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

Here we provide RD Sharma Class 12 Ex 12.1 Solutions Chapter 12 Higher Order Derivatives for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 12.1 Solutions Chapter 12 Higher Order Derivatives book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter12
Exercise12.1
CategoryRD 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).  x+ tanx

Solution:

Let us considered

f(x) = x+ tanx

On differentiating both sides w.r.t x,

f'(x) = 3x+ 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/dx= 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/dx= -cosec2x

Question 1(iv). exsin5x

Solution:

Let us considered

y = exsin5x

On differentiating both sides w.r.t x,

(dy/dx) = exsin5x + 5excos5x

Again differentiating both sides w.r.t x,

d2y/dx= exsin5x + 5excos5x + 5(excos5x – 5exsin5x)

d2y/dx= -24exsin5x + 10excos5x

d2y/dx2 = 2ex(5cos5x – 12sinx)

Question 1(v). e6xcos3x

Solution:

Let us considered

y = e6xcos3x

On differentiating both sides w.r.t x,

(dy/dx) = 6e6xcos3x – 3e6xsin3x

Again differentiating both sides w.r.t x,

d2y/dx= 6(6e6xcos3x – 3e6xsin3x) – 3(6e6xsin3x + 3e6xcos3x)

d2y/dx= 36e6xcos3x – 18e6xsin3x – 18e6xsin3x – 9e6xcos3x

d2y/dx= 27e6xcos3x – 36e6xsin3x

d2y/dx= 9e6x(3cos3x – 4sin3x)

Question 1(vi). x3logx

Solution:

Let us considered

y = x3logx

On differentiating both sides w.r.t x,

(dy/dx) = logx.3x+ x3(1/x)

(dy/dx) = logx.3x+ x2

(dy/dx) = x2(1 + 3logx)

Again differentiating both sides w.r.t x,

d2y/dx= (1 + 3logx).2x + x2(3/x)

d2y/dx= 2x + 6xlogx + 3x

d2y/dx= 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/dx= (-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/dx= -sinx – (sinx + xcosx)

d2y/dx= -2sinx – xcosx

d2y/dx= -(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/dx= (-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/dx= 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/dx= -(-e-x.cosx – e-x.sinx) – (-e-x.sinx + e-x.cosx)

d2y/dx= e-x.cosx – e-x.cosx + e-x.sinx + e-x.sinx

d2y/dx= 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/dx= 0 + (2secx)(secx.tanx)

d2y/dx= 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.3x+ x3(1/x)

(dy/dx) = logx.3x+ x2

(dy/dx) = x2(1 + 3logx)

Again differentiating both sides w.r.t x,

d2y/dx= (1 + 3logx).2x + x2(3/x)

d2y/dx2 = 2x + 6xlogx + 3x

d2y/dx= 5x + 6xlogx

Again differentiating both sides w.r.t x,

d3y/dx= 5 + 6[logx + (x/x)]

d3y/dx= 11 + 6logx

Again differentiating both sides w.r.t x,

d4y/dx= (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/dx= -cosec2x

Again differentiating both sides w.r.t x,

d3y/dx= -2cosecx.(-cosesx.cotx)

d3y/dx= 2cosec2x.cotx

d3y/dx= 2cosec2x.(cosx/sinx)

d3y/dx= 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/dx= -2sinx – 3cosx

d2y/dx= -(2sinx + 3cosx)

d2y/dx= -y

d2y/dx+ 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/dx= [-x – 2x(1 – logx)]/x4

d2y/dx= (2xlogx – 3x)/x4

d2y/dx= (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/dx= -(b/a2).(1/tan3θ)

d2y/dx= -(b/a2tan3θ).(b3/b3)

d2y/dx= -(b4/a2y3)

Hence Proved

Question 9. If x = a(cost + tsint) and y = a(sint – tcost), prove that d2y/dx= 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 = excosx, prove that d2y/dx= 2excos(x + π/2).

