第 12 类 RD Sharma 解决方案 - 第 22 章微分方程 - 练习 22.10 |设置 2

求解以下微分方程：

问题 24. (2x – 10y ³ )(dy/dx) + y = 0

解决方案：

We have,

(2x – 10y³)(dy/dx) + y = 0

(2x – 10y³)(dy/dx) = -y

(dx/dy) = -(2x – 10y³)/y

(dx/dy) + 2x/y = 10y²  ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 2/y, Q = 10y² 

So,

I.F = e^∫Pdy

= e^∫(2/y)dy

= e^2log|y|

= y²

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(y²) = ∫(10y²)(y²)dy + c

xy² = 10(y⁵/5) + c

x = 2y³ + cy^-2

This is the required solution.

问题 25. (x + tany)dy = sin2ydx

解决方案：

We have,

(x + tany)dy = sin2ydx

(dx/dy) = (x + tany)/sin2y

(dx/dy) – cosec2y.x = tany/sin2y ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = -cosec2y, Q = tany/sin2y

So, I.F = e^∫Pdy

= e^{∫-cosec2ydy}

= $e^{-\frac{1}{2}log|tany|}$

= $e^{log|\sqrt{coty}|}$

= √coty

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(√coty) = ∫(tany/sin2y).(√coty)dy + c

$x\sqrt{coty}=∫\frac{\sqrt{tany}}{\frac{2tany}{1+tan^2y}}dy+c$

$x\frac{1}{\sqrt{tany}}=∫\frac{1+tan^2y}{2\sqrt{tany}}dy+c$

$x\frac{1}{\sqrt{tany}}=∫\frac{sec^2x}{2\sqrt{tany}}dy+c$

Let, tany = z

On differentiating both side we have,

sec²ydx = dz

(x/√tany) = (1/2)∫dz/√z + c

(x/√tany) = (1/2)(2√z) + c

x = (√tany)(√tany) + c(√tany)

x = tany + c(√tany)

This is the required solution.

问题 26. dx + xdy = e ^-y sec ² ydy

解决方案：

We have,

dx + xdy = e^-ysec²ydy

(x – e^-ysec²y)dy = -dx

(dx/dy) = (e^-ysec²y-x) ………..(i)

The above equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1, Q = e^-ysec²y

So, I.F = e^∫Pdy

= e^∫dy

= e^y

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(e^y) = ∫e^-ysce²ye^ydy + c

x(e^y) = ∫sec²ydy + c

x(e^y) = tany + c

x = (tany + c)e^-y

This is the required solution.

问题 27. (dy/dx) = ytanx – 2sinx

解决方案：

We have,

(dy/dx) = ytanx – 2sinx

(dy/dx) – ytanx = -2sinx   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -tanx, Q = sinx

So,

I.F = e^∫Pdx

= e^∫-tanxdx

= e^-log|secx|

= 1/secx

= cosx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(cosx) = -2∫sinx.(cosx)dx + c

ycosx = -∫2sinx.cosxdx + c

Let, sinx = z

On differentiating both sides we have,

cosxdx = dz

ycosx = -2∫zdz + c

ycosx = -2(z²/2) + c

ycosx = -sin²xdx + c

y = secx(-sin²xdx + c)

This is the required solution.

问题 28. (dy/dx) + ycosx = sinx.cosx

解决方案：

We have,

(dy/dx) + ycosx = sinx.cosx  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cosx, Q = sinx.cosx

So,

I.F = e^∫Pdx

= e^∫cosxdx

= e^sinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^sinx) = ∫(e^sinx)(sinx.cosx)dx + c

Let, sinx = z

Differentating both sides we get,

cosxdx = dz

y(e^z) = ∫ze^zdz + c

y(e^z) = z∫e^zdz – {(dz/dz)∫e^zdz}dz

y(e^z) = ze^z – ∫e^zdz + c

y(e^z) = e^z(z – 1) + c

y = (z – 1) + ce^-z

y = (sinx – 1) + ce^-sinx

This is the required solution.

