Class 12 RD Sharma解决方案–第12章高阶导数–练习12.1 |套装1

查找以下函数的二阶导数

问题1(i)。 x ³ + tanx

解决方案：

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

问题1(ii)。罪(logx)

解决方案：

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

问题1(iii)。日志(sinx)

解决方案：

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

问题1(iv)。 e ^x sin5x

解决方案：

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)

问题1(v)。 e ^6x cos3x

解决方案：

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)

问题1(vi)。 x ³ logx

解决方案：

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)

问题1(vii)。棕褐色^-1 x

解决方案：

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

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

问题1(viii)。 x.cosx

解决方案：

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)

问题1(ix)。日志(logx)

解决方案：

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)

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

问题2.如果y = e ^-x .cosx，则表明d ² y / dx ² = 2e ^-x .sinx。

解决方案：

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

问题3.如果y = x + tanx，则表明cos ² x(d ² y / dx ² )– 2y + 2x = 0。

解决方案：

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

问题4.如果y = x ³ logx，则证明(d ⁴ y / dx ⁴ )=(6 / x)。

解决方案：

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

问题5.如果y = log(sinx)，则证明(d ³ y / dx ³ )= 2cosx.cosec ³ x。

解决方案：

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

问题6.如果y = 2sinx + 3cosx，则表明(d ² y / dx ² )+ y = 0。

解决方案：

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

问题7。如果y =(logx / x)，则表明(d ² y / dx ² )=(2logx – 3)/ x ³

解决方案：

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

问题8。如果x = asecθ，y = btanθ，则表明(d ² y / dx ² )= -b ⁴ / a ² y ³

解决方案：

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

问题9。如果x = a(成本+ tsint)，y = a(sint-tcost)，则证明d ² y / dx ² = sec ³ t /在0 π/ 2处。

解决方案：

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

问题10.如果y = e ^x cosx，则证明d ² y / dx ² = 2e ^x cos(x +π/ 2)。

解决方案：

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)

问题11。如果x = acosθ，y = bsinθ，则表明(d ² y / dx ² )= -b ⁴ / a ² y ³

解决方案：

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

问题12.如果X = A(1 – COS ^3θ)，Y = A罪^3θ，表明(d ² Y / DX ²⁾⁼ 32 / 27A，在θ=π/ 6。

解决方案：

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

问题13。如果x = a(θ+sinθ)，y = a(1 +cosθ)，则证明(d ² y / dx ² )=-(a / y ² )。

解决方案：

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

问题14。如果x = a(θ–sinθ)，y = a(1 +cosθ)，则求出(d ² y / dx ² )。

解决方案：

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)]

问题15。如果x = a(1 –cosθ)，y = a(θ+sinθ)，则证明在θ=π/ 2时(d ² y / dx ² )= -1 / a。

解决方案：

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

问题16。如果x = a(1 +cosθ)，y = a(θ+sinθ)，则证明在θ=π/ 2时，(d ² y / dx ² )= -1 / a。

解决方案：

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

问题17如果x =COSθ，Y = ³罪θ，证明Y(d ² Y / DX ^2)+(DY / ^DX)2 = ² 3sinθ(5cos ^2θ – 1)。

解决方案：

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

问题18.如果y = sin(sinx)，则证明(d ² y / dx ² )+ tanx。(dy / dx)+ ycos ² x = 0

解决方案：

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

问题19：如果x = sin t，y = sin pt，则证明(1 – x ² )(d ² y / dx ² )– x。(dy / dx)+ p ² y = 0

解决方案：

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

问题20.如果y =(sin ^-1 x) ² ，则证明(1 – x ² )(d ² y / dx ² )– x。(dy / dx)+ p ² y = 0。

解决方案：

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

问题21：如果y = $e^{tan^{-1}x}$ ，证明(1 + x ² )y ₂ +(2x – 1)y ₁ = 0。

解决方案：

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

问题22.如果y = 3cos(logx)+ 4sin(logx)，则证明x ² y ₂ + xy ₁ + y = 0。

解决方案：

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

问题23。如果y = e ^2x (ax + b)，则表明y ₂ – 4y ₁ + 4y = 0。

解决方案：

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

问题24.如果x = sin(logy / a)，则表明(1 – x ² )y ₂ – xy ₁ – a ² y = 0。

解决方案：

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

问题25.如果逻辑= tan ^-1 x，则表明(1 + x ² )y ₂ +(2x – 1)y ₁ = 0。

解决方案：

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

问题26.如果y = tan ^-1 x，则证明(1 + x ² )(d ² y / dx ² )+ 2x(dy / dx)= 0。

解决方案：

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