SOLUTION The given equation is linear since it has the form of Equation 1 with ..... 4. 4e 1/2. 1.57 A. I(t). 4. 4e 5t y. [Cx4. 2/(5x)] 1/2. Click here for solutions. S ...

Linear Differential Equations A first-order linear differential equation is one that can be put into the form dy Pxy Qx dx

1

where P and Q are continuous functions on a given interval. This type of equation occurs frequently in various sciences, as we will see. An example of a linear equation is xy y 2x because, for x 0, it can be written in the form y

2

1 y2 x

Notice that this differential equation is not separable because it’s impossible to factor the expression for y as a function of x times a function of y. But we can still solve the equation by noticing, by the Product Rule, that xy y xy and so we can rewrite the equation as xy 2x If we now integrate both sides of this equation, we get xy x 2 C

or

C x

yx

If we had been given the differential equation in the form of Equation 2, we would have had to take the preliminary step of multiplying each side of the equation by x. It turns out that every first-order linear differential equation can be solved in a similar fashion by multiplying both sides of Equation 1 by a suitable function Ix called an integrating factor. We try to find I so that the left side of Equation 1, when multiplied by Ix, becomes the derivative of the product Ixy: Ixy Pxy Ixy

3

If we can find such a function I , then Equation 1 becomes Ixy IxQx Integrating both sides, we would have Ixy y IxQx dx C so the solution would be 4

yx

1 Ix

y

IxQx dx C

To find such an I, we expand Equation 3 and cancel terms: Ixy IxPxy Ixy Ixy Ixy IxPx Ix 1

2 ■ LINEAR DIFFERENTIAL EQUATIONS

This is a separable differential equation for I , which we solve as follows:

y

dI y Px dx I

ln I y Px dx I Ae x Px dx where A e C. We are looking for a particular integrating factor, not the most general one, so we take A 1 and use Ix e x Px dx

5

Thus, a formula for the general solution to Equation 1 is provided by Equation 4, where I is given by Equation 5. Instead of memorizing this formula, however, we just remember the form of the integrating factor. To solve the linear differential equation y Pxy Qx, multiply both sides by the integrating factor Ix e x Px dx and integrate both sides.

EXAMPLE 1 Solve the differential equation

dy 3x 2 y 6x 2. dx

SOLUTION The given equation is linear since it has the form of Equation 1 with

Px 3x 2 and Qx 6x 2. An integrating factor is Ix e x 3x

2

dx

ex

3

3

Multiplying both sides of the differential equation by e x , we get Figure 1 shows the graphs of several members of the family of solutions in Example 1. Notice that they all approach 2 as x l . ■ ■

ex

3

dy 3 3 3x 2e x y 6x 2e x dx

6 C=2

d x3 3 e y 6x 2e x dx

or

C=1 C=0

Integrating both sides, we have

C=_1 _1.5

1.8 3

3

3

e x y y 6x 2e x dx 2e x C

C=_2 _3

y 2 Cex

FIGURE 1

3

EXAMPLE 2 Find the solution of the initial-value problem

x 2 y xy 1

x0

y1 2

SOLUTION We must first divide both sides by the coefficient of y to put the differential equation into standard form: 6

y

1 1 y 2 x x

x0

The integrating factor is Ix e x 1x dx e ln x x

LINEAR DIFFERENTIAL EQUATIONS ■ 3

Multiplication of Equation 6 by x gives 1 x

xy y

■ ■ The solution of the initial-value problem in Example 2 is shown in Figure 2.

ln x C x

y

and so

5

1 x

1 dx ln x C x

xy y

Then

xy

or

Since y1 2, we have

(1, 2) 0

2

4

ln 1 C C 1

Therefore, the solution to the initial-value problem is _5

ln x 2 x

y

FIGURE 2

EXAMPLE 3 Solve y 2xy 1. SOLUTION The given equation is in the standard form for a linear equation. Multiplying by the integrating factor

e x 2x dx e x we get

2

2

e x y 2xe x y e x 2

(e x y) e x

or Even though the solutions of the differential equation in Example 3 are expressed in terms of an integral, they can still be graphed by a computer algebra system (Figure 3). ■ ■

2

2

2

2

2

e x y y e x dx C

Therefore 2

Recall from Section 5.8 that x e x dx can’t be expressed in terms of elementary functions. Nonetheless, it’s a perfectly good function and we can leave the answer as

2.5

y ex

C=2 _2.5

2.5

2

ye

dx Cex

2

Another way of writing the solution is

C=_2

y ex

2

y

_2.5

FIGURE 3

x2

x

0

2

e t dt Cex

2

(Any number can be chosen for the lower limit of integration.) Application to Electric Circuits R

E

L

switch

In Section 7.2 we considered the simple electric circuit shown in Figure 4: An electromotive force (usually a battery or generator) produces a voltage of Et volts (V) and a current of It amperes (A) at time t . The circuit also contains a resistor with a resistance of R ohms () and an inductor with an inductance of L henries (H). Ohm’s Law gives the drop in voltage due to the resistor as RI . The voltage drop due to the inductor is LdIdt. One of Kirchhoff’s laws says that the sum of the voltage drops is equal to the supplied voltage Et. Thus, we have

FIGURE 4 7

L

dI RI Et dt

which is a first-order linear differential equation. The solution gives the current I at time t .

4 ■ LINEAR DIFFERENTIAL EQUATIONS

EXAMPLE 4 Suppose that in the simple circuit of Figure 4 the resistance is 12 and the inductance is 4 H. If a battery gives a constant voltage of 60 V and the switch is closed when t 0 so the current starts with I0 0, find (a) It, (b) the current after 1 s, and (c) the limiting value of the current. SOLUTION The differential equation in Example 4 is both linear and separable, so an alternative method is to solve it as a separable equation (Example 4 in Section 7.3). If we replace the battery by a generator, however, we get an equation that is linear but not separable (Example 5). ■ ■

(a) If we put L 4, R 12, and Et 60 in Equation 7, we obtain the initial-value problem 4

or

dI 12I 60 dt

I0 0

dI 3I 15 dt

I0 0

Multiplying by the integrating factor e x 3 dt e 3t, we get e 3t

dI 3e 3tI 15e 3t dt d 3t e I 15e 3t dt e 3tI y 15e 3t dt 5e 3t C It 5 Ce3t

■ ■ Figure 5 shows how the current in Example 4 approaches its limiting value.