问题 29. (1 + x ² )(dy/dx) – 2xy = (x ² + 2)(x ² + 1)

解决方案：

We have,

(1 + x²)(dy/dx) – 2xy = (x² + 2)(x² + 1)

(dy/dx) – 2xy/(1 + x²) = (x² + 2)  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -2x/(1 + x²), Q = (x² + 1)

So, 

I.F = e^∫Pdx

= $e^{-∫\frac{2x}{1+x^2}dx}$

$=e^{-log|x^2+1|}$

= 1/(x²+1)

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.[1/(x² + 1)] = ∫[(x² + 2)/(x² + 1)]dx + c

y/(x² + 1) = ∫[1 + 1/(x² + 1)]dx + c

y/(x² + 1) = x + tan^-1x + c

y = (x² + 1)(x + tan^-1x + c)

This is the required solution.

问题 30. sinx(dy/dx) + ycosx = 2sin ² xcosx

解决方案：

We have,

sinx(dy/dx) + ycosx = 2sin²xcosx

(dy/dx) + ycotx = 2sinx.cosx  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = 2sinx.cosx

So,

I.F = e^∫Pdx

= e^∫cotxdx

= e^log|sinx|

= sinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(sinx) = ∫(2sinx.cosx)sinxdx + c

Let, sinx = z

On differentiating both sides we have,

cosxdx = dz

y.z = 2∫z² + c

y.z = (2/3)z³ + c

y.sinx = (2/3)sin³x + c

This is the required solution.

问题 31. (x ² – 1)(dy/dx) + 2(x + 2)y = 2(x + 1)

解决方案：

We have,

(x² – 1)(dy/dx) + 2(x + 2)y = 2(x + 1)

(dy/dx) + 2(x + 2)y/(x² – 1) = 2(x + 1)/(x² – 1)

(dy/dx) + 2(x + 2)y/(x² – 1) = 2/(x – 1)   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2(x + 2)/(x² – 1), Q = 2/(x – 1)

So,

I.F = e^∫Pdx

= $e^{∫\frac{2(x+2)}{x^2-1}dx}$

$=e^{∫\frac{2x}{x^2-1}dx+∫\frac{4}{x^2-1}dx}$

$=e^{log|x^2-1|+\frac{4}{2}log|\frac{x-1}{x+1}|}$

$=e^{log(|x^2-1||\frac{x-1}{x+1}|^2)}$

$=e^{log|\frac{(x-1)^3}{x+1}|}$

= (x – 1)³/(x + 1)

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.[(x – 1)³/(x + 1)] = ∫[(x – 1)³/(x + 1)[{2/(x – 1)]dx + c

$\frac{(x-1)^3}{(x+1)}y=2∫\frac{(x-1)^2}{(x+1)}+c$

$\frac{(x-1)^3}{(x+1)}y=2∫\frac{(x+1)^2-4x}{(x+1)}+c$

$\frac{(x-1)^3}{(x+1)}y=2∫[(x+1)-4\frac{x}{(x+1)}]+c$

$\frac{(x-1)^3}{(x+1)}y=∫[2x+2-8\frac{x}{x+1}]dx$

y.[(x – 1)³/(x + 1)] = (x² – 6x + 8log|x + 1|) + c

$y=\frac{x+1}{(x-1)^3}(x^2-6x+8log|x+1|+c)$

This is the required solution.

问题 32. (dy/dx) + (2y/x) = cosx

解决方案：

We have,

(dy/dx) + (2y/x) = cosx   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2/x, Q = cosx

So,

I.F = e^∫Pdx

= e^∫(2/x)dx

= e^2log|x|

= x² 

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x²) = ∫(x²).(cosx)dx + c

x²(y) = x²∫cosxdx – ∫{(d/dx)x²∫cosxdx}dx + c

x²y = x²sinx – 2∫xsinxdx + c

x²y = x²sinx – 2x∫sinxdx + 2∫{(dx/dx)∫sinxdx}dx + c

x²y = x²sinx + 2xcosx – 2∫cosxdx + c

x²y = x²sinx + 2xcosx – 2sinx + c

This is the required solution.

