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Class 12 RD Sharma Solutions – Chapter 12 Higher Order Derivatives – Exercise 12.1 | Set 2
  • Last Updated : 28 Mar, 2021

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 = ae2x + be-x, show that (d2y/dx2) – (dy/dx) – 2y = 0.

Solution:

We have,

y = ae2x + be-x

On differentiating both sides w.r.t t,

(dy/dx) = 2ae2x – be-x

Again differentiating both sides w.r.t x,

(d2y/dx2) = 4ae2x + be-x

(d2y/dx2) = 2ae2x – be-x + 2(ae2x + be-x)

(d2y/dx2) = (dy/dx) + 2y

(d2y/dx2) – (dy/dx) – 2y = 0

Hence Proved

Question 32. If y = ex(sinx + cosx), prove that d2y/dx2 – 2(dy/dx) + 2y = 0.

Solution:

We have,

y = ex(sinx + cosx)

On differentiating both sides w.r.t x,

(dy/dx) = ex(sinx + cosx) + ex(cosx – sinx)

(dy/dx) = 2excosx

Again differentiating both sides w.r.t x,

(d2y/dx2) = 2excosx – 2exsinx

Lets take L.H.S,

= d2y/dx2 – 2(dy/dx) + 2y

= 2excosx – 2exsinx – 2(2excosx) + 2ex(sinx + cosx)

= 4excosx – 4excosx – 2exsinx + 2exsinx

= 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 = 500e7x + 600e-7x, show that d2y/dx2 = 49y.

Solution:

We have,

y = 500e7x + 600e-7x

On differentiating both sides w.r.t θ,

(dy/dx) = 7 × (500e7x – 600e-7x)

Again differentiating both sides w.r.t x,

(d2y/dx2) = 49 × (500e7x + 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,

\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 = 3e2x + 2e3x, prove that d2y/dx2 – 5(dy/dx) + 6y = 0.

Solution:

We have,

y = 3e2x + 2e3x

On differentiating both sides w.r.t x,

(dy/dx) = 6e2x + 6e3x

(dy/dx) = 6(e2x + e3x)

Again differentiating both sides w.r.t x,

d2y/dx2 = 6(2e2x + 3e3x)

d2y/dx2 = 12e2x + 18e3x

d2y/dx2 = 5(6e2x + 6e3x) – 6(3e2x + 2e3x)

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) = nxn-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) + nxn-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) = nx(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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