t]

(dy/dx) = pcos pt/cos t

Again differentiating both sides w.r.t x,

$\frac{d^2y}{dx^2}=\frac{-p^2sin pt.cos t+pcos pt.sint}{cos^2t}.\frac{dt}{dx}$

d²y/dx²= (-p²sin pt.cos t + pcos pt.sin t)/cos³t

d²y/dx²= -(p²sin pt)/cos²t + (pcos pt.sin t)/cos³t

$\frac{d^2y}{dx^2}=-\frac{p^2y}{cos^2t}+\frac{x\frac{dy}{dx}}{cos^2y}$

cos²t(d²y/dx²) = -p²y + x(dy/dx)

(1 – sin²x)(d²y/dx²) + p²y – x(dy/dx) = 0

(1 – y²)(d²y/dx²) + p²y – x(dy/dx) = 0

Question 20. If y = (sin^-1x)², prove that (1 – x²)(d²y/dx²) – x.(dy/dx) + p²y = 0.

Solution:

We have,

y = (sin^-1x)²,

On differentiating both sides w.r.t t,

$\frac{dy}{dx}=2sin^{-1}x.\frac{1}{\sqrt{1-x^2}}$

Again differentiating both sides w.r.t x,

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

$\frac{d^2y}{dx^2}=\frac{2xsin^{-1}x}{(1-x^2)\sqrt{1-x^2}}+\frac{2}{1-x^2}$

d²y/dx²= [x/(1 – x²)](dy/dx) + 2/(1 – x²)

(1 – x²)d²y/dx²= x(dy/dx) + 2

(1 – x²)d²y/dx²– x(dy/dx) – 2 = 0

Hence Proved

Question 21. If y = $e^{tan^{-1}x}$ , prove that (1 + x²)y₂+ (2x – 1)y₁= 0.

Solution:

We have,
y = $e^{tan^{-1}x}$
On differentiating both sides w.r.t t,
y₁= $e^{tan^{-1}x}$ × [1/(1 + x²)]
Again differentiating both sides w.r.t x,
y₂= $e^{tan^{-1}x}\frac{1}{(1+x^2)^2}+e^{tan^{-1}x}\frac{-2x}{(1+x^2)^2}$
(1 + x²)y₂= $e^{tan^{-1}x}$ /(1 + x²) – 2x $e^{tan^{-1}x}$ /(1 + x²)
(1 + x²)y₂= (dy/dx) – 2x(dy/dx)
(1 + x²)y₂– (dy/dx) + 2x(dy/dx) = 0
(1 + x²)y₂+ (2x – 1)(dy/dx) = 0
Hence Proved

Question 22. If y = 3cos(logx) + 4sin(logx), prove that x²y₂+ xy₁+ y = 0.

Solution:

We have,
y = 3cos(logx) + 4sin(logx)
On differentiating both sides w.r.t x,
y₁ = -3sin(logx) × (1/x) + 4cos(logx) × (1/x)
xy₁= -3sin(logx) + 4cos(logx)
Again differentiating both sides w.r.t x,
xy₂+ y₁= -3cos(logx)×(1/x) – 4sin(logx) × (1/x)
x²y₂+ xy₁= -[3cos(logx) + 4sin(logx)]
x²y₂+ xy₁= -y
x²y₂+ xy₁+ y = 0
Hence Proved

Question 23. If y = e^2x(ax + b), show that y₂– 4y₁+ 4y = 0.

Solution:

We have,
y = e^2x(ax + b)
On differentiating both sides w.r.t θ,
y₁= 2e^2x(ax + b) + a.e^2x
Again differentiating both sides w.r.t x,
y₂= 4e^2x(ax + b) + 2ae^2x+ 2a.e^2x
y₂= 4e^2x(ax + b) + 4a.e^2x
Lets take L.H,S,
= y₂– 4y₁+ 4y
= 4e^2x(ax + b) + 4a.e^2x– 4[2e^2x(ax + b) + a.e^2x] + 4[e^2x(ax + b)]
= 8e^2x(ax + b) – 8e^2x(ax + b) + 4a.e^2x– 4a.e^2x
= 0
= R.H.S
L.H.S = R.H.S
Hence Proved

Question 24. If x = sin(logy/a), show that (1 – x²)y₂– xy₁– a²y = 0.

Solution:

We have,
x = sin(logy/a)
(logy/a) = sin^-1x
logy = asin^-1x
On differentiating both sides w.r.t x,
(1/y)y₁= a/√(1 – x²)
y₁= ay/√(1 – x²)
Again differentiating both sides w.r.t x,
y₂ $=a[\frac{\sqrt{1-x^2}\frac{dy}{dx}-\frac{2xy}{2\sqrt{1-x^2}}}{1-x^2}]$
(1 – x²)y₂= a√(1 – x²) × y₁+ axy/√(1 – x²)
(1 – x²)y₂= a√(1 – x²) × [ay/√(1 – x²)] + x[ay/√(1 – x²)]
(1 – x²)y₂= a²p + xy
(1 – x²)y₂– a²p – xy₁= 0
Hence Proved

Question 25. If logy = tan^-1x, show that (1 + x²)y₂+ (2x – 1)y₁= 0.