6

Since I0 0, we have 5 C 0, so C 5 and It 51 e3t (b) After 1 second the current is

y=5

I1 51 e3 4.75 A lim It lim 51 e3t

(c)

tl

5 5 lim e3t

2.5

0

tl

tl

505

FIGURE 5

EXAMPLE 5 Suppose that the resistance and inductance remain as in Example 4 but, instead of the battery, we use a generator that produces a variable voltage of Et 60 sin 30t volts. Find It. SOLUTION This time the differential equation becomes Figure 6 shows the graph of the current when the battery is replaced by a generator. ■ ■

4

dI 12I 60 sin 30t dt

or

dI 3I 15 sin 30t dt

The same integrating factor e 3t gives

2

dI d 3t e I e 3t 3e 3tI 15e 3t sin 30t dt dt 0

2.5

Using Formula 98 in the Table of Integrals, we have e 3tI y 15e 3t sin 30t dt 15

_2

FIGURE 6

e 3t 3 sin 30t 30 cos 30t C 909

5 I 101 sin 30t 10 cos 30t Ce3t

LINEAR DIFFERENTIAL EQUATIONS ■ 5

Since I0 0, we get 50 101 C0

It 101 sin 30t 10 cos 30t 101 e3t 5

so

50

Exercises A Click here for answers. 1–4

S

Observe that, if n 0 or 1, the Bernoulli equation is linear. For other values of n, show that the substitution u y 1n transforms the Bernoulli equation into the linear equation

Click here for solutions.

Determine whether the differential equation is linear.

1. y e x y x 2 y 2 3. xy ln x x 2 y 0 ■

■

5–14

■

■

du 1 nPxu 1 nQx dx

2. y sin x x 3y 4. y cos y tan x

■

■

■

■

■

■

■

24–26 ■

Solve the differential equation.

5. y 2y 2e

24. xy y xy 2

6. y x 5y

x

7. xy 2y x 2

■

10. 1 xy xy

dy x sin 2x y tan x, dx

■

■

2 x 2

15–20

■

17.

■

■

■

■

■

■

■

■

Solve the initial-value problem.

15. y x y, 16. t

;

■

■

■

■

■

■

■

■

■

of Et 40 sin 60t volts, the inductance is 1 H, the resistance is 20 , and I0 1 A. (a) Find It. (b) Find the current after 0.1 s. (c) Use a graphing device to draw the graph of the current function. a capacitor with a capacitance of C farads (F), and a resistor with a resistance of R ohms (). The voltage drop across the

t 0,

y1 0 C

dv 2 2tv 3t 2e t , v0 5 dt

18. 2xy y 6x,

x 0,

19. xy y x 2 sin x,

dy y x, dx x1

20. x ■

■

■

■

■

E

y4 20

R

y 0 y1 0, ■

■

x0 ■

■

■

■

■

■

capacitor is QC, where Q is the charge (in coulombs), so in this case Kirchhoff’s Law gives

; 21–22

Solve the differential equation and use a graphing calculator or computer to graph several members of the family of solutions. How does the solution curve change as C varies?

21. xy y x cos x, ■

■

29. The figure shows a circuit containing an electromotive force,

y0 2

dy 2y t 3, dt

■

28. In the circuit shown in Figure 4, a generator supplies a voltage

dr r te t dt ■

■

voltage of 40 V, the inductance is 2 H, the resistance is 10 , and I0 0. (a) Find It. (b) Find the current after 0.1 s.

du u 1 t, t 0 13. 1 t dt 14. t ln t

2 y3 y 2 x x

27. In the circuit shown in Figure 4, a battery supplies a constant

dy 2xy x 2 11. dx 12.

25. y

26. y y xy 3

8. x 2 y 2xy cos 2 x

9. xy y sx

Use the method of Exercise 23 to solve the differential

equation.

■

■

■

■

■

x0 ■

RI

But I dQdt (see Example 3 in Section 3.3), so we have

22. y cos xy cos x ■

■

■

■

■

23. A Bernoulli differential equation (named after James

Bernoulli) is of the form dy Pxy Qxy n dx

■

Q Et C

■

R

dQ 1 Q Et dt C

Suppose the resistance is 5 , the capacitance is 0.05 F, a battery gives a constant voltage of 60 V, and the initial charge is Q0 0 C. Find the charge and the current at time t.

■

6 ■ LINEAR DIFFERENTIAL EQUATIONS

30. In the circuit of Exercise 29, R 2 , C 0.01 F, Q0 0,

and Et 10 sin 60t. Find the charge and the current at time t.

31. Let Pt be the performance level of someone learning a skill

as a function of the training time t. The graph of P is called a learning curve. In Exercise 13 in Section 7.1 we proposed the differential equation dP kM Pt dt as a reasonable model for learning, where k is a positive constant. Solve it as a linear differential equation and use your solution to graph the learning curve. 32. Two new workers were hired for an assembly line. Jim

processed 25 units during the first hour and 45 units during the second hour. Mark processed 35 units during the first hour and 50 units the second hour. Using the model of Exercise 31 and assuming that P0 0, estimate the maximum number of units per hour that each worker is capable of processing. 33. In Section 7.3 we looked at mixing problems in which the

volume of fluid remained constant and saw that such problems give rise to separable equations. (See Example 6 in that section.) If the rates of flow into and out of the system are different, then the volume is not constant and the resulting differential equation is linear but not separable. A tank contains 100 L of water. A solution with a salt concentration of 0.4 kgL is added at a rate of 5 Lmin. The solution is kept mixed and is drained from the tank at a rate of 3 Lmin. If yt is the amount of salt (in kilograms) after

t minutes, show that y satisfies the differential equation 3y dy 2 dt 100 2t Solve this equation and find the concentration after 20 minutes. 34. A tank with a capacity of 400 L is full of a mixture of water

and chlorine with a concentration of 0.05 g of chlorine per liter. In order to reduce the concentration of chlorine, fresh water is pumped into the tank at a rate of 4 Ls. The mixture is kept stirred and is pumped out at a rate of 10 Ls. Find the amount of chlorine in the tank as a function of time. 35. An object with mass m is dropped from rest and we assume

that the air resistance is proportional to the speed of the object. If st is the distance dropped after t seconds, then the speed is v st and the acceleration is a vt. If t is the acceleration due to gravity, then the downward force on the object is mt cv, where c is a positive constant, and Newton’s Second Law gives dv m mt cv dt (a) Solve this as a linear equation to show that v

mt 1 ectm c

(b) What is the limiting velocity? (c) Find the distance the object has fallen after t seconds. 36. If we ignore air resistance, we can conclude that heavier

objects fall no faster than lighter objects. But if we take air resistance into account, our conclusion changes. Use the expression for the velocity of a falling object in Exercise 35(a) to find dvdm and show that heavier objects do fall faster than lighter ones.