问题 33. (dy/dx) – y = xe ^x

解决方案：

We have,

(dy/dx) – y = xe^x  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1, Q = xe^x

So,

I.F = e^∫Pdx

= e^∫-dx

= e^-x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^-x) = ∫(e^-x)(xe^x)dx + c

ye^-x = ∫xdx + c

ye^-x = (x²/2) + c

y = [(x²/2) + c].e^x 

This is the required solution.

问题 34. (dy/dx) + 2y = xe ^4x

解决方案：

We have,

 (dy/dx) + 2y = xe^4x   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = xe^4x

So,

I.F = e^∫Pdx

= e^2∫dx

= e^2x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^2x) = ∫(e^2x).(xe^4x)dx + c

y(e^2x) = ∫xe^6xdx + c

y(e^2x) = x∫e^6xdx – ∫{(dx/dx)∫e^6xdx}dx + c

e^2xy = (xe^6x)/6 – ∫(e^6x/6)dx + c

e^2xy = (xe^6x)/6 – e^6x/36 + c

y = (xe^4x)/6 – e^4x/36 + ce^-2x

This is the required solution.

问题 35. (x + 2y ² )(dy/dx) = y，假设当 x = 2 时，y = 1

解决方案：

We have,

(x + 2y²)(dy/dx) = y

(dx/dy) = (x + 2y²)/y

(dx/dy) – (x/y) = 2y       ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1/y, Q = 2y

So,

I.F = e^∫Pdy

= e^∫-dy/y

= e^-log|y|

= 1/y

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(1/y) = ∫(1/y)(2y)dy + c

(x/y) = 2∫dy + c

(x/y) = 2y + c

x = 2y² + cy

Given that when x = 2, y = 1

2 = 2 + c

c = 0

x = 2y²

This is the required solution.

问题 36(i)。求下列微分方程 (dy/dx) + 3y = e ^mx的单参数解曲线族，m 是给定的实数

解决方案：

We have,

(dy/dx) + 3y = e^mx      ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 3, Q = e^mx 

So,

I.F = e^∫Pdx

= e^∫3dx

= e^3x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^3x) = ∫(e^3x).(e^mx)dx + c

y(e^3x) = ∫e^(3+m)xdx + c

y(e^3x) = e^(m+3)x/(m + 3) + c

y = e^mx/(m + 3) + c

This is the required solution.

问题 36(ii)。求下列微分方程的单参数解曲线族 (dy/dx) – y = cos2x

解决方案：

We have,

(dy/dx) – y = cos2x     ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1, Q = cos2x

So,

I.F = e^∫Pdx

= e^-∫dx

= e^-x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^-x) = ∫(e^-x).(cos2x)dx + c

y(e^-x) = ∫e^-xcos2xdx + c

Let, 

A = ∫e^-xcos2xdx

= e^-x∫cos2xdx – {(d/dx)e^-x∫cos2xdx}dx

= (e^-x/2)sin2x + ∫(e^-x/2)sin2xdx

=$e^{-x}sin2x+\frac{1}{2}e^{-x}∫sin2x-\frac{1}{2}∫[\frac{d}{dx}e^{-x}∫sin2xdx]$

$=\frac{1}{2}e^{-x}sin2x-\frac{1}{4}e^{-x}cos2x-\frac{1}{4}∫e^{-x}cos2xdx$

$A=e^{-x}sin2x-\frac{1}{4}e^{-x}cos2x-\frac{1}{4}A$

(5/4)A = (e^-x/2)(2sin2x – cos2x)

$A=\frac{2}{5}e^{-x}(2sin2x-cos2x)$

$ye^{-x}=\frac{2}{5}e^{-x}(2sin2x-cos2x)+c$

$y=\frac{2}{5}(2sin2x-cos2x)+ce^x$

This is the required solution.