Solution:

We have,
logy = tan^-1x
On differentiating both sides w.r.t θ,
(1/y)y₁= 1/(1 + x²)
(1 + x²)y₁= y
Again differentiating both sides w.r.t x,
2xy₁+ (1 + x²)y₂= y₁
(1 + x²)y₂+ (2x – 1)y₁= 0
Hence Proved

Question 26. If y = tan^-1x, show that (1 + x²)(d²y/dx²) + 2x(dy/dx) = 0.

Solution:

We have,
y = tan^-1x
On differentiating both sides w.r.t x,
(dy/dx) = 1/(1 + x²)
Again differentiating both sides w.r.t x,
d²y/dx²= [-1/(1 + x²)²] × (2x)
d²y/dx²= [-2x/(1 + x²)²]
(1 + x²)(d²y/dx²) = -2x/(1 + x²)
(1 + x²)(d²y/dx²) = -2x(dy/dx)
(1 + x²)(d²y/dx²) + 2x(dy/dx) = 0
Hence Proved
Question 27. If y = [log{x+(√x²+1)}]², show that (1 + x²)(d²y/dx²) + x(dy/dx) = 2.
Solution:

We have,
y = [log{x + (√x²+ 1)}]²
On differentiating both sides w.r.t x,
$\frac{dy}{dx}=2log(x+\sqrt{x^2+1}).\frac{d}{dx}(x+\sqrt{x^2+1})$
$\frac{dy}{dx}=\frac{2log(x+\sqrt{x^2+1})}{(x+\sqrt{x^2+1})}(1+\frac{2x}{2\sqrt{x^2+1}})$
$\frac{dy}{dx}=\frac{2log(x+\sqrt{x^2+1})}{(x+\sqrt{x^2+1})}(\frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}})$
dy/dx = 2[log{x + (√x²+ 1)}]/(√x²+ 1)
Again differentiating both sides w.r.t x,
$\frac{d^2y}{dx^2}=\frac{2-\frac{2xlog(x+\sqrt{x^2+1}}{x^2+1}}{x^2+1}$
$\frac{d^2y}{dx^2}=\frac{2-x\frac{dy}{dx}}{x^2+1}$
(x²+ 1)(d²y/dx²) = 2 – x(dy/dx)
(x²+ 1)(d²y/dx²) + x(dy/dx) = 2
Hence Proved
Question 28. If y = (tan^-1x)², then prove that (1 + x²)²y₂+ 2x(1 + x²)y₁= 2
Solution:
We have,
y = (tan^-1x)²
On differentiating both sides w.r.t x,
y₁= 2(tan^-1x)[1/(1 + x²)]
(1 + x²)y₁= 2(tan^-1x)
Again differentiating both sides w.r.t x,
(1 + x²)y₂+ 2xy₁= 2/(1 + x²)
(1 + x²)²y₂+ 2x(1 + x²)y₁= 2
Hence Proved
Question 29. If y = cotx, prove that (d²y/dx²) + 2y(dy/dx) = 0.
Solution:
We have,
y = cotx
On differentiating both sides w.r.t x,
(dy/dx) = -cosec²x
Again differentiating both sides w.r.t x,
d²y/dx²= -(2cosec x) × (-cosec x.cot x)
d²y/dx²= 2cosec²x.cot x
d²y/dx²= 2(cot x)(cosec²x)
d²y/dx²= -2y(dy/dx)
d²y/dx²+ 2y(dy/dx) = 0
Hence Proved
Question 30. Find d²y/dx² where y = log(x²/e²).
Solution:
We have,
y = log(x²/e²)
On differentiating both sides w.r.t x,
$\frac{dy}{dx}=\frac{1}{\frac{x^2}{e^2}}×(\frac{2x}{e^2})$
(dy/dx) = (2/x)
Again differentiating both sides w.r.t x,
d²y/dx²= -(2/x²)
Hence Proved
Question 31. If y = ae^2x+ be^-x, show that (d²y/dx²) – (dy/dx) – 2y = 0.
Solution:
We have,
y = ae^2x+ be^-x
On differentiating both sides w.r.t t,
(dy/dx) = 2ae^2x– be^-x
Again differentiating both sides w.r.t x,
(d²y/dx²) = 4ae^2x+ be^-x
(d²y/dx²) = 2ae^2x– be^-x+ 2(ae^2x+ be^-x)
(d²y/dx²) = (dy/dx) + 2y
(d²y/dx²) – (dy/dx) – 2y = 0
Hence Proved
Question 32. If y = e^x(sinx + cosx), prove that d²y/dx²– 2(dy/dx) + 2y = 0.
Solution:
We have,
y = e^x(sinx + cosx)
On differentiating both sides w.r.t x,
(dy/dx) = e^x(sinx + cosx) + e^x(cosx – sinx)
(dy/dx) = 2e^xcosx
Again differentiating both sides w.r.t x,
(d²y/dx²) = 2e^xcosx – 2e^xsinx
Lets take L.H.S,
= d²y/dx²– 2(dy/dx) + 2y
= 2e^xcosx – 2e^xsinx – 2(2e^xcosx) + 2e^x(sinx + cosx)
= 4e^xcosx – 4e^xcosx – 2e^xsinx + 2e^xsinx
= 0
L.H.S = R.H.S
Hence Proved
