Section 9.5 Linear Equations

Section 9.5 Linear Equations

30. Solve the second-order equation xy⁰⁰+ 2y⁰= 12x² by making the substitution u = y⁰. Solution:

SECTION 9.5 LINEAR EQUATIONS ¤ 841 21. ⁰+ 2 = ^ ⇒ ⁰+2

 =^

.

() = ^{(2) }= ^{2 ln||}=

^ln||2

= ||²= ².

Multiplying by () gives ²⁰+ 2 = ^ ⇒ (²)⁰= ^ ⇒

² =

^ = ( − 1)^+  [by parts] ⇒

 = [( − 1)^+ ]². The graphs for  = −5, −3, −1, 1, 3, 5, and 7 are shown.  = 1 is a transitional value. For   1, there is an inflection point and for   1, there is a local minimum. As || gets larger, the “branches” get further from the origin.

22. ⁰= ²+ 2 ⇔ ⁰− 2 = ² ⇔ ⁰−2

 = .

() = ^{−2 }= ^{−2 ln||}= (^ln||)⁻²= ||⁻²= 1

². Multiplying by

()gives 1

²⁰− 2

³ = 1

 ⇒

 1

²

₀

= 1

 ⇒ 1

² =

 1

 ⇒ 1

² = ln || +  ⇒  = (ln || + )². For all values of , as || → 0,

 → 0, and as || → ∞,  → ∞. As || increases from 0, the function decreases and attains an absolute minimum.

The inflection points, absolute minimums, and -intercepts all move farther from the origin as  decreases.

23. Setting  = ^1−, 

 = (1 − ) ^−

or 

 = ^ 1 − 



 = ^{(1−)} 1 − 



. Then the Bernoulli differential equation becomes^{(1−)}

1 − 



+  ()^{1(1−)}= ()^{(1−)}or

+ (1 − ) () = ()(1 − ).

24. Here ⁰+  = −² ⇒ ⁰+

= −², so  = 2,  () = 1

and () = −1. Setting  = ⁻¹,  satisfies

⁰−1

 = 1. Then () = ^{(−1)}= 1

 (for   0) and  =  1

 + 



= (ln || + ). Thus,

 = 1

( + ln ||).

25. Here ⁰+2

 = ³

², so  = 3,  () = 2

and () = 1

². Setting  = ⁻²,  satisfies ⁰−4

 = −2

². Then () = ^{(−4) }= ⁻⁴and  = ⁴

−2

⁶ + 



= ⁴

 2 5⁵ + 



= ⁴+ 2 5. Thus,  = ±



⁴+ 2 5

−12

.

26. ⁰⁰+ 2⁰ = 12²and  = ⁰ ⇒ ⁰+ 2 = 12² ⇒ ⁰+2

 = 12.

() = ^{(2) }= ^{2 ln||}=

^ln||2

= ||²= ². Multiplying the last differential equation by ²gives

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SECTION 9.5 LINEAR EQUATIONS ¤ 841 21. ⁰+ 2 = ^ ⇒ ⁰+2

 =^

.

() = ^{(2) }= ^{2 ln||}=

^ln||2

= ||²= ².

Multiplying by () gives ²⁰+ 2 = ^ ⇒ (²)⁰= ^ ⇒

² =

^ = ( − 1)^+  [by parts] ⇒

 = [( − 1)^+ ]². The graphs for  = −5, −3, −1, 1, 3, 5, and 7 are shown.  = 1 is a transitional value. For   1, there is an inflection point and for   1, there is a local minimum. As || gets larger, the “branches” get further from the origin.

22. ⁰= ²+ 2 ⇔ ⁰− 2 = ² ⇔ ⁰−2

 = .

() = ^{−2 }= ^{−2 ln||}= (^ln||)⁻²= ||⁻²= 1

². Multiplying by

()gives 1

²⁰− 2

³ = 1

 ⇒

 1

²

0

= 1

 ⇒ 1

² =

 1

 ⇒ 1

² = ln || +  ⇒  = (ln || + )². For all values of , as || → 0,

 → 0, and as || → ∞,  → ∞. As || increases from 0, the function decreases and attains an absolute minimum.

The inflection points, absolute minimums, and -intercepts all move farther from the origin as  decreases.

23. Setting  = ^1−, 

 = (1 − ) ^−

or 

 = ^ 1 − 



 = ^{(1−)} 1 − 



. Then the Bernoulli differential equation becomes^{(1−)}

1 − 



+  ()^{1(1−)}= ()^{(1−)}or

+ (1 − ) () = ()(1 − ).

24. Here ⁰+  = −² ⇒ ⁰+

= −², so  = 2,  () = 1

and () = −1. Setting  = ⁻¹,  satisfies

⁰−1

 = 1. Then () = ^{(−1)}= 1

 (for   0) and  = 

 1

 + 



= (ln || + ). Thus,

 = 1

( + ln ||).

25. Here ⁰+2

 = ³

², so  = 3,  () = 2

and () = 1

². Setting  = ⁻²,  satisfies ⁰−4

 = −2

². Then () = ^{(−4) }= ⁻⁴and  = ⁴



−2

⁶ + 



= ⁴

 2 5⁵ + 



= ⁴+ 2 5. Thus,  = ±



⁴+ 2 5

−12

.

26. ⁰⁰+ 2⁰ = 12²and  = ⁰ ⇒ ⁰+ 2 = 12² ⇒ ⁰+2

 = 12.

() = ^{(2) }= ^{2 ln||}=

^ln||2

= ||²= ². Multiplying the last differential equation by ²gives

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35. Let P (t) 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 9.1.27 we proposed the differential equation

dP

dt = k[M − P (t)]

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.

