Class 12 RD Sharma解决方案–第12章高阶导数–练习12.1 |套装2

问题27如果y = [日志{X ^+(√x2 +1)}] ^2，表明：(1 + X ^{2)(d 2} Y / DX ²⁾⁺ X(DY / DX)= 2。

解决方案：

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

问题28.如果y =(tan ^-1 x) ² ，则证明(1 + x ² ) ² y ₂ + 2x(1 + x ² )y ₁ = 2

解决方案：

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

问题29.如果y = cotx，则证明(d ² y / dx ² )+ 2y(dy / dx)= 0。

解决方案：

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

问题30.求d ² y / dx ² ，其中y = log(x ² / e ² )。

解决方案：

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

问题31.如果y = ae ^2x +是^-x ，则表明(d ² y / dx ² )–(dy / dx)– 2y = 0。

解决方案：

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

问题32.如果y = e ^x (sinx + cosx)，则证明d ² y / dx ² – 2(dy / dx)+ 2y = 0。

解决方案：

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

问题33.如果y = cos ^-1 x，则仅根据y^{求d 2} y / dx ^2。

解决方案：

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

问题34.如果y = $e^{acos^{-1}x}$ ，证明(1 – x ² )(d ² y / dx ² )– x(dy / dx)– a ² y = 0。

解决方案：

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

问题35.如果y = 500e ^7x + 600e ^-7x ，则表明d ² y / dx ² = 49y。

解决方案：

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

问题36.如果x = 2COS吨- COS 2T，Y = 2sin吨-罪2T，在t =π/ 2找到d ² Y / DX ^2。

解决方案：

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)

问题37.如果x = 4z ² + 5，y = 6z ² + 7z + 3，则求出d ² y / dx ² 。

解决方案：

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

问题38.如果y = log(1 + cosx)，则证明d ³ y / dx ³ +(d ² y / dx ² )。(dy / dx)= 0。

解决方案：

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

问题39.如果y = sin(logx)，则证明x ² (d ² y / dx ² )+ x(dy / dx)+ y = 0。

解决方案：

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

问题40.如果y = 3e ^2x + 2e ^3x ，则证明d ² y / dx ² – 5(dy / dx)+ 6y = 0。

解决方案：

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

问题41.如果y =(cot ^-1 x) ² ，则证明y ₂ (x ² +1) ² + 2x(x ² +1)y ₁ = 2。

解决方案：

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

问题42.如果y = cosec ^-1 x，则表明x(x ² – 1)d ² y / dx ² –(2x ² – 1)(dy / dx)= 0。

解决方案：

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

问题43.如果x = cos t + log(tant / 2)，y = sin t，则仅以y表示，在t =π/ 4处^{找到d 2} y / dt ²和d ² y / dx ^2的值。

解决方案：

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

问题44.如果x = asin t，y = a [cos t + log(tant / 2)]，则求出d ² y / dx ² 。

解决方案：

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)

问题45.如果x = a(cos t + tsin t)，并且y = a(sin t – tcos t)，则在t =π/ 4处^{找到d 2} y / dx ^2的值。

解决方案：

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)

问题46.如果x = a [cos t + log(tant / 2)]，y = asin t，则在t =π/ 3时^{求出(d 2} y / dx ^2)。

解决方案：

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)

问题47.如果x = a(cos2t + 2tsin2t)，而y = a(sin2t – 2tcos2t)，则求出d ² y / dx ² 。

解决方案：

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)

问题48.如果x = asin t – bcos t，y = acos t + bsin t，则证明(d ² y / dx ² )=-(x ² + y ² )/ y ³

解决方案：

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

问题49.找到A和B，使y = Asin3x + Bcos3x，满足方程d ² y / dx ² + 4(dy / dx)+ 3y = 10cos3x。

解决方案：

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)

问题50.如果y = Ae ^-kt cos(pt + c)，则证明(d ² y / dt ² )+ 2k(dy / dt)+ n ² y = 0，其中n ² = p ² + k ²

解决方案：

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

问题51.如果y = x ⁿ {acos(logx)+ bsin(logx)}，则证明x ² (d ² y / dt ² )+(1 – 2n)x(dy / dt)+(1 + n ² y = 0。

解决方案：

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

问题52。y = $a[x + \sqrt{x^2+1}]^n+b[x - \sqrt{x^2+1}]^{-n}$ ，证明(x ² +1)d ² y / d ² x + xdy / dx – ny = 0。

解决方案：

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