Question 33. If y = cos^-1x, find d²y/dx² in terms of y alone.
Solution:
We have,
y = cos^-1x
On differentiating both sides w.r.t x,
(dy/dx) = -1/√(1-x²)
Again differentiating both sides w.r.t x,
$\frac{d^2y}{dx^2}=\frac{-2x}{(2\sqrt{1-x^2})\frac{3}{2}}$ …(i)
y = cos^-1x
x = cosy
On putting the value of x in equation (i), we get
$\frac{d^2y}{dx^2}=\frac{-cosy}{(\sqrt{1-cos^2y})\frac{3}{2}}$
$\frac{d^2y}{dx^2}=\frac{-cosy}{(sin^2y)\frac{3}{2}}$
d²y/dx²= -cosy/sin³y
d²y/dx²= -cot y cosec²y
Question 34. If y = $e^{acos^{-1}x}$ , prove that (1 – x²)(d²y/dx²) – x(dy/dx) – a²y = 0.
Solution:
We have,
y = $e^{acos^{-1}x}$
Taking log both sides
logy = acos^-1x.loge
logy = acos^-1x
On differentiating both sides w.r.t x,
(1/y)(dy/dx) = a×[-1/√(1-x²)]
(dy/dx) = -ay/√(1-x²)
On squaring both sides, we have
(dy/dx)²= a²y²/(1 – x²)
(1 – x²)(dy/dx)²= a²y²
Again differentiating both sides w.r.t x,
2(1 – x²)(dy/dx)(d²y/dx²) – 2x(dy/dx)²= 2a²y(dy/dx)
(1 – x²)(d²y/dx²) – x(dy/dx) = a²y
(1 – x²)(d²y/dx²) – x(dy/dx) – a²y = 0
Hence Proved
Question 35. If y = 500e^7x+ 600e^-7x, show that d²y/dx²= 49y.
Solution:
We have,
y = 500e^7x+ 600e^-7x
On differentiating both sides w.r.t θ,
(dy/dx) = 7 × (500e^7x– 600e^-7x)
Again differentiating both sides w.r.t x,
(d²y/dx²) = 49 × (500e^7x+ 600e^-7x)
(d²y/dx²) = 49y
Hence Proved
Question 36. If x = 2cos t – cos 2t, y = 2sin t – sin 2t, find d²y/dx²at t = π/2.
Solution:
We have,
x = 2cos t – cos 2t, and y = 2sin t – sin 2t
On differentiating both sides w.r.t t,
(dx/dt) = -2sin t + 2sin 2t, (dy/dt) = 2cos t – 2cos 2t
(dy/dx) = (dy/dt) × (dt/dx)
(dy/dx) = (2cos t – 2cos 2t)/(-2sin t + 2sin 2t)
(dy/dx) = (cos t – cos 2t)/(-sin t + sin 2t)
Again differentiating both sides w.r.t x,
$\frac{d^2y}{dx^2}=\frac{(-sin t+2sin 2t)(-sin t+sin 2t)-(cos t-cos 2t)(-cos t+2cos 2t)}{(-sin t+sin 2t)^2}.\frac{dt}{dx}$
$\frac{d^2y}{dx^2}=\frac{(-sin t+2sin 2t)(-sin t+sin 2t)-(cos t-cos 2t)(-cos t+2cos 2t)}{(-sin t+sin 2t)^2(-2sin t+2sin 2t)}$
At t = π/2
$\frac{d^2y}{dx^2}=\frac{(-1+0)(-1+0)-(0+1)(0-2)}{(-1+0)^2(-2+0)}$
d²y/dx²= (1 + 2)/-2
d²y/dx²= -(3/2)
Question 37. If x = 4z²+ 5, y = 6z²+ 7z + 3, find d²y/dx².
Solution:
We have,
x = 4z²+ 5, and y = 6z²+ 7z + 3
On differentiating both sides w.r.t z,
(dx/dz) = 8z, and (dy/dz) = 12z + 7
(dy/dx) = (dy/dz) × (dz/dx)
(dy/dx) = (12z + 7)/8z
Again differentiating both sides w.r.t x,
$\frac{d^2y}{dx^2}=\frac{12×8z-8(12z+7)}{64z^2}.\frac{dz}{dx}$
$\frac{d^2y}{dx^2}=\frac{96z-96z-56}{64z^2}.\frac{1}{8z}$
(d²y/dx²) = -7/64z³
Hence Proved
Question 38. If y = log(1 + cosx), prove that d³y/dx³+ (d²y/dx²).(dy/dx) = 0.
Solution:
We have,
y = log(1 + cosx)
On differentiating both sides w.r.t x,
(dy/dx) = -sinx/(1 + cosx)
Again differentiating both sides w.r.t x,
$\frac{d^2y}{dx^2}=\frac{-cosx(1+cosx)-(-sinx)(-cosx)}{(1+cosx)^2}$
d²y/dx²= (-cosx – cos²x – sin²x)/(1 + cosx)²
d²y/dx²= -(1 + cosx)/(1 + cosx)²
d²y/dx²= -1/(1 + cosx)
Again differentiating both sides w.r.t x,
d³y/dx³= -sinx/(1 + cosx)²
d³y/dx³+ [-1/(1 + cosx)][-sinx/(1 + cosx)] = 0
d³y/dx³+ (d²y/dx²).(dy/dx) = 0
Hence Proved