LINEAR DIFFERENTIAL EQUATIONS ■ 7

Answers

1. No

5. y 3 e x Ce2x 2

3. Yes

7. y x 2 ln x Cx 2 11. 13. 15. 19. 21.

25. 27. 29. 31.

Click here for solutions.

S

x

2

9. y 3 sx Cx 2

1 x 2

2

y Cx 4 25x 12 (a) It 4 4e5t (b) 4 4e12 1.57 A 4t Qt 31 e , It 12e4t P(t) Pt M Cekt M

2

y x Ce e x e dx u t 2 2t 2C2t 1 2 2 y x 1 3e x 17. v t 3e t 5e t y x cos x x y sin x cos xx Cx 6 1 2

x

P(0) 0

33. y 5 100 2t 40,000100 2t32; 0.2275 kgL

C=2 C=1 C=0.2

0

35. (b) mtc 10

C=_1 C=_2 _4

2

(c) mtct mcectm m 2tc 2

t

8 ■ LINEAR DIFFERENTIAL EQUATIONS

Solutions: Linear Differential Equations 1. y 0 + ex y = x2 y 2 is not linear since it cannot be put into the standard linear form (1), y 0 + P (x) y = Q(x). 3. xy0 + ln x − x2 y = 0 ⇒ xy 0 − x2 y = − ln x ⇒

y0 + (−x) y = −

ln x , which is in the standard linear x

form (1), so this equation is linear. 5. Comparing the given equation, y0 + 2y = 2ex , with the general form, y0 + P (x)y = Q(x), we see that P (x) = 2 R

R

and the integrating factor is I(x) = e P (x)dx = e 2 dx = e2x . Multiplying the differential equation by I(x) gives ¡ ¢0 R e2x y 0 + 2e2x y = 2e3x ⇒ e2x y = 2e3x ⇒ e2x y = 2e3x dx ⇒ e2x y = 23 e3x + C ⇒ y = 23 ex + Ce−2x . 0

2

7. xy − 2y = x I(x) = e

R

[divide by x] ⇒

P (x) dx

=e

R

(−2/x) dx

equation (∗) by I(x) gives

µ

2 y + − x 0

¶

y = x (∗). −2

= e−2 ln|x| = eln|x|

2 1 1 0 y − 3y= x2 x x

⇒

y = x2 (ln |x| + C ) = x2 ln |x| + Cx2 .

µ

2

= eln(1/x ) = 1/x2 . Multiplying the differential ¶0 1 1 1 y = y = ln |x| + C ⇒ ⇒ x2 x x2

9. Since P (x) is the derivative of the coefficient of y0 [P (x) = 1 and the coefficient is x], we can write the √ √ differential equation xy 0 + y = x in the easily integrable form (xy)0 = x ⇒ xy = 23 x3/2 + C ⇒ √ y = 23 x + C/x. R

2

= ex . Multiplying the differential equation y 0 + 2xy = x2 by I(x) gives ³ 2 ´0 2 2 2 2 ex y 0 + 2xex y = x2 ex ⇒ ex y = x2 ex . Thus

11. I(x) = e

2

y = e−x

13. (1 + t)

2x dx

hR

i h R 2 2 2 x2 ex dx + C = e−x 12 xex −

1 x2 e 2

du + u = 1 + t, t > 0 [divide by 1 + t] dt

⇒

u0 + P (t) u = Q(t). The integrating factor is I(t) = e

R

i 2 2 R dx + C = 12 x + Ce−x − e−x

1 x2 e 2

dx.

du 1 + u = 1 (∗), which has the form dt 1+t P (t) dt

=e

R

[1/(1+t)] dt

= eln(1+t) = 1 + t.

Multiplying (∗) by I(t) gives us our original equation back. We rewrite it as [(1 + t) u]0 = 1 + t. Thus, (1 + t) u =

15. y 0 = x + y ⇒

R

(1 + t) dt = t + 12 t2 + C

t + 12 t2 + C t2 + 2t + 2C or u = . 1+t 2 (t + 1)

R

y0 + (−1)y = x. I(x) = e (−1) dx = e−x . Multiplying by e−x gives e−x y 0 − e−x y = xe−x R (e−x y)0 = xe−x ⇒ e−x y = xe−x dx = −xe−x − e−x + C [integration by parts with u = x, ⇒

dv = e−x dx]

17.

⇒ u=

⇒ y = −x − 1 + Cex . y(0) = 2

⇒ −1 + C = 2 ⇒ C = 3, so y = −x − 1 + 3ex .

R 2 2 dv − 2tv = 3t2 et , v (0) = 5. I(t) = e (−2t)dt = e−t . Multiply the differential equation by I(t) to get dt ³ 2 ´0 R 2 dv 2 2 2 2 e−t − 2te−t v = 3t2 ⇒ e−t v = 3t2 ⇒ e−t v = 3t2 dt = t3 + C ⇒ v = t3 et + Cet . dt 2

2

5 = v(0) = 0 · 1 + C · 1 = C, so v = t3 et + 5et .