问题 36(iii)。求下列微分方程 x(dy/dx) – y = (x + 1)e ^-x的单参数解曲线族

解决方案：

We have,

x(dy/dx) – y = (x + 1)e^-x

(dy/dx) – y/x = [(x + 1)/x]e^-x   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1/x, Q = [(x + 1)/x]e^-x

So,

I.F = e^∫Pdx

= e^-∫dx/x

= e^-log|x|

= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x) = ∫[(x+1)/x]e^-x(1/x)dx + c

y/x = ∫[1/x+1/x²]e^-xdx + c

Let, (1/x)e^-x = z

On differentiating both sides we have

-[1/x + 1/x2]e^-xdx = dz

y/x = -∫dz + c

y/x = -z + c

y/x = -(e^-x/x) + c

y = -e^-x + cx

This is the required solution.

问题 36(iv)。求下列微分方程 x(dy/dx) + y = x ⁴的单参数解曲线族

解决方案：

We have,

 x(dy/dx) + y = x⁴ 

(dy/dx) + y/x = x³       ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/x, Q = x³

So,

I.F = e^∫Pdx

= e^∫dx/x

= e^log|x|

= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x) = ∫(x)(x³)dx + c

xy = ∫x⁴ + c

xy = (x⁵/5) + c

y = (x⁴/5) + c/x

This is the required solution.

问题 36(v)。求下列微分方程 (xlogx)(dy/dx) + y = logx 的单参数解曲线族

解决方案：

We have,

(xlogx)(dy/dx) + y = logx

(dy/dx) + y/xlogx = 1/x   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/xlogx, Q = 1/x

So,

I.F = e^∫Pdx

= e^∫dx/xlogx

Let, logx = z

On differentiating both sides we have

dx/x = dz

= e^∫dz/z

= e^log|z|

= z

=l ogx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(logx) = ∫(1/x)(logx)dx + c

y(logx) = ∫zdz + c (Let, logx=z and differentiating both sides)

y(logx) = (z²/2) + c

y(logx) = (logx)²/2 + c

y = logx/2 + c/logx

This is the required solution.

问题 36(vi)。求下列微分方程的单参数解曲线族 (dy/dx) – 2xy/(1 + x ² ) = x ² + 2

解决方案：

We have,

(dy/dx) – 2xy/(1 + x²) = x² + 2   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -2x/(1 + x²), Q = x² + 2

So,

I.F = e^∫Pdx

= e^{-∫2xdx/(1+x2)}

= e^-log|1+x2|

= 1/(1+x²)

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y[1/(1 + x²)] = ∫[1/(1 + x²)](x² + 2)dx + c

y/(1 + x²) = ∫[(x² + 2)/(x² + 1)]dx + c

$\frac{y}{x^2+1}=∫\frac{x^2+1+1}{x^2+1}dx+c$

y/(1 + x²) = ∫dx + ∫dx/(x² + 1) + c

y/(x² + 1) = x + tan^-1x + c

y = (x + tan^-1x + c)(x² + 1)

This is the required solution.

问题 36(vii)。求下列微分方程的单参数解曲线族 (dy/dx) + ycosx = e ^sinx cosx

解决方案：

We have,

(dy/dx) + ycosx = e^sinxcosx    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cosx, Q = e^sinxcosx

So,

I.F = e^∫Pdx

= e∫cosxdx

= e^sinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y[(e^sinx) = ∫(e^sinx)(e^sinxcosx)dx + c

Let, sinx = z

ON differentiating both sides we have,

cosxdx = dz

ye^z = ∫e^2zdz + c

ye^z = (e^2z/2) + c

y = ez/2 + ce – z

y = (e^sinx/2) + ce^-sinx

This is the required solution.