Question 39. If y = sin(logx), prove that x²(d²y/dx²) + x(dy/dx) + y = 0.
Solution:
We have,
y = sin(logx)
On differentiating both sides w.r.t x,
(dy/dx) = cos(logx).(1/x)
x(dy/dx) = cos(logx)
Again differentiating both sides w.r.t x,
x(d²y/dx²) + (dy/dx) = -sin(logx).(1/x)
x²(d²y/dx²) + x(dy/dx) = -sin(logx)
x²(d²y/dx²) + x(dy/dx) = -y
x²(d²y/dx²) + x(dy/dx) + y = 0
Hence Proved
Question 40. If y = 3e^2x+ 2e^3x, prove that d²y/dx²– 5(dy/dx) + 6y = 0.
Solution:
We have,
y = 3e^2x+ 2e^3x
On differentiating both sides w.r.t x,
(dy/dx) = 6e^2x+ 6e^3x
(dy/dx) = 6(e^2x+ e^3x)
Again differentiating both sides w.r.t x,
d²y/dx²= 6(2e^2x+ 3e^3x)
d²y/dx²= 12e^2x+ 18e^3x
d²y/dx²= 5(6e^2x+ 6e^3x) – 6(3e^2x+ 2e^3x)
d²y/dx²= 5(dy/dx) – 6y
d²y/dx²– 5(dy/dx) + 6y = 0
Hence Proved
Question 41. If y = (cot^-1x)², prove that y₂(x²+ 1)²+ 2x(x²+ 1)y₁= 2.
Solution:
We have,
y = (cot^-1x)²
On differentiating both sides w.r.t x,
y₁= 2(cot^-1x) × [-1/(1 + x²)]
(1 + x²)y₁= -2cot^-1x
Again differentiating both sides w.r.t x,
(1 + x²)y₂+ 2xy₁= 2/(1 + x²)
(1 + x²)²y₂+ 2x(1 + x²)y₁= 2
Hence Proved
Question 42. If y = cosec^-1x, then show that x(x²– 1)d²y/dx²– (2x²– 1)(dy/dx) = 0.
Solution:
We have,
y = cosec^-1x
On differentiating both sides w.r.t x,
(dy/dx) = -1/x√(x²– 1)
On squaring both sides,
(dy/dx)²= 1/x²(x²– 1)
x²(x²– 1)(dy/x)²= 1
(x⁴– x²)(dy/dx)²= 1
2(dy/dx)(d²y/dx²)(x⁴– x²) + (dy/dx)²(4x³– 2x) = 0
2x²(x²– 1)(dy/dx)(d²y/dx²) + 2x(2x²– 1)(dy/dx)²= 0
x(x²– 1)(d²y/dx²) + (2x²– 1)(dy/dx) = 0
Hence Proved
Question 43. If x = cos t + log(tant/2), y = sin t, then find the value of d²y/dt² and d²y/dx² at t = π/4 in terms of y alone.
Solution:
We have,
y = sin t
On differentiating both sides w.r.t t,
(dy/dt) = cos t
Again differentiating both sides w.r.t x,
(d²y/dx²) = -sin t
At t = π/4
(d²y/dx²)_t=π/4= -sin(π/4)
= -1/√2
x = cos t + log(tant/2)
On differentiating both sides w.r.t t,
$\frac{dx}{dt}=-sin t+\frac{1}{tan\frac{t}{2}}.sec^2\frac{t}{2}.\frac{1}{2}$
$\frac{dx}{dt}=-sin t+\frac{1}{2sin\frac{t}{2}cos\frac{t}{2}}$
(dx/dt) = -sin t + (1/sin t)
(dx/dt) = (-sin²t + 1)/sin t
(dx/dt) = cos²t/sint
(dx/dt) = cos t × cot t
(dy/dx) = (dy/dt) × (dt/dx)
(dy/dx) = [cos t] × [1/cos t × cot t]
(dy/dx) = tan t
Again differentiating both sides w.r.t x,
(d²y/dx²) = sec²t × (dt/dx)
(d²y/dx²) = sec²t × [1/cos t × cot t]
(d²y/dx²) = sin t/cos⁴t
(d²y/dx²)_t=π/4= sin(π/4)/cos⁴(π/4)
(d²y/dx²) = 2√2
At t = π/4, (d²y/dx²) = -1/√2 and (d²y/dx²) = 2√2
Question 44. If x = asin t, y = a[cos t + log(tant/2)], find d²y/dx².
Solution:
We have,
x = asin t, and y = a[cos t + log(tant/2)]
On differentiating both sides w.r.t t,
(dx/dt) = acos t and $\frac{dy}{dt}=-asin t+a\frac{1}{tan\frac{t}{2}}.sec^2\frac{t}{2}.\frac{1}{2}$
$\frac{dy}{dt}=-asin t+a\frac{1}{2sin\frac{t}{2}cos\frac{t}{2}}$
(dy/dt) = a[-sin t + (1/sin t)]
(dy/dt) = a[(-sin²t + 1)/sin t]
(dy/dt) = a[cos²t/sint]
(dy/dt) = acos t × cot t
(dy/dx) = (dy/dt) × (dt/dx)
(dy/dx) = [acos t × cot t] × [1/acos t]
(dy/dx) = cot t
Again differentiating both sides w.r.t x,
(d²y/dx²)=-cosec²t × (dt/dx)
(d²y/dx²) = -cosec²t × [1/acos t]