Solution:

We have,

y = excosx

On differentiating both sides w.r.t x,

(dy/dx) = excosx – exsinx

Again differentiating both sides w.r.t x,

d2y/dx= (excosx – exsinx) – (exsinx + excosx)

d2y/dx= excosx – excosx – exsinx – exsinx

d2y/dx= -2exsinx

d2y/dx= 2excos(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/dx= -(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/dx= sec4(π/6)/3asin(π/6)

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

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

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

At θ = π/2,

d2y/dx= -(1 + 0)/a

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

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

At θ = π/2,

d2y/dx= -(1 + 0)/a

d2y/dx= -(1/a)

Hence Proved

Question 17. If x = cosθ, y = sin3θ, prove that y(d2y/dx2) + (dy/dx)= 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/dx= -3[sinθ(-sinθ) + cosθ.cosθ](dθ/dx)

d2y/dx= (3sin2θ – 3cos2θ)/-sinθ

d2y/dx= -(3sin2θ – 3cos2θ)/sinθ

L.H.S,

y(d2y/dx2) + (dy/dx)= -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/dx= -sin(sinx).cosx.cosx – cos(sinx).sinx

d2y/dx= -sin(sinx).cos2x – cos(sinx).sinx

d2y/dx= -sin(sinx).cos2x – cos(sinx).cosx.tanx

d2y/dx= -ycos2x – (dy/dx)tanx

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

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

(1 – x2)d2y/dx= x(dy/dx) + 2

(1 – x2)d2y/dx– x(dy/dx) – 2 = 0

Hence Proved

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

Solution:

We have,

y = e^{tan^{-1}x}

On differentiating both sides w.r.t t,

ye^{tan^{-1}x} × [1/(1 + x2)]

Again differentiating both sides w.r.t x,

ye^{tan^{-1}x}\frac{1}{(1+x^2)^2}+e^{tan^{-1}x}\frac{-2x}{(1+x^2)^2}

(1 + x2)ye^{tan^{-1}x}/(1 + x2) – 2xe^{tan^{-1}x}/(1 + x2)

(1 + x2)y= (dy/dx) – 2x(dy/dx)

(1 + x2)y– (dy/dx) + 2x(dy/dx) = 0

(1 + x2)y+ (2x – 1)(dy/dx) = 0

Hence Proved

Question 22. If y = 3cos(logx) + 4sin(logx), prove that x2y+ xy+ 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)

xy= -3sin(logx) + 4cos(logx)

Again differentiating both sides w.r.t x,

xy+ y= -3cos(logx)×(1/x) – 4sin(logx) × (1/x)

x2y+ xy= -[3cos(logx) + 4sin(logx)]

x2y+ xy= -y

x2y+ xy+ y = 0

Hence Proved

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

Solution:

We have,

y = e2x(ax + b)

On differentiating both sides w.r.t θ,

y= 2e2x(ax + b) + a.e2x

Again differentiating both sides w.r.t x,

y= 4e2x(ax + b) + 2ae2x + 2a.e2x

y= 4e2x(ax + b) + 4a.e2x

Lets take L.H,S,

= y– 4y+ 4y

= 4e2x(ax + b) + 4a.e2x – 4[2e2x(ax + b) + a.e2x] + 4[e2x(ax + b)]

= 8e2x(ax + b) – 8e2x(ax + b) + 4a.e2x – 4a.e2x

= 0

= R.H.S

L.H.S = R.H.S

Hence Proved

Question 24. If x = sin(logy/a), show that (1 – x2)y– xy– 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)y= a/√(1 – x2)

y= ay/√(1 – x2)

Again differentiating both sides w.r.t x,

y=a[\frac{\sqrt{1-x^2}\frac{dy}{dx}-\frac{2xy}{2\sqrt{1-x^2}}}{1-x^2}]

(1 – x2)y= a√(1 – x2) × y+ axy/√(1 – x2)

(1 – x2)y= a√(1 – x2) × [ay/√(1 – x2)] + x[ay/√(1 – x2)]

(1 – x2)y= a2p + xy

(1 – x2)y– a2p – xy= 0

Hence Proved

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

Solution:

We have,

logy = tan-1x

On differentiating both sides w.r.t θ,

(1/y)y= 1/(1 + x2)