问题 36(viii)。求下列微分方程 (x + y)(dy/dx) = 1 的单参数解曲线族

解决方案：

We have,

(x + y)(dy/dx) = 1

(dy/dx) = 1/(x + y)

(dx/dy) = (x + y)

(dx/dy) – x = y    ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = -1, Q = y

So,

I.F = e^∫Pdy

= e^-∫dy

= e^-y

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(e^-y) = ∫(e^-y)(y)dy + c

xe^-y = y∫e^-ydy – ∫{(dy/dy)∫e^-ydy}dy + c

xe^-y = -ye^-y + ∫e^-y + c

xe^-y = -ye^-y – e^-y + c

xe^-y + ye^-y + e^-y = c

e^-y(x + y + 1) = c

(x + y + 1) = ce^y

This is the required solution.

问题 36(ix)。求下列微分方程的单参数解曲线族 cos ² x(dy/dx) = (tanx – y)

解决方案：

We have,

cos²x(dy/dx) = (tanx – y)

(dy/dx) = (tanx – y)/cos²x

(dy/dx) = tanx.sec²x – ysec²x

(dy/dx) + ysec²x = tanx.sec²x    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = sec²x, Q = tanx.sec²x

So,

I.F = e∫Pdx

$=e^{∫sec^2xdx}$

= e^tanx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^tanx) = ∫(e^tanx)(tanx.sec²x)dx + c

Let, tanx = z

On differentiating both sides we have,

sec²xdx = dz

y(e^z) = ∫ze^zdz + c

y(e^z) = z∫e^zdz – ∫{(dz/dz)∫e^zdz}dz

y(e^z) = ze^z – ∫e^zdz + c

y(e^z) = ze^z – e^z + c

y = (z – 1) + c.e^-z

y = (tanx – 1) + c.e^-tanx

This is the required solution.

问题 36(x)。求下列微分方程的单参数解曲线族 e ^-y sec ² ydy = dx + xdy

解决方案：

We have,

dx + xdy = e^-ysec²ydy

(x – e^-ysec²y)dy = -dx

(dx/dy) = (e^-ysec²y – x)

(dx/dy) + x = e^-ysec²y        ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1, Q = e^-ysec²y

So, I.F = e^∫Pdy

= e^∫dy

= e^y

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(e^y) = ∫e^-ysce²ye^ydy + c

x(e^y) = ∫sec²ydy + c

x(e^y) =tany + c

x = (tan y + c)e^-y

This is the required solution.

问题 36(xi)。求下列微分方程 (xlogx)(dy/dx) + y = 2logx 的单参数解曲线族

解决方案：

We have,

(xlogx)(dy/dx) + y = 2logx

(dy/dx) + y/xlogx = 2/x   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/xlogx, Q = 2/x

So, 

I.F = e^∫Pdx

= e^∫dx/xlogx

Let, logx = z

On differentiating both sides we have

dx/x = dz

= e^∫dz/z

= e^log|z|

= z

= logx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(logx) = ∫(2/x)(logx)dx + c

y(logx) = 2∫zdz + c (Let, logx = z and differentiating both sides)

y(logx) = 2(z²/2) + c

y(logx) = (logx)² + c

y = logx + c/logx

This is the required solution.

问题 36(xii)。求下列微分方程 x(dy/dx) + 2y = x ² logx 的单参数解曲线族

解决方案：

We have,

x(dy/dx) + 2y = x²logx

(dy/dx) + 2y/x = xlogx   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2/x, Q = xlogx

So,

I.F = e∫Pdx

= e^2∫dx/x

= e^2logx

= x²

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x²) = ∫(x²)(xlogx)dx + c

x²y = ∫x³logxdx + c

x²y = logx∫x³dx + ∫{(d/dx)logx∫x³dx}dx + c

x²y = (1/4)x⁴logx – (1/4)∫x³dx + c

x²y = (1/4)x⁴logx – (1/16)x⁴ + c 

y = (x² /16)(4logx – 1) + c/x²

This is the required solution.

