Class 12 RD Sharma Solutions – Chapter 12 Higher Order Derivatives – Exercise 12.1 | Set 2

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,

dy/dx = 2[log{x + (√x² + 1)}]/(√x² + 1)

Again differentiating both sides w.r.t x,

(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,

(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,

        …(i)

y = cos^-1x

x = cosy

On putting the value of x in equation (i), we get

d²y/dx² = -cosy/sin³y

d²y/dx² = -cot y cosec²y

Question 34. If y = , prove that (1 – x²)(d²y/dx²) – x(dy/dx) – a²y = 0.

Solution:

We have,

y = 

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/dx² at 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,

At t = π/2

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,

(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,

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,

(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 

(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

(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,

(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,

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 = , prove that (x²+1)d²y/d²x + xdy/dx – ny = 0.

Solution:

We have y = 

On differentiating both sides w.r.t x,

dy/dx = 

dy/dx = 

dy/dx = 

xdy/dx = 

Again differentiating both sides w.r.t x,

d²y/dx² = 

d²y/dx² = 

d²y/dx² = 

(x²+1)d²y/d²x = 

Now put all these values in this equation 

(x²+1)d²y/d²x + xdy/dx – ny

Hence Proved