LINEAR DIFFERENTIAL EQUATIONS ■ 9

R −1 1 1 y = x sin x. I(x) = e (−1/x) dx = e− ln x = eln x = . x x ¶0 µ 1 1 1 1 1 y = sin x ⇒ y = − cos x + C Multiplying by gives y0 − 2 y = sin x ⇒ x x x x x

19. xy0 = y + x2 sin x ⇒

y0 −

⇒

y = −x cos x + Cx. y(π) = 0 ⇒ −π · (−1) + Cπ = 0 ⇒ C = −1, so y = −x cos x − x. R 1 y = cos x (x 6= 0), so I(x) = e (1/x)dx = eln|x| = x (for x x > 0). Multiplying the differential equation by I(x) gives

21. y 0 +

xy0 + y = x cos x ⇒ (xy)0 = x cos x. Thus, ·Z ¸ 1 1 y= x cos x dx + C = [x sin x + cos x + C] x x = sin x +

C cos x + x x

The solutions are asymptotic to the y-axis (except for C = −1). In fact, for C > −1, y → ∞ as x → 0+ , whereas for C < −1, y → −∞ as x → 0+ . As x gets larger, the solutions approximate y = sin x more closely. The graphs for larger C lie above those for smaller C. The distance between the graphs lessens as x increases. 23. Setting u = y 1−n ,

equation becomes

dy du dy yn du un/(1−n) du = (1 − n) y −n or = = . Then the Bernoulli differential dx dx dx 1 − n dx 1 − n dx un/(1−n) du du + P (x)u1/(1−n) = Q(x)un/(1−n) or + (1 − n)P (x)u = Q(x)(1 − n). 1 − n dx dx

y3 1 2 2 2 4u y = 2 . Here n = 3, P (x) = , Q(x) = 2 and setting u = y −2 , u satisfies u0 − = − 2. x x x x x x µZ µ ¶ ¶ R 2 2 2 Then I(x) = e (−4/x)dx = x−4 and u = x4 − 6 dx + C = x4 + C = Cx4 + . x 5x5 5x

25. y 0 +

µ ¶−1/2 2 Thus, y = ± Cx4 + . 5x 27. (a) 2

R dI dI + 10I = 40 or + 5I = 20. Then the integrating factor is e dt dt

equation by the integrating factor gives e5t I(t) = e−5t

£R

dI + 5Ie5t = 20e5t dt

5 dt

¡

⇒

= e5t . Multiplying the differential e5t I

¢0

= 20e5t

⇒

¤ 20e5t dt + C = 4 + Ce−5t . But 0 = I(0) = 4 + C, so I(t) = 4 − 4e−5t .

(b) I(0.1) = 4 − 4e−0.5 ≈ 1.57 A 29. 5

R dQ + 20Q = 60 with Q(0) = 0 C. Then the integrating factor is e dt

equation by the integrating factor gives e4t Q(t) = e−4t

£R

dQ + 4e4t Q = 12e4t dt

4 dt

⇒

¡

= e4t , and multiplying the differential e4t Q

¢0

= 12e4t

⇒

¢ ¤ ¡ 12e4t dt + C = 3 + Ce−4t . But 0 = Q(0) = 3 + C so Q(t) = 3 1 − e−4t is the charge at time

t and I = dQ/dt = 12e−4t is the current at time t.

10 ■ LINEAR DIFFERENTIAL EQUATIONS

31.

R dP + kP = kM , so I(t) = e k dt = ekt . Multiplying the differential dt dP equation by I(t) gives ekt + kP ekt = kM ekt ⇒ dt ¡ kt ¢0 e P = kM ekt ⇒ ¡R ¢ P (t) = e−kt kM ekt dt + C = M + Ce−kt , k > 0. Furthermore, it is

reasonable to assume that 0 ≤ P (0) ≤ M , so −M ≤ C ≤ 0. µ

33. y(0) = 0 kg. Salt is added at a rate of

0.4

kg L

¶µ

5

L min

¶

=2

kg . Since solution is drained from the tank at a min

rate of 3 L/min, but salt solution is added at a rate of 5 L/min, the tank, which starts out with 100 L of water, contains (100 + 2t) L of liquid after t min. Thus, the salt concentration at time t is µ

y(t) kg 100 + 2t L

¶µ

¶

y(t) kg . Salt therefore 100 + 2t L

3y kg . Combining the rates at which salt enters 100 + 2t min µ ¶ dy 3y dy 3 and leaves the tank, we get =2− . Rewriting this equation as + y = 2, we see that dt 100 + 2t dt 100 + 2t µZ ¶ ¢ ¡ 3 dt it is linear. I(t) = exp = exp 32 ln(100 + 2t) = (100 + 2t)3/2 . Multiplying the differential 100 + 2t leaves the tank at a rate of

equation by I(t) gives (100 + 2t)3/2 h

(100 + 2t)3/2 y

i0

L 3 min

=

dy + 3(100 + 2t)1/2 y = 2(100 + 2t)3/2 dt

= 2(100 + 2t)3/2

⇒

⇒

(100 + 2t)3/2 y = 25 (100 + 2t)5/2 + C

⇒

1 y = 25 (100 + 2t) + C(100 + 2t)−3/2 . Now 0 = y(0) = 25 (100) + C · 100−3/2 = 40 + 1000 C ⇒ i h C = −40,000, so y = 25 (100 + 2t) − 40,000(100 + 2t)−3/2 kg. From this solution (no pun intended), we

y(t) calculate the salt concentration at time t to be C(t) = = 100 + 2t

"

−40,000

5/2

(100 + 2t)

2 + 5

#

kg . In particular, L

−40,000 2 kg + ≈ 0.2275 and y(20) = 25 (140) − 40,000(140)−3/2 ≈ 31.85 kg. 1405/2 5 L R c dv + v = g and I(t) = e (c/m)dt = e(c/m)t , and multiplying the differential equation by I(t) gives 35. (a) dt m h i0 dv vce(c/m)t e(c/m)t + = ge(c/m)t ⇒ e(c/m)t v = ge(c/m)t . Hence, dt i hRm v(t) = e−(c/m)t ge(c/m)t dt + K = mg/c + Ke−(c/m)t . But the object is dropped from rest, so v(0) = 0 i h and K = −mg/c. Thus, the velocity at time t is v(t) = (mg/c) 1 − e−(c/m)t . C(20) =

(b) lim v(t) = mg/c t→∞

h i v(t) dt = (mg/c) t + (m/c)e−(c/m)t + c1 where c1 = s(0) − m2 g/c2 . s(0) is the initial position, i h so s(0) = 0 and s(t) = (mg/c) t + (m/c)e−(c/m)t − m2 g/c2 .