问题 37. 使用初始值问题解决以下问题：-

（一世）。 y' + y = e ^x , y(0) = (1/2)

解决方案：

We have,

y’ + y = e^x

dy/dx + y = e^x    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1, Q = e^x

So, I.F = e^∫Pdx

= e^∫dx

= e^x

The solution of differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^x) = ∫e^x.e^xdx + c

y(e^x) = (1/2)e^2x + c

At t = 0, y = (1/2)

(1/2)e⁰ = (1/2)e⁰ + c

c = 0

y(e^x) = (1/2)e^2x

y = (e^x/2)

This is the required solution.

(二)。 x(dy/dx) – y = logx, y(1) = 0

解决方案：

We have,

x(dy/dx) – y = logx

(dy/dx) – y/x = logx/x   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1/x, Q = logx/x

So,

I.F = e^∫Pdx

= e^-∫dx/x

= e^-log|x|

= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x) = ∫(1/x)(logx/x)dx + c

(y/x) = ∫(logx/x²)dx + c

(y/x) = logx∫(dx/x²) – ∫{(d/dx)logx∫(dx/x²)}dx + c

(y/x) = -(logx/x) + ∫(dx/x²) + c

(y/x) = -(logx/x) – (1/x) + c

At x = 1, y = 0

0 = -0 – 1 + c

c = 1

(y/x) = -(logx/x) – (1/x) + 1

y = x – 1 – logx

This is the required solution.

(iii)。 (dy/dx) + 2y = e ^-2x sinx, y(0) = 0

解决方案：

We have,

(dy/dx) + 2y = e^-2xsinx    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = e^-2xsinx

So,

I.F = e^∫Pdx

= e^∫2dx

= e^2x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^2x) = ∫e^-2xsinx.(e^2x)dx + c

y(e^2x) = ∫sinxdx + c

y(e^2x) = -cosx + c

At x = 0, y = 0

0 = -1 + c

c = 1

y(e^2x) = 1 – cosx

This is the required solution.

(四)。 x(dy/dx) – y = (x + 1)e ^-x , y(1) = 0

解决方案：

We have,

x(dy/dx) – y = (x + 1)e^-x

(dy/dx) – (y/x) = [(x + 1)/x]e^-x   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -(1/x), Q = [(x + 1)/x]e^-x

So,

I.F = e^∫Pdx

= e^-∫(dx/x)

= e^-log(x)

= e^log(1/x)

= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

$y(\frac{1}{x})=∫(\frac{x+1}{x})e^{-x}.(\frac{dx}{x})+c$

(y/x) = ∫[(1/x) + (1/x²)]e^-x + c

Since, -∫[f(x) + f'(x)]e^-xdx = f(x)e^-x + c

(y/x) = -e^-x/x + c

At x = 1, y = 0

0 = -e^-1 + c

c = e^-1

(y/x) = -e^-x/x + e^-1

y = xe^-1 – e^-x

This is the required solution.

(五)。 (1 + y ² )dx + (x – $e^{-tan^{-1}y}$ )dx = 0, y(0) = 0

解决方案：

We have,

(1 + y²)(dx/dy) + x = $e^{-tan^{-1}y}$

(dy/dx) + [1/(y² + 1)]x = $e^{-tan^{-1}y}$/(y² + 1)   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Px = Q

Where, P = 1/(y² + 1), Q = $e^{-tan^{-1}y}$/(y² + 1)

So,

I.F = e^∫Pdy

$=e^{∫\frac{1}{y^2+1}dy}$

= e^tan-1y

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x($e^{-tan^{-1}y}$) = ∫[$e^{-tan^{-1}y}$/(y² + 1)]$e^{-tan^{-1}y}$dx + c

x($e^{-tan^{-1}y}$) = ∫dy/(1 + y²) + c

x(e^tan-1y) = tan^-1y + c

At x = 0, y = 0

0*e⁰ = 0 + c

c = 0

x($e^{-tan^{-1}y}$) = tan^-1y

This is the required solution.