(d²y/dx²) = -(1/asin²t × cos t)
Question 45. If x = a(cos t + tsin t), and y = a(sin t – tcos t), then find the value of d²y/dx² at t = π/4.
Solution:
We have,
x = a(cos t + tsin t), and y = a(sin t – tcos t)
On differentiating both sides w.r.t t,
(dx/dt) = a(-sin t + sin t + tcos t)
(dx/dt) = atcos t
y = a(sin t – tcos t)
On differentiating both sides w.r.t t,
(dy/dx) = a(cos t – cos t + tsin t)
(dy/dx) = atsin t
(dy/dx) = (dy/dt) × (dt/dx)
(dy/dx) = [atsin t] × [1/atcos t]
(dy/dx) = tan t
Again differentiating both sides w.r.t x,
(d²y/dx²) = sec²x × (dt/dx)
(d²y/dx²) = sec²x × (1/atcos t)
(d²y/dx²) = 1/atcos³t
$\frac{d^2y}{dx^2}=\frac{1}{a.\frac{π}{4}.cos^3\frac{π}{4}}$
(d²y/dx²) = (8√2/aπ)
Question 46. If x = a[cos t + log(tant/2)], y = asin t, evaluate (d²y/dx²) at t = π/3.
Solution:
We have,
y = asin t
On differentiating both sides w.r.t t,
(dy/dt) = acos t
x = a[cos t + log(tant/2)]
On differentiating both sides w.r.t t,
$\frac{dx}{dt}=a[-sin t+\frac{1}{tan\frac{t}{2}}.sec^2\frac{t}{2}.\frac{1}{2}]$
$\frac{dx}{dt}=a[-sin t+\frac{1}{2sin\frac{t}{2}cos\frac{t}{2}}]$
(dx/dt) = a[-sin t + (1/sin t)]
(dx/dt) = a[(-sin²t + 1)/sin t]
(dx/dt) = a[cos²t/sint]
(dx/dt) = acos t × cot t
(dy/dx) = (dy/dt) × (dt/dx)
(dy/dx) = [cos t] × [1/cos t × cot t]
(dy/dx) = tan t
Again differentiating both sides w.r.t x,
(d²y/dx²) = sec²t × (dt/dx)
(d²y/dx²) = sec²t × [1/acos t × cot t]
(d²y/dx²) = sin t/acos⁴t
(d²y/dx²)_t=π/3= sin(π/3)/acos⁴(π/3)
(d²y/dx²) = (8√3/a)
Question 47. If x = a(cos2t + 2tsin2t), and y = a(sin2t – 2tcos2t), then find d²y/dx².
Solution:
We have,
x = a(cos2t + 2tsin2t), and y = a(sin2t – 2tcos2t)
On differentiating both sides w.r.t t,
(dx/dt) = a(-2sin2t + 2sin2t + 4tcos2t), and (dy/dt) = a(2cos2t – 2cos2t + 4tsin2t)
(dy/dt) = a(4tcos2t), and (dy/dt)=a(4tsin2t)
(dy/dx) = (dy/dz) × (dz/dx)
(dy/dx) = a(4tsin2t)/a(4tcos2t)
(dy/dx) = tan2t
Again differentiating both sides w.r.t x,
(d²y/dx²) = 2sec²2t.(dt/dx)
(d²y/dx²) = 2sec²2t/4atcos2t
(d²y/dx²) = 1/2atcos³2t
(d²y/dx²) = (1/2at) × (sec³x)
Question 48. If x = asin t – bcos t, y = acos t + bsin t, prove that (d²y/dx²) = -(x²+ y²)/y³
Solution:
We have,
x = asin t – bcos t
On differentiating both sides w.r.t t,
(dx/dt) = acos t + bsin t
y = acos t + b sin t
On differentiating both sides w.r.t t,
(dy/dt) = -asin t + bcos t
(dy/dx) = (dy/dt) × (dt/dx)
(dy/dx) = [-asin t + bcos t] × [1/(acos t + bsin t)]
Again differentiating both sides w.r.t x,
$\frac{d^2y}{dx^2}=\frac{(acost+bsint)(-asint-bcost)-(-asint+bcost)(-asint+bcost)}{(acost+bsint)^2}.\frac{dt}{dx}$
$\frac{d^2y}{dx^2}=\frac{-(acost+bsint)^2-(-asint+bcost)^2}{(acost+bsint)^3}$
$\frac{d^2y}{dx^2}=\frac{-(acost+bsint)^2-(asint-bcost)^2}{(acost+bsint)^3}$
d²y/dx²= (-y²– x²)/y³
d²y/dx²= -(x²+ y²)/y³
Hence Proved
Question 49. Find A and B so that y = Asin3x + Bcos3x, satisfies the equation d²y/dx²+ 4(dy/dx) + 3y = 10cos3x.
Solution:
We have,
y = Asin3x + Bcos3x,
On differentiating both sides w.r.t x,
(dy/dx) = 3Acos3x – 3Bsin3x
Again differentiating both sides w.r.t x,
d²y/dx²= -9Asin3x – 9Bcos3x
