Multiply the integrating factor to the differential equation, we have: √xy0+ 1 2√ xy = 3√ x i.e

18.

Sol.

2xy⁰+ y = 6x ⇒ y⁰+ 1 2xy = 3 The integrating factor:

I(x) = e^{R P (x)dx}

= e^{R 2x1dx}

= e^{12ln |x|}

= e^{12ln x}

=√ x.

Multiply the integrating factor to the differential equation, we have:

√xy⁰+ 1 2√

xy = 3√ x

i.e. (√

xy)⁰ = 3√ x

⇒√ xy =

Z 3√

xdx

= 3 · (2

3x³²) + C

= 2x³² + C Therefore, we get:

y(x) = 2x + C

√x y(4) = 8 + 2

C ⇒ C = 24 y(x) = 2x + 24 1

√x

Q.E.D.

20.

Sol.

Let u(x) = y(x) − 1 , we have du dx = dy

dx y(x) = u(x) + 1 We have

(x²+ 1)du

dx + 3x · u(x) = 0, u(0) = 1 The integrating factor is

e

R _3x

x2+1dx

= e^ln(x2+1)

3

2 = (x²+ 1)³² Therefore,

(x²+ 1)³²du

dx + 3x(x²+ 1)¹²u(x) = 0 ((x²+ 1)³²u(x))⁰ = 0

⇒ (x²+ 1)³²u(x) = C u(x) = C · (x²+ 1)⁻³² = C

(√

x²+ 1)³

Since u(0) = 1 ⇒ C = 1

u(x) = 1

(√

x²+ 1)³

⇒ y(x) = u(x) + 1 = 1 (√

x²+ 1)³ + 1

Q.E.D.

22.

Sol.

The integrating factor

e^{R cos xdx} = e^{sin x} So we have

e^{sin x}y⁰+ cos xe^{sin x}y = cos xe^{sin x}

i.e. (e^{sin x}y)⁰ = cos xe^{sin x}

⇒ e^{sin x}y = Z

cos xe^{sin x}dx = e^{sin x}+ C

Therefore, y = y(x) = 1 + C ˙e^{− sin x} , C is a constant. Q.E.D.

24.

Sol.

xy⁰+ y = −xy² ⇒ y⁰+ 1

xy = −y² Let u = u(x) = y¹⁻²= y⁻¹ , we have

du dx +−1

x u = 1 The integrating factor is

e^{R −1x dx} = e^{− ln |x|}= 1 x

So we have the differential equation:

1 x

du dx − 1

x²u = 1 x

⇒ (1

xu)⁰ = 1 x

⇒ u(x) = x(

Z 1

xdx) = x(ln |x| + C)

⇒ y(x) = 1

u(x) = 1

x(ln |x| + C)

Q.E.D.

26.

Sol.

Let u(x) = y⁰(x) , then we have

x · du

dx + 2u = 12x² ⇒ du dx + 2

xu = 12x The integrating factor is e^{R 2xdx} = e^{2 ln |x|} = x² , so we have

x²du

dx + 2xu = 12x³ i.e. (x²u)⁰ = 12x³

⇒ x²u =R 12x³dx = 3x⁴+ C₁

⇒ u = u(x) = 3x²+ C₁x⁻²

⇒ y(x) = R u(x)dx = R (3x²+ C₁x⁻²)dx = x³− C₁· x⁻¹+ C₂ . where C₁ , C₂ are constants.

Q.E.D.

28.

Sol.

Use the equation 4 in textbook, we have a differential equation:

1 ·dI

dt + 20 · I = 40 sin(60t) where I = I(t) is a function of t.

And the integrating factor is e^{R 2dt} = e^20t , we have

e^20tdI

dt + 20e^20tI = 40e^20tsin(60t) i.e. (e^20tI)⁰ = 40e^20tsin(60t) By using the technique of integration by parts, we have

e^20tI = 1

5e^20tsin(60t) − 3

5e^20tcos(60t) + C

⇒ I = I(t) = 1

5sin(60t) −3

5cos(60t) + C · e^−20t Since I(0) = ⁻³₅ + C = 1 , C = ⁸₅

⇒ I(t) = ¹₅sin(60t) − ³₅cos(60t) +⁸₅ · e^−20t .