(c) s(t) =

R

1

where P and Q are continuous functions on a given interval. This type of equation occurs frequently in various sciences, as we will see. An example of a linear equation is xy y 2x because, for x 0, it can be written in the form y

2

1 y2 x

Notice that this differential equation is not separable because it’s impossible to factor the expression for y as a function of x times a function of y. But we can still solve the equation by noticing, by the Product Rule, that xy y xy and so we can rewrite the equation as xy 2x If we now integrate both sides of this equation, we get xy x 2 C

or

C x

yx

If we had been given the differential equation in the form of Equation 2, we would have had to take the preliminary step of multiplying each side of the equation by x. It turns out that every first-order linear differential equation can be solved in a similar fashion by multiplying both sides of Equation 1 by a suitable function Ix called an integrating factor. We try to find I so that the left side of Equation 1, when multiplied by Ix, becomes the derivative of the product Ixy: Ixy Pxy Ixy

3

If we can find such a function I , then Equation 1 becomes Ixy IxQx Integrating both sides, we would have Ixy y IxQx dx C so the solution would be 4

yx

1 Ix

y

IxQx dx C

To find such an I, we expand Equation 3 and cancel terms: Ixy IxPxy Ixy Ixy Ixy IxPx Ix 1

2 ■ LINEAR DIFFERENTIAL EQUATIONS

This is a separable differential equation for I , which we solve as follows:

y

dI y Px dx I

ln I y Px dx I Ae x Px dx where A e C. We are looking for a particular integrating factor, not the most general one, so we take A 1 and use Ix e x Px dx

5

Thus, a formula for the general solution to Equation 1 is provided by Equation 4, where I is given by Equation 5. Instead of memorizing this formula, however, we just remember the form of the integrating factor. To solve the linear differential equation y Pxy Qx, multiply both sides by the integrating factor Ix e x Px dx and integrate both sides.

EXAMPLE 1 Solve the differential equation

dy 3x 2 y 6x 2. dx

SOLUTION The given equation is linear since it has the form of Equation 1 with

Px 3x 2 and Qx 6x 2. An integrating factor is Ix e x 3x

2

dx

ex

3

3

Multiplying both sides of the differential equation by e x , we get Figure 1 shows the graphs of several members of the family of solutions in Example 1. Notice that they all approach 2 as x l . ■ ■

ex

3

dy 3 3 3x 2e x y 6x 2e x dx

6 C=2

d x3 3 e y 6x 2e x dx

or

C=1 C=0

Integrating both sides, we have

C=_1 _1.5

1.8 3

3

3

e x y y 6x 2e x dx 2e x C

C=_2 _3

y 2 Cex

FIGURE 1

3

EXAMPLE 2 Find the solution of the initial-value problem

x 2 y xy 1

x0

y1 2

SOLUTION We must first divide both sides by the coefficient of y to put the differential equation into standard form: 6

y

1 1 y 2 x x

x0

The integrating factor is Ix e x 1x dx e ln x x

LINEAR DIFFERENTIAL EQUATIONS ■ 3

Multiplication of Equation 6 by x gives 1 x

xy y

■ ■ The solution of the initial-value problem in Example 2 is shown in Figure 2.

ln x C x

y

and so

5

1 x

1 dx ln x C x

xy y

Then

xy

or

Since y1 2, we have

(1, 2) 0

2

4

ln 1 C C 1

Therefore, the solution to the initial-value problem is _5

ln x 2 x

y

FIGURE 2

EXAMPLE 3 Solve y 2xy 1. SOLUTION The given equation is in the standard form for a linear equation. Multiplying by the integrating factor

e x 2x dx e x we get

2

2

e x y 2xe x y e x 2

(e x y) e x

or Even though the solutions of the differential equation in Example 3 are expressed in terms of an integral, they can still be graphed by a computer algebra system (Figure 3). ■ ■

2

2

2

2

2

e x y y e x dx C

Therefore 2

Recall from Section 5.8 that x e x dx can’t be expressed in terms of elementary functions. Nonetheless, it’s a perfectly good function and we can leave the answer as

2.5

y ex

C=2 _2.5

2.5

2

ye

dx Cex

2

Another way of writing the solution is

C=_2

y ex

2

y

_2.5

FIGURE 3

x2

x

0

2

e t dt Cex

2

(Any number can be chosen for the lower limit of integration.) Application to Electric Circuits R

E

L

switch

In Section 7.2 we considered the simple electric circuit shown in Figure 4: An electromotive force (usually a battery or generator) produces a voltage of Et volts (V) and a current of It amperes (A) at time t . The circuit also contains a resistor with a resistance of R ohms () and an inductor with an inductance of L henries (H). Ohm’s Law gives the drop in voltage due to the resistor as RI . The voltage drop due to the inductor is LdIdt. One of Kirchhoff’s laws says that the sum of the voltage drops is equal to the supplied voltage Et. Thus, we have

FIGURE 4 7

L

dI RI Et dt

which is a first-order linear differential equation. The solution gives the current I at time t .

4 ■ LINEAR DIFFERENTIAL EQUATIONS

EXAMPLE 4 Suppose that in the simple circuit of Figure 4 the resistance is 12 and the inductance is 4 H. If a battery gives a constant voltage of 60 V and the switch is closed when t 0 so the current starts with I0 0, find (a) It, (b) the current after 1 s, and (c) the limiting value of the current. SOLUTION The differential equation in Example 4 is both linear and separable, so an alternative method is to solve it as a separable equation (Example 4 in Section 7.3). If we replace the battery by a generator, however, we get an equation that is linear but not separable (Example 5). ■ ■

(a) If we put L 4, R 12, and Et 60 in Equation 7, we obtain the initial-value problem 4

or

dI 12I 60 dt

I0 0

dI 3I 15 dt

I0 0

Multiplying by the integrating factor e x 3 dt e 3t, we get e 3t

dI 3e 3tI 15e 3t dt d 3t e I 15e 3t dt e 3tI y 15e 3t dt 5e 3t C It 5 Ce3t

■ ■ Figure 5 shows how the current in Example 4 approaches its limiting value.