d²y/dx²+ 4(dy/dx) + 3y = (-9Asin3x – 9Bcos3x) + 4(3Acos3x – 3Bsin3x) + 3(Asin3x + Bcos3x)
= -9Asin3x – 9Bcos3x + 12Acos3x – 12Bsin3x + 3Asin3x + 3Bcos3x
= -6Asin3x – 12Bsin3x – 6Bcos3x + 12Acos3x
= (-6A – 12B)sin3x + (-6B + 12A)cos3x …(i)
Given that
d²y/dx²+ 4(dy/dx) + 3y = 10cos3x …(ii)
On comparing the coefficients, we get
(-6A – 12B) = 0 and (-6B + 12A) = 10
Solving equation,
A = (2/3) and B = -(1/3)
Question 50. If y = Ae^-ktcos(pt + c), prove that (d²y/dt²) + 2k(dy/dt) + n²y = 0, where n²= p²+ k²
Solution:
We have,
y = Ae^-ktcos(pt + c)
On differentiating both sides w.r.t t,
(dy/dt) = -kAe^-ktcos(pt + c) – pAe^-ktsin(pt + c)
(dy/dt) = -ky – pAe^-ktsin(pt + c)
Again differentiating both sides w.r.t t,
(d²y/dt²) = -k(dy/dt) + pAke^-ktsin(pt + c) – p²Ake^-ktcos(pt + c)
(d²y/dt²) = -k(dy/dt) + k(-ky – dy/dx) – p²y
(d²y/dt²) = -k(dy/dt) – k²y – k(dy/dt) – p²y
(d²y/dt²) + 2k(dy/dt) + (k²+ p²)y = 0
(d²y/dt²) + 2k(dy/dt) + n²y = 0
Hence Proved
Question 51. If y = xⁿ{acos(logx) + bsin(logx)}, prove that x²(d²y/dt²) + (1 – 2n) x (dy/dt) + (1 + n²)y = 0.
Solution:
We have,
y = xⁿ{acos(logx) + bsin(logx)} …(i)
On differentiating both sides w.r.t x,
(dy/dx) = nx^n-1{acos(logx) + bsin(logx)} + xⁿ{-asin(logx).(1/x) + bcos(logx).(1/x)}
x(dy/dx) = nxⁿ{acos(logx) + bsin(logx)} + xⁿ{-asin(logx) + bcos(logx)}
x(dy/dx) = ny + xⁿ{-asin(logx) + bcos(logx)} …(ii)
xⁿ{-asin(logx) + bcos(logx)} = x(dy/dx) – ny …(iii)
Again differentiating both sides w.r.t x,
x(d²y/dx²) + (dy/dx) = n(dy/dx) + nx^n-1{-asin(logx) + bcos(logx)} + xⁿ{-acos(logx).(1/x) – bsin(logx).(1/x)}
x²(d²y/dx²) + (dy/dx) = nx(dy/dx) + nxⁿ{-asin(logx) + bcos(logx)} – xⁿ{acos(logx) + bsin(logx)}
x²(d²y/dx²) = n^x(dy/dx) + n{x(dy/dx) – ny} – y – (dy/dx) [From equation (ii) and (iii)]
x²(d²y/dx²) = nx(dy/dx) + nx(dy/dx) – (dy/dx) – n²y – y
x²(d²y/dx²) = (dy/dx) x [2n – 1] – (n²+ 1)y
x²(d²y/dt²) + (1 – 2n) x (dy/dt) + (1 + n²)y = 0
Hence Proved
Question 52. y = $a[x + \sqrt{x^2+1}]^n+b[x - \sqrt{x^2+1}]^{-n}$ , prove that (x²+1)d²y/d²x + xdy/dx – ny = 0.
Solution:
We have y = $a[x + \sqrt{x^2+1}]^n+b[x - \sqrt{x^2+1}]^{-n}$
On differentiating both sides w.r.t x,
dy/dx = $na[x + \sqrt{x^2+1}]^{n-1}[1+x(x^2+1)]^{\frac{-1}{2}}+nb[x - \sqrt{x^2+1}]^{-n-1}[1-x(x^2+1)]^{\frac{-1}{2}}$
dy/dx = $\frac{na}{\sqrt{x^2+1}}[x + \sqrt{x^2+1}]^{n}+\frac{nb}{\sqrt{x^2+1}}[x - \sqrt{x^2+1}]^{-n}$
dy/dx = $\frac{n}{\sqrt{x^2+1}}[a[x + \sqrt{x^2+1}]^{n}+b[x - \sqrt{x^2+1}]^{-n}$
xdy/dx = $\frac{nx}{\sqrt{x^2+1}}y$
Again differentiating both sides w.r.t x,
d²y/dx²= $\frac{nx}{\sqrt{x^2+1}}\frac{dy}{dx}+y[\frac{\sqrt{x^2+1}-x^2(x^2+1)^{-\frac{1}{2}}}{x^2+1}]$
d²y/dx²= $\frac{n^2x^2}{x^2+1}+y[\frac{1}{(x^2+1)(\sqrt{x^2+1})}]$
d²y/dx²= $\frac{n^2x^2(\sqrt{x^2+1})+y}{(x^2+1)(\sqrt{x^2+1})}$
(x²+1)d²y/d²x = $\frac{n^2x^4(\sqrt{x^2+1})+x^2y}{(x^2+1)(\sqrt{x^2+1})}-\frac{n^2x^2(\sqrt{x^2+1})+y}{(x^2+1)(\sqrt{x^2+1})}$
Now put all these values in this equation
(x²+1)d²y/d²x + xdy/dx – ny
$\frac{n^2x^4(\sqrt{x^2+1})+x^2y}{(x^2+1)(\sqrt{x^2+1})}-\frac{n^2x^2(\sqrt{x^2+1})+y}{(x^2+1)(\sqrt{x^2+1})}+\frac{nx}{\sqrt{x^2+1}}y-ny=0$
Hence Proved