(1 + x2)y= y

Again differentiating both sides w.r.t x,

2xy+ (1 + x2)y= y1

(1 + x2)y+ (2x – 1)y= 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/dx= [-1/(1 + x2)2] × (2x)

d2y/dx= [-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 + (√x+ 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 + (√x+ 1)}]/(√x+ 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}
(x+ 1)(d2y/dx2) = 2 – x(dy/dx)
(x+ 1)(d2y/dx2) + x(dy/dx) = 2
Hence Proved
Question 28. If y = (tan-1x)2, then prove that (1 + x2)2y+ 2x(1 + x2)y= 2
Solution:
We have,
y = (tan-1x)2
On differentiating both sides w.r.t x,
y= 2(tan-1x)[1/(1 + x2)]
(1 + x2)y= 2(tan-1x)
Again differentiating both sides w.r.t x,
(1 + x2)y+ 2xy= 2/(1 + x2)
(1 + x2)2y+ 2x(1 + x2)y= 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/dx= -(2cosec x) × (-cosec x.cot x)
d2y/dx= 2cosec2x.cot x
d2y/dx= 2(cot x)(cosec2x)
d2y/dx= -2y(dy/dx)
d2y/dx+ 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/dx= -(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/dx– 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/dx– 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/dx= -cosy/sin3y
d2y/dx= -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)= a2y2/(1 – x2)
(1 – x2)(dy/dx)= a2y2
Again differentiating both sides w.r.t x,
2(1 – x2)(dy/dx)(d2y/dx2) – 2x(dy/dx)= 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/dx= 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/dxat 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/dx= (1 + 2)/-2
d2y/dx= -(3/2)
Question 37. If x = 4z+ 5, y = 6z+ 7z + 3, find d2y/dx2.
Solution:
We have,
x = 4z+ 5, and y = 6z+ 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/dx+ (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/dx= (-cosx – cos2x – sin2x)/(1 + cosx)2
d2y/dx= -(1 + cosx)/(1 + cosx)2
d2y/dx= -1/(1 + cosx)
Again differentiating both sides w.r.t x,
d3y/dx= -sinx/(1 + cosx)2
d3y/dx+ [-1/(1 + cosx)][-sinx/(1 + cosx)] = 0
d3y/dx+ (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/dx– 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/dx= 6(2e2x + 3e3x)
d2y/dx= 12e2x + 18e3x
d2y/dx= 5(6e2x + 6e3x) – 6(3e2x + 2e3x)
d2y/dx= 5(dy/dx) – 6y
d2y/dx– 5(dy/dx) + 6y = 0
Hence Proved
Question 41. If y = (cot-1x)2, prove that y2(x+ 1)+ 2x(x+ 1)y= 2.
Solution:
We have,
y = (cot-1x)2
On differentiating both sides w.r.t x,
y= 2(cot-1x) × [-1/(1 + x2)]
(1 + x2)y= -2cot-1x
Again differentiating both sides w.r.t x,
(1 + x2)y+ 2xy= 2/(1 + x2)
(1 + x2)2y+ 2x(1 + x2)y= 2
Hence Proved
Question 42. If y = cosec-1x, then show that x(x– 1)d2y/dx– (2x– 1)(dy/dx) = 0.
Solution:
We have,
y = cosec-1x
On differentiating both sides w.r.t x,
(dy/dx) = -1/x√(x– 1)
On squaring both sides,
(dy/dx)= 1/x2(x– 1)
x2(x– 1)(dy/x)= 1
(x– x2)(dy/dx)= 1
2(dy/dx)(d2y/dx2)(x– x2) + (dy/dx)2(4x– 2x) = 0
2x2(x– 1)(dy/dx)(d2y/dx2) + 2x(2x– 1)(dy/dx)= 0
x(x– 1)(d2y/dx2) + (2x– 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) = -(x+ 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/dx= (-y– x2)/y3
d2y/dx= -(x+ y2)/y3
Hence Proved
Question 49. Find A and B so that y = Asin3x + Bcos3x, satisfies the equation d2y/dx+ 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/dx= -9Asin3x – 9Bcos3x
d2y/dx+ 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/dx+ 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 n= p+ 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) + (k+ 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] – (n+ 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/dx\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/dx\frac{n^2x^2}{x^2+1}+y[\frac{1}{(x^2+1)(\sqrt{x^2+1})}]
d2y/dx\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

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

Leave a Comment

Your email address will not be published.