㈥。 (dy/dx) + ytanx = x ² tanx + 2x, y(0) = 1

解决方案：

We have,

(dy/dx) + ytanx = x²tanx + 2x   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = tanx, Q = x²tanx+2x

So,

I.F = e^∫Pdx

= e^∫tanxdx

= e^log|secx|

= secx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(secx) = ∫(x²tanx + 2x)secxdx + c

y(secx) = ∫(x²tanxsecx + 2xsecx)dx + c

y(secx) = ∫x²tanxsecxdx + 2∫xsecxdx + c

y(secx) = ∫x²tanxsecxdx + 2secx∫xdx – 2∫{(d/dx)secx∫xdx}dx + c

y(secx) = ∫x²tanxsecxdx + x².secx – ∫x²tanxsecxdx + c

y(secx) = x²,(secx)+c

At, x = 0, y = 1

1 = 0 + c

c = 1

y = x² + cosx

This is the required solution.

(七)。 x(dy/dx) + y = xcosx + sinx, y(π/2) = 1

解决方案：

We have,

x(dy/dx) + y = xcosx + sinx

(dy/dx) + (y/x) = cosx + sinx/x  ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/x, Q = cosx + sinx/x

So,

I.F = e^∫Pdx

= e^∫dx/x

= e^log|x|

= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x) = ∫(cosx + sinx/x)(x)xdx + c

y(x) = ∫xcosxdx + ∫sinxdx + c

xy = x∫cosxdx – ∫{(dx/dx)∫cosxdx}dx – cosx + c

xy = xsinx – ∫sinxdx – cosx + c

xy = xsinx + cosx – cosx + c

xy = xsinx + c

At x = π/2, y = 1

π/2 = π/2sin(π/2) + c

c = 0

y = sinx

This is the required solution.

(八)。 (dy/dx) + ycotx = 4xcosecx, y(π/2) = 0

解决方案：

We have,

(dy/dx) + ycotx = 4xcosecx   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = 4xcosecx

So,

I.F = e^∫Pdx

= e^∫cotx.dx

= e^log|sinx|

= sinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(sinx) = 4∫(xcosecx)(sinx)xdx + c

y(sinx) = 4∫xdx + c

y(sinx) = 4(x²/2) + c

y(sinx) = 2x² + c

At x = π/2, y = 0,

0 = 2(π/2)² + c

c = -π²/2

y(sinx) = 2x² – π²/2

This is the required solution.

(九)。 (dy/dx) + 2ytanx = sinx，当 x = π/3 时，y = 0

解决方案：

We have,

(dy/dx) + 2ytanx = sinx   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2tanx, Q = sinx

So, I.F = e^∫Pdx

= e^∫2tanxdx

= e^2log|secx|

= sec²x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.sec²x = ∫sinx.sec²xdx + c

y.sec²x = ∫tanx.secxdx + c

y.sec²x = secx+c

At x = π/3, y = 0,

0 = sec²(π/3) + c

c = -2

y.sec²x = secx – 2

y = cosx – 2cos²x

This is the required solution.

（X）。 (dy/dx) – 3ycotx = sin2x, y = 2 当 x = π/2

解决方案：

We have,

(dy/dx) – 3ycotx = sin2x        ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -3cotx, Q = sin2x

So, I.F = e^∫Pdx

= e^∫-3cotxdx

= e^-3log|sinx|

= cosec³x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.(cosec³x) = 2∫(cosec³x).(sin2x)dx + c

y.(cosec³x) = 2∫cotx.cosecxdx + c

y.(cosec³x) = -2cosesx +c

y = -2sin²x + c.sin³x

At x = π/2, y = 2.

2 = -2sin²(π/2) + c.sin³(π/2)

c = 4

y = c.sin³x – 2sin²x

This is the required solution.