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RD Sharma Class 12 Ex 12.1 Solutions Chapter 12 Higher Order Derivatives

Find the second order derivative of following function

Question 1(i). x3 + tanx

Question 1(ii). sin(logx)

Question 1(iii). log(sinx)

Question 1(iv). exsin5x

Question 1(v). e6xcos3x

Question 1(vi). x3logx

Question 1(vii). tan-1x

Question 1(viii). x.cosx

Question 1(ix). log(logx)

Question 2. If y = e-x.cosx, show that d2y/dx2 = 2e-x.sinx.

Question 3. If y = x + tanx, Show that cos2x(d2y/dx2) – 2y + 2x = 0.

Question 4. If y = x3logx, prove that (d4y/dx4) = (6/x).

Question 5. If y = log(sinx), prove that (d3y/dx3) = 2cosx.cosec3x.

Question 6. If y = 2sinx + 3cosx, show that (d2y/dx2) + y = 0.

Question 7. If y = (logx/x), show that (d2y/dx2) = (2logx – 3)/x3

Question 8. If x = a secθ, y = b tanθ, show that (d2y/dx2) = -b4/a2y3

Question 9. If x = a(cost + tsint) and y = a(sint – tcost), prove that d2y/dx2 = sec3t/ at 0 < t < π/2.

Question 10. If y = excosx, prove that d2y/dx2 = 2excos(x + π/2).

Question 11. If x = a cosθ, y = b sinθ, show that (d2y/dx2) = -b4/a2y3

Question 12. If x = a(1 – cos3θ), y = a sin3θ, show that (d2y/dx2) = 32/27a, at θ = π/6.