6

Since I0 0, we have 5 C 0, so C 5 and It 51 e3t (b) After 1 second the current is

y=5

I1 51 e3 4.75 A lim It lim 51 e3t

(c)

tl

5 5 lim e3t

2.5

0

tl

tl

505

FIGURE 5

EXAMPLE 5 Suppose that the resistance and inductance remain as in Example 4 but, instead of the battery, we use a generator that produces a variable voltage of Et 60 sin 30t volts. Find It. SOLUTION This time the differential equation becomes Figure 6 shows the graph of the current when the battery is replaced by a generator. ■ ■

4

dI 12I 60 sin 30t dt

or

dI 3I 15 sin 30t dt

The same integrating factor e 3t gives

2

dI d 3t e I e 3t 3e 3tI 15e 3t sin 30t dt dt 0

2.5

Using Formula 98 in the Table of Integrals, we have e 3tI y 15e 3t sin 30t dt 15

_2

FIGURE 6

e 3t 3 sin 30t 30 cos 30t C 909

5 I 101 sin 30t 10 cos 30t Ce3t

LINEAR DIFFERENTIAL EQUATIONS ■ 5

Since I0 0, we get 50 101 C0

It 101 sin 30t 10 cos 30t 101 e3t 5

so

50

Exercises A Click here for answers. 1–4

S

Observe that, if n 0 or 1, the Bernoulli equation is linear. For other values of n, show that the substitution u y 1n transforms the Bernoulli equation into the linear equation

Click here for solutions.

Determine whether the differential equation is linear.

1. y e x y x 2 y 2 3. xy ln x x 2 y 0 ■

■

5–14

■

■

du 1 nPxu 1 nQx dx

2. y sin x x 3y 4. y cos y tan x

■

■

■

■

■

■

■

24–26 ■

Solve the differential equation.

5. y 2y 2e

24. xy y xy 2

6. y x 5y

x

7. xy 2y x 2

■

10. 1 xy xy

dy x sin 2x y tan x, dx

■

■

2 x 2

15–20

■

17.

■

■

■

■

■

■

■

■

Solve the initial-value problem.

15. y x y, 16. t

;

■

■

■

■

■

■

■

■

■

of Et 40 sin 60t volts, the inductance is 1 H, the resistance is 20 , and I0 1 A. (a) Find It. (b) Find the current after 0.1 s. (c) Use a graphing device to draw the graph of the current function. a capacitor with a capacitance of C farads (F), and a resistor with a resistance of R ohms (). The voltage drop across the

t 0,

y1 0 C

dv 2 2tv 3t 2e t , v0 5 dt

18. 2xy y 6x,

x 0,

19. xy y x 2 sin x,

dy y x, dx x1

20. x ■

■

■

■

■

E

y4 20

R

y 0 y1 0, ■

■

x0 ■

■

■

■

■

■

capacitor is QC, where Q is the charge (in coulombs), so in this case Kirchhoff’s Law gives

; 21–22

Solve the differential equation and use a graphing calculator or computer to graph several members of the family of solutions. How does the solution curve change as C varies?

21. xy y x cos x, ■

■

29. The figure shows a circuit containing an electromotive force,

y0 2

dy 2y t 3, dt

■

28. In the circuit shown in Figure 4, a generator supplies a voltage

dr r te t dt ■

■

voltage of 40 V, the inductance is 2 H, the resistance is 10 , and I0 0. (a) Find It. (b) Find the current after 0.1 s.

du u 1 t, t 0 13. 1 t dt 14. t ln t

2 y3 y 2 x x

27. In the circuit shown in Figure 4, a battery supplies a constant

dy 2xy x 2 11. dx 12.

25. y

26. y y xy 3

8. x 2 y 2xy cos 2 x

9. xy y sx

Use the method of Exercise 23 to solve the differential

equation.

■

■

■

■

■

x0 ■

RI

But I dQdt (see Example 3 in Section 3.3), so we have

22. y cos xy cos x ■

■

■

■

■

23. A Bernoulli differential equation (named after James

Bernoulli) is of the form dy Pxy Qxy n dx

■

Q Et C

■

R

dQ 1 Q Et dt C

Suppose the resistance is 5 , the capacitance is 0.05 F, a battery gives a constant voltage of 60 V, and the initial charge is Q0 0 C. Find the charge and the current at time t.

■

6 ■ LINEAR DIFFERENTIAL EQUATIONS

30. In the circuit of Exercise 29, R 2 , C 0.01 F, Q0 0,

and Et 10 sin 60t. Find the charge and the current at time t.

31. Let Pt be the performance level of someone learning a skill

as a function of the training time t. The graph of P is called a learning curve. In Exercise 13 in Section 7.1 we proposed the differential equation dP kM Pt dt as a reasonable model for learning, where k is a positive constant. Solve it as a linear differential equation and use your solution to graph the learning curve. 32. Two new workers were hired for an assembly line. Jim

processed 25 units during the first hour and 45 units during the second hour. Mark processed 35 units during the first hour and 50 units the second hour. Using the model of Exercise 31 and assuming that P0 0, estimate the maximum number of units per hour that each worker is capable of processing. 33. In Section 7.3 we looked at mixing problems in which the

volume of fluid remained constant and saw that such problems give rise to separable equations. (See Example 6 in that section.) If the rates of flow into and out of the system are different, then the volume is not constant and the resulting differential equation is linear but not separable. A tank contains 100 L of water. A solution with a salt concentration of 0.4 kgL is added at a rate of 5 Lmin. The solution is kept mixed and is drained from the tank at a rate of 3 Lmin. If yt is the amount of salt (in kilograms) after

t minutes, show that y satisfies the differential equation 3y dy 2 dt 100 2t Solve this equation and find the concentration after 20 minutes. 34. A tank with a capacity of 400 L is full of a mixture of water

and chlorine with a concentration of 0.05 g of chlorine per liter. In order to reduce the concentration of chlorine, fresh water is pumped into the tank at a rate of 4 Ls. The mixture is kept stirred and is pumped out at a rate of 10 Ls. Find the amount of chlorine in the tank as a function of time. 35. An object with mass m is dropped from rest and we assume

that the air resistance is proportional to the speed of the object. If st is the distance dropped after t seconds, then the speed is v st and the acceleration is a vt. If t is the acceleration due to gravity, then the downward force on the object is mt cv, where c is a positive constant, and Newton’s Second Law gives dv m mt cv dt (a) Solve this as a linear equation to show that v

mt 1 ectm c

(b) What is the limiting velocity? (c) Find the distance the object has fallen after t seconds. 36. If we ignore air resistance, we can conclude that heavier

objects fall no faster than lighter objects. But if we take air resistance into account, our conclusion changes. Use the expression for the velocity of a falling object in Exercise 35(a) to find dvdm and show that heavier objects do fall faster than lighter ones.