(十一)。 (dy/dx) + ycotx = 2cosx, y(π/2) = 0

解决方案：

We have,

(dy/dx) + ycotx = 2cosx        ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = 2cosx 

So, I.F = e^∫Pdx

= e^∫cotxdx

= e^log|sinx|

= sinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.(sinx) = ∫(sinx).(2cosx)dx + c

y.(sinx) = 2∫sinx.cosxdx + c

y.(sinx) = ∫sin2xdx + c

y.(sinx) = -(cos2x/2) + c

At x = π/2, y = 0

0 = -cos(π)/2 + c

c = -(1/2)

y.(sinx) = -(cos2x/2) – (1/2)

2y(sinx) = -(1 + cos2x)

2y(sinx) = -2cos²x

y = -cotx.cosx

This is the required solution.

(十二)。 dy = cosx(2 – ycosecx)dx,

解决方案：

We have,

dy = cosx(2 – ycosecx)dx

(dy/dx) = -ycotx + 2cosx

(dy/dx) + ycotx = 2cosx    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = 2cosx

So,

I.F = e^∫Pdx

= e^∫cotxdx

= e^log|sinx|

= sinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(sinx) = 2∫cosx.(sinx)dx + c

ysinx = ∫2cosx.sinxdx + c

ysinx = ∫sin2x + c

ysinx = -(cos2x/2) + c

This is the required solution.

问题 38. x(dy/dx) + 2y = x ²

解决方案：

We have,

x(dy/dx) + 2y = x²

(dy/dx) + 2y/x = x    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2/x, Q = x

So, I.F = e^∫Pdx

= e^2∫dx/x

= e^2logx

= x²

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

yx² = ∫x2.xdx + c

yx² = ∫x³dx + c

x²y = (x⁴/4) + c

y = (x1/4) + c.x^-2

This is the required solution.

问题 39. (dy/dx) – y = cosx

解决方案：

We have,

(dy/dx) – y = cosx   ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1, Q = cosx

So, I.F = e^∫Pdx

= e^-∫dx

= e^-x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^-x) = ∫cosx.e^-xdx + c

Let, I = ∫cosx.e^-xdx

I = e^-x∫cosxdx – ∫{(d/dx)e^-x∫cosxdx}dx

I = e^-xsinx + ∫e^-xsinxdx

I = e^-xsinx + e^-x∫sinxdx-∫{(d/dx)e^-x∫sinxdx}dx

I = e^-xsinx – e^-xcosx-∫e^-xcosxdx

2I = e^-x(sinx – cosx)

I = (e^-x/2)(sinx – cosx)

y(e^-x) = (e^-x/2)(sinx – cosx) + c

y = (1/2)(sinx-cosx) + ce^x

This is the required solution.

问题 40. (y + 3x ² )(dx/dy) = x

解决方案：

We have,

(y + 3x²)(dx/dy) = x

(dy/dx) = (y + 3x²)/x

(dy/dx) – y/x = 3x    ………..(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -1/x, Q = 3x

So, I.F = e^∫Pdx

= e^-∫dx/x

= e^-logx

= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x) = 3∫x.(1/x)dx + c

y(1/x) = 3∫dx + c

y/x = 3x + c

This is the required solution.

问题 41. 求微分方程 (dx/dy) + xcoty = y ² coty + 2y 的特解，假设 x = 0，当 y = π/2

解决方案：

We have,

(dx/dy) + xcoty = y²coty + 2y     ………..(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = coty, Q = y²coty + 2y

So,

I.F = e^∫Pdy

= e^∫cotydy

= e^log|siny|

= siny

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(siny) = ∫(y²coty + 2y)sinydy + c

x(siny) = ∫(y²cosy + 2xsiny)dy + c

x(siny) = y²∫cosydx – {(d/dy)y²∫cosydy}dy + ∫2ysinydy + c

x(siny) = y²siny – ∫2ysinydy + ∫2ysinydy + c

x(siny) = y²siny + c

At x = 0, y = π/2

0 = (π/2)²sin(π/2) + c

c = -π²/4

x(siny) = y²siny – π²/4

This is the required solution.