Question 13. If x = a(θ + sinθ), y = a(1 + cosθ), prove that (d2y/dx2) = -(a/y2).

Question 14. If x = a(θ – sinθ), y = a(1 + cosθ), find (d2y/dx2).

Question 15. If x = a(1 – cosθ), y = a(θ + sinθ), prove that (d2y/dx2) = -1/a at θ = π/2.

Question 16. If x = a(1 + cosθ), y = a(θ + sinθ), prove that (d2y/dx2) = -1/a at θ = π/2.

Question 17. If x = cosθ, y = sin3θ, prove that y(d2y/dx2) + (dy/dx)2 = 3sin2θ(5cos2θ – 1).

Question 18. If y = sin(sinx), prove that (d2y/dx2) + tanx.(dy/dx) + ycos2x = 0

Question 19. If x = sin t, y = sin pt, prove that (1 – x2)(d2y/dx2) – x.(dy/dx) + p2y = 0

Question 20. If y = (sin-1x)2, prove that (1 – x2)(d2y/dx2) – x.(dy/dx) + p2y = 0.

Question 21. If y = , prove that (1 + x2)y2 + (2x – 1)y1 = 0.

Question 22. If y = 3cos(logx) + 4sin(logx), prove that x2y2 + xy1 + y = 0.

Question 23. If y = e2x(ax + b), show that y2 – 4y1 + 4y = 0.

Question 24. If x = sin(logy/a), show that (1 – x2)y2 – xy1 – a2y = 0.

Question 25. If logy = tan-1x, show that (1 + x2)y2 + (2x – 1)y1 = 0.

Question 26. If y = tan-1x, show that (1 + x2)(d2y/dx2) + 2x(dy/dx) = 0.

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Question 1(i). x³+ tanx

Question 1(iv). e^xsin5x

Question 1(v). e^6xcos3x

Question 1(vi). x³logx

Question 1(vii). tan^-1x

Question 2. If y = e^-x.cosx, show that d²y/dx²= 2e^-x.sinx.

Question 3. If y = x + tanx, Show that cos²x(d²y/dx²) – 2y + 2x = 0.

Question 4. If y = x³logx, prove that (d⁴y/dx⁴) = (6/x).

Question 5. If y = log(sinx), prove that (d³y/dx³) = 2cosx.cosec³x.

Question 6. If y = 2sinx + 3cosx, show that (d²y/dx²) + y = 0.

Question 7. If y = (logx/x), show that (d²y/dx²) = (2logx – 3)/x³

Question 8. If x = a secθ, y = b tanθ, show that (d²y/dx²) = -b⁴/a²y³

Question 9. If x = a(cost + tsint) and y = a(sint – tcost), prove that d²y/dx²= sec³t/ at 0 < t < π/2.

Question 10. If y = e^xcosx, prove that d²y/dx²= 2e^xcos(x + π/2).

Question 11. If x = a cosθ, y = b sinθ, show that (d²y/dx²) = -b⁴/a²y³

Question 12. If x = a(1 – cos³θ), y = a sin³θ, show that (d²y/dx²) = 32/27a, at θ = π/6.

Question 13. If x = a(θ + sinθ), y = a(1 + cosθ), prove that (d²y/dx²) = -(a/y²).

Question 14. If x = a(θ – sinθ), y = a(1 + cosθ), find (d²y/dx²).

Question 15. If x = a(1 – cosθ), y = a(θ + sinθ), prove that (d²y/dx²) = -1/a at θ = π/2.

Question 16. If x = a(1 + cosθ), y = a(θ + sinθ), prove that (d²y/dx²) = -1/a at θ = π/2.

Question 17. If x = cosθ, y = sin³θ, prove that y(d²y/dx²) + (dy/dx)²= 3sin²θ(5cos²θ – 1).

Question 18. If y = sin(sinx), prove that (d²y/dx²) + tanx.(dy/dx) + ycos²x = 0

Question 19. If x = sin t, y = sin pt, prove that (1 – x²)(d²y/dx²) – x.(dy/dx) + p²y = 0

Question 20. If y = (sin^-1x)², prove that (1 – x²)(d²y/dx²) – x.(dy/dx) + p²y = 0.

Question 21. If y = $e^{tan^{-1}x}$ , prove that (1 + x²)y₂+ (2x – 1)y₁= 0.

Question 22. If y = 3cos(logx) + 4sin(logx), prove that x²y₂+ xy₁+ y = 0.

Question 23. If y = e^2x(ax + b), show that y₂– 4y₁+ 4y = 0.

Question 24. If x = sin(logy/a), show that (1 – x²)y₂– xy₁– a²y = 0.

Question 25. If logy = tan^-1x, show that (1 + x²)y₂+ (2x – 1)y₁= 0.

Question 26. If y = tan^-1x, show that (1 + x²)(d²y/dx²) + 2x(dy/dx) = 0.