LINEAR DIFFERENTIAL EQUATIONS ■ 7

Answers

1. No

5. y 3 e x Ce2x 2

3. Yes

7. y x 2 ln x Cx 2 11. 13. 15. 19. 21.

25. 27. 29. 31.

Click here for solutions.

S

x

2

9. y 3 sx Cx 2

1 x 2

2

y Cx 4 25x 12 (a) It 4 4e5t (b) 4 4e12 1.57 A 4t Qt 31 e , It 12e4t P(t) Pt M Cekt M

2

y x Ce e x e dx u t 2 2t 2C2t 1 2 2 y x 1 3e x 17. v t 3e t 5e t y x cos x x y sin x cos xx Cx 6 1 2

x

P(0) 0

33. y 5 100 2t 40,000100 2t32; 0.2275 kgL

C=2 C=1 C=0.2

0

35. (b) mtc 10

C=_1 C=_2 _4

2

(c) mtct mcectm m 2tc 2

t

8 ■ LINEAR DIFFERENTIAL EQUATIONS

Solutions: Linear Differential Equations 1. y 0 + ex y = x2 y 2 is not linear since it cannot be put into the standard linear form (1), y 0 + P (x) y = Q(x). 3. xy0 + ln x − x2 y = 0 ⇒ xy 0 − x2 y = − ln x ⇒

y0 + (−x) y = −

ln x , which is in the standard linear x

form (1), so this equation is linear. 5. Comparing the given equation, y0 + 2y = 2ex , with the general form, y0 + P (x)y = Q(x), we see that P (x) = 2 R

R

and the integrating factor is I(x) = e P (x)dx = e 2 dx = e2x . Multiplying the differential equation by I(x) gives ¡ ¢0 R e2x y 0 + 2e2x y = 2e3x ⇒ e2x y = 2e3x ⇒ e2x y = 2e3x dx ⇒ e2x y = 23 e3x + C ⇒ y = 23 ex + Ce−2x . 0

2

7. xy − 2y = x I(x) = e

R

[divide by x] ⇒

P (x) dx

=e

R

(−2/x) dx

equation (∗) by I(x) gives

µ

2 y + − x 0

¶

y = x (∗). −2

= e−2 ln|x| = eln|x|

2 1 1 0 y − 3y= x2 x x

⇒

y = x2 (ln |x| + C ) = x2 ln |x| + Cx2 .

µ

2

= eln(1/x ) = 1/x2 . Multiplying the differential ¶0 1 1 1 y = y = ln |x| + C ⇒ ⇒ x2 x x2

9. Since P (x) is the derivative of the coefficient of y0 [P (x) = 1 and the coefficient is x], we can write the √ √ differential equation xy 0 + y = x in the easily integrable form (xy)0 = x ⇒ xy = 23 x3/2 + C ⇒ √ y = 23 x + C/x. R

2

= ex . Multiplying the differential equation y 0 + 2xy = x2 by I(x) gives ³ 2 ´0 2 2 2 2 ex y 0 + 2xex y = x2 ex ⇒ ex y = x2 ex . Thus

11. I(x) = e

2

y = e−x

13. (1 + t)

2x dx

hR

i h R 2 2 2 x2 ex dx + C = e−x 12 xex −

1 x2 e 2

du + u = 1 + t, t > 0 [divide by 1 + t] dt

⇒

u0 + P (t) u = Q(t). The integrating factor is I(t) = e

R

i 2 2 R dx + C = 12 x + Ce−x − e−x

1 x2 e 2

dx.

du 1 + u = 1 (∗), which has the form dt 1+t P (t) dt

=e

R

[1/(1+t)] dt

= eln(1+t) = 1 + t.

Multiplying (∗) by I(t) gives us our original equation back. We rewrite it as [(1 + t) u]0 = 1 + t. Thus, (1 + t) u =

15. y 0 = x + y ⇒

R

(1 + t) dt = t + 12 t2 + C

t + 12 t2 + C t2 + 2t + 2C or u = . 1+t 2 (t + 1)

R

y0 + (−1)y = x. I(x) = e (−1) dx = e−x . Multiplying by e−x gives e−x y 0 − e−x y = xe−x R (e−x y)0 = xe−x ⇒ e−x y = xe−x dx = −xe−x − e−x + C [integration by parts with u = x, ⇒

dv = e−x dx]

17.

⇒ u=

⇒ y = −x − 1 + Cex . y(0) = 2

⇒ −1 + C = 2 ⇒ C = 3, so y = −x − 1 + 3ex .

R 2 2 dv − 2tv = 3t2 et , v (0) = 5. I(t) = e (−2t)dt = e−t . Multiply the differential equation by I(t) to get dt ³ 2 ´0 R 2 dv 2 2 2 2 e−t − 2te−t v = 3t2 ⇒ e−t v = 3t2 ⇒ e−t v = 3t2 dt = t3 + C ⇒ v = t3 et + Cet . dt 2

2

5 = v(0) = 0 · 1 + C · 1 = C, so v = t3 et + 5et .