Q.E.D.

30.

Sol.

Use the result in Exercise 29, we have a differential equation:

RdQ(t) dt + 1

CQ(t) = E(t)

Since R = 2 , C = 0.01 , E(t) = 10 sin(60t) , the corresponding equation is

2 ·dQ(t) dt + 1

0.01Q(t) = 10 sin(60t)

⇒ dQ(t)

dt + 50Q(t) = 5 sin(60t) Solve it by using the integrating factor e^{R 50dt} = e^50t, e^{50t dQ}_dt + 50e^50tQ(t) = 5e^50tsin(60t)

i.e.

(e^50tQ(t))⁰ = 5e^50tsin(60t)

⇒ e^50tQ(t) = Z

5e^50tsin(60t)dt

= 1

122(5 sin(60t) − 6 cos(60t))e^50t+ C

⇒ Q(t) = 1

122(5 sin(60t) − 6 cos(60t)) + Ce^−50t

Since Q(0) = ₁₂₂⁻⁶ + C = 0 ⇒ C = ₁₂₂⁶ , we solve the charge function:

Q(t) = 1

122(5 sin(60t) − 6 cos(60t)) + 6 122e^−50t And the current I(t) = dQ(t)

dt = 150 cos(60t) + 180 sin(60t) − 150e^−50t

61 .

Q.E.D.

32.

Sol.

Solve the equation in Exercise 31 to get the solution:

P (t) = M + Ce^−kt, k > 0 For P (0) = 0 , P (t) = M [1 − e^−kt] .

For the first worker, P1(1) = 25 , P1(2) = 45 , i.e.

M₁[1 − e^−k1] = 25 M₁[1 − e^−2k1] = 45

⇒ 1 − e^−k1 1 − e^−2k1 = 25

45 = 5 9

⇒ e^−k1 = 4 5

Then we can evaluate M₁ = 25· 1

1 − ⁴₅ = 125 , which is the maximum number of units per hour the first man can achieve.

Similarly, we can get M₂ = 61.25 .

Q.E.D.

34.

Sol.

Let y(t) be the amount of chlorine in the tank at time t , and then y(0) = 20 (g). The amount of liquid in the tank at time t is (400 − 6t) (L), thus the concentration of chlorine at time t is y(t)

400 − 6t (_L^g) . The rate of chlorine leaving the tank at time t is y(t)

400 − 6t · 10 since there’s 10 (L) liquid leaves the tank per second.

Therefore, we have a differential equation:

dy

dt = −10y(t)

400 − 6t = −5y(t)

200 − 3t Rearrange the differential equation, we have

dy

y = −5dt 200 − 3t

Integrate both side of the equation separately, we have ln y = 5

3ln(200 − 3t) + C

⇒ y(t) = e⁵³ln(200−3t)+C

= e^C · (200 − 3t)⁵³

Since y(0) = e^C· 200⁵³ = 20 , e^C = 20 200⁵³ We have the solution:

y(t) = 20

200⁵³ · (200 − 3t)⁵³ .

Q.E.D.

36.

Sol.

Use the equation in Exercise 35(a) : v = mg

c (1 − e^−ctm ) Differentiate it with respect to m ,we have:

dv dm = g

c − (g

ce^−ctm +gt me^−ctm )

= g

c(1 − e^−ctm − ct me^−ctm )

= g

c(1 − e^−u− ue^−u)

= g

c(1 −1 + u e^u ) where u = ct

m > 0, ∀m > 0, t > 0 Since e^u > 1 + u, ∀u > 0 , 1 + u

e^u < 1, ∀u > 0

⇒ (1 − 1 + u e^u ) > 0

⇒ dv dm = g

c(1 −1 + u

e^u ) > 0, ∀u > 0, t > 0 That is, v increases as m increases.

Q.E.D.