LINEAR DIFFERENTIAL EQUATIONS ■ 9

R −1 1 1 y = x sin x. I(x) = e (−1/x) dx = e− ln x = eln x = . x x ¶0 µ 1 1 1 1 1 y = sin x ⇒ y = − cos x + C Multiplying by gives y0 − 2 y = sin x ⇒ x x x x x

19. xy0 = y + x2 sin x ⇒

y0 −

⇒

y = −x cos x + Cx. y(π) = 0 ⇒ −π · (−1) + Cπ = 0 ⇒ C = −1, so y = −x cos x − x. R 1 y = cos x (x 6= 0), so I(x) = e (1/x)dx = eln|x| = x (for x x > 0). Multiplying the differential equation by I(x) gives

21. y 0 +

xy0 + y = x cos x ⇒ (xy)0 = x cos x. Thus, ·Z ¸ 1 1 y= x cos x dx + C = [x sin x + cos x + C] x x = sin x +

C cos x + x x

The solutions are asymptotic to the y-axis (except for C = −1). In fact, for C > −1, y → ∞ as x → 0+ , whereas for C < −1, y → −∞ as x → 0+ . As x gets larger, the solutions approximate y = sin x more closely. The graphs for larger C lie above those for smaller C. The distance between the graphs lessens as x increases. 23. Setting u = y 1−n ,

equation becomes

dy du dy yn du un/(1−n) du = (1 − n) y −n or = = . Then the Bernoulli differential dx dx dx 1 − n dx 1 − n dx un/(1−n) du du + P (x)u1/(1−n) = Q(x)un/(1−n) or + (1 − n)P (x)u = Q(x)(1 − n). 1 − n dx dx

y3 1 2 2 2 4u y = 2 . Here n = 3, P (x) = , Q(x) = 2 and setting u = y −2 , u satisfies u0 − = − 2. x x x x x x µZ µ ¶ ¶ R 2 2 2 Then I(x) = e (−4/x)dx = x−4 and u = x4 − 6 dx + C = x4 + C = Cx4 + . x 5x5 5x

25. y 0 +

µ ¶−1/2 2 Thus, y = ± Cx4 + . 5x 27. (a) 2

R dI dI + 10I = 40 or + 5I = 20. Then the integrating factor is e dt dt

equation by the integrating factor gives e5t I(t) = e−5t

£R

dI + 5Ie5t = 20e5t dt

5 dt

¡

⇒

= e5t . Multiplying the differential e5t I

¢0

= 20e5t

⇒

¤ 20e5t dt + C = 4 + Ce−5t . But 0 = I(0) = 4 + C, so I(t) = 4 − 4e−5t .

(b) I(0.1) = 4 − 4e−0.5 ≈ 1.57 A 29. 5

R dQ + 20Q = 60 with Q(0) = 0 C. Then the integrating factor is e dt

equation by the integrating factor gives e4t Q(t) = e−4t

£R

dQ + 4e4t Q = 12e4t dt

4 dt

⇒

¡

= e4t , and multiplying the differential e4t Q

¢0

= 12e4t

⇒

¢ ¤ ¡ 12e4t dt + C = 3 + Ce−4t . But 0 = Q(0) = 3 + C so Q(t) = 3 1 − e−4t is the charge at time

t and I = dQ/dt = 12e−4t is the current at time t.

10 ■ LINEAR DIFFERENTIAL EQUATIONS

31.

R dP + kP = kM , so I(t) = e k dt = ekt . Multiplying the differential dt dP equation by I(t) gives ekt + kP ekt = kM ekt ⇒ dt ¡ kt ¢0 e P = kM ekt ⇒ ¡R ¢ P (t) = e−kt kM ekt dt + C = M + Ce−kt , k > 0. Furthermore, it is

reasonable to assume that 0 ≤ P (0) ≤ M , so −M ≤ C ≤ 0. µ

33. y(0) = 0 kg. Salt is added at a rate of

0.4

kg L

¶µ

5

L min

¶

=2

kg . Since solution is drained from the tank at a min

rate of 3 L/min, but salt solution is added at a rate of 5 L/min, the tank, which starts out with 100 L of water, contains (100 + 2t) L of liquid after t min. Thus, the salt concentration at time t is µ

y(t) kg 100 + 2t L

¶µ

¶

y(t) kg . Salt therefore 100 + 2t L

3y kg . Combining the rates at which salt enters 100 + 2t min µ ¶ dy 3y dy 3 and leaves the tank, we get =2− . Rewriting this equation as + y = 2, we see that dt 100 + 2t dt 100 + 2t µZ ¶ ¢ ¡ 3 dt it is linear. I(t) = exp = exp 32 ln(100 + 2t) = (100 + 2t)3/2 . Multiplying the differential 100 + 2t leaves the tank at a rate of

equation by I(t) gives (100 + 2t)3/2 h

(100 + 2t)3/2 y

i0

L 3 min

=

dy + 3(100 + 2t)1/2 y = 2(100 + 2t)3/2 dt

= 2(100 + 2t)3/2

⇒

⇒

(100 + 2t)3/2 y = 25 (100 + 2t)5/2 + C

⇒

1 y = 25 (100 + 2t) + C(100 + 2t)−3/2 . Now 0 = y(0) = 25 (100) + C · 100−3/2 = 40 + 1000 C ⇒ i h C = −40,000, so y = 25 (100 + 2t) − 40,000(100 + 2t)−3/2 kg. From this solution (no pun intended), we

y(t) calculate the salt concentration at time t to be C(t) = = 100 + 2t

"

−40,000

5/2

(100 + 2t)

2 + 5

#

kg . In particular, L

−40,000 2 kg + ≈ 0.2275 and y(20) = 25 (140) − 40,000(140)−3/2 ≈ 31.85 kg. 1405/2 5 L R c dv + v = g and I(t) = e (c/m)dt = e(c/m)t , and multiplying the differential equation by I(t) gives 35. (a) dt m h i0 dv vce(c/m)t e(c/m)t + = ge(c/m)t ⇒ e(c/m)t v = ge(c/m)t . Hence, dt i hRm v(t) = e−(c/m)t ge(c/m)t dt + K = mg/c + Ke−(c/m)t . But the object is dropped from rest, so v(0) = 0 i h and K = −mg/c. Thus, the velocity at time t is v(t) = (mg/c) 1 − e−(c/m)t . C(20) =

(b) lim v(t) = mg/c t→∞

h i v(t) dt = (mg/c) t + (m/c)e−(c/m)t + c1 where c1 = s(0) − m2 g/c2 . s(0) is the initial position, i h so s(0) = 0 and s(t) = (mg/c) t + (m/c)e−(c/m)t − m2 g/c2 .

(c) s(t) =

R