Solution: SECTION 9.5 LINEAR EQUATIONS ¤ 843

31. 

 +  = , so () = ^{ }= ^. Multiplying the differential equation by () gives ^

 +  ^=  ^ ⇒ (^ )⁰=  ^ ⇒

 () = ^−

 ^ + 

=  + ^−,   0. Furthermore, it is reasonable to assume that 0 ≤  (0) ≤ , so − ≤  ≤ 0.

32. Since  (0) = 0, we have  () = (1 − ^−). If 1()is Jim’s learning curve, then 1(1) = 25and 1(2) = 45. Hence,

25 = 1(1 − ^−)and 45 = 1(1 − ^−2), so 1 − 25¹ = ^−or  = − ln

 1 − 25

1



= ln

 1

1− 25

 . But

45 = 1(1 − ^−2)so 45 = 1

 1 −

1− 25

1

2

or 45 = 501− 625

1

. Thus, 1= 125is the maximum number of

units per hour Jim is capable of processing. Similarly, if 2()is Mark’s learning curve, then 2(1) = 35and 2(2) = 50.

So  = ln

 2

2− 35



and 50 = 2

 1 −

2− 35

2

2

or 2= 6125. Hence the maximum number of units per hour

for Mark is approximately 61. Another approach would be to use the midpoints of the intervals so that 1(05) = 25and

1(15) = 45. Doing so gives us 1≈ 526 and ²≈ 518.

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

 04kg

L



5 L min



= 2 kg

minSince solution is drained from the tank at a rate of 3L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 + 2) L of liquid after  min. Thus, the salt concentration at time  is ()

100 + 2

kg

L. Salt therefore leaves the tank at a rate of

 () 100 + 2

kg L



3 L min



= 3

100 + 2

kg

min. Combining the rates at which salt enters and leaves the tank, we get



 = 2 − 3

100 + 2. Rewriting this equation as

 +

 3

100 + 2



 = 2, we see that it is linear.

() = exp

  3 

100 + 2



= exp₃

2ln(100 + 2)

= (100 + 2)^32

Multiplying the differential equation by () gives (100 + 2)^32

 + 3(100 + 2)^12 = 2(100 + 2)^32 ⇒ [(100 + 2)^32]⁰ = 2(100 + 2)^32 ⇒ (100 + 2)^32 = ²₅(100 + 2)^52+  ⇒

 = ²₅(100 + 2) + (100 + 2)^−32. Now 0 = (0) = ²₅(100) +  · 100^−32= 40 +₁₀₀₀¹  ⇒  = −40,000, so

 =

2

5(100 + 2) − 40,000(100 + 2)^−32

kg. From this solution (no pun intended), we calculate the salt concentration

at time  to be () = () 100 + 2=

 −40,000 (100 + 2)^52+2

5

 kg

L. In particular, (20) = −40,000 140^52 +2

5 ≈ 02275kg L and (20) = ²₅(140) − 40,000(140)^−32≈ 3185 kg.

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41. (a) Show that the substitution z = 1/P transforms the logistic differential equation P⁰ = kP (1 − P/M ) into the linear differential equation

z⁰+ kz = k M

(b) Solve the linear differential equation in part (a) and thus obtain an expression for P (t). Compare with Equation 9.4.7.

Solution:

844 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS

34.Let () denote the amount of chlorine in the tank at time  (in seconds). (0) = (005 gL) (400 L) = 20 g. The amount of liquid in the tank at time  is (400 − 6) L since 4 L of water enters the tank each second and 10 L of liquid leaves the tank each second. Thus, the concentration of chlorine at time  is ()

400 − 6

g

L. Chlorine doesn’t enter the tank, but it leaves at a rate of

 () 400 − 6

g L



10L s



= 10 () 400 − 6

g

s = 5 () 200 − 3

g

s. Therefore,

 = − 5

200 − 3 ⇒

 

 =

 −5 

200 − 3 ⇒ ln  = ⁵₃ln(200 − 3) +  ⇒  = exp₅

3ln(200 − 3) + 

= ^(200 − 3)^53. Now 20 = (0) = ^· 200^53 ⇒

^= 20

200^53, so () = 20(200 − 3)^53

200^53 = 20(1 − 0015)^53g for 0 ≤  ≤ 66²3s, at which time the tank is empty.

35. (a) 

 + 

 = and () = ^{() } = ^{()}, and multiplying the differential equation by

()gives ^{()}

 +^{()}

 = ^{()} ⇒ 

^{()}0

= ^{()}. Hence,

() = ^{−()}

^{()} + 

=  + ^{−()}. But the object is dropped from rest, so (0) = 0 and

 = −. Thus, the velocity at time  is () = ()[1 − ^{−()}].

(b) lim

→∞() = 

(c) () =

()  = ()[ + ()^{−()}] + 1where 1= (0) − ²².

(0)is the initial position, so (0) = 0 and () = ()[ + ()^{−()}] − ²². 36. = ()(1 − ^{−}) ⇒



= 





0 − ^{−}· 

²

 +

(1 − ^{−}) · 1 = −

^{−}+

−

^{−}

= 





1 − ^{−}− 

^{−}



⇒







 = 1 −

 1 +





^{−}= 1 −1 + 

^ = 1 −1 + 

^ , where  = 

 ≥ 0. Since ^ 1 + for all   0, it follows that   0 for   0. In other words, for all   0,  increases as  increases.

37. (a)  = 1

 ⇒  =1

 ⇒ ⁰= −⁰

². Substituting into ⁰=  (1 − ) gives us −⁰

² = 1



 1 − 1





⇒

⁰= −

 1 − 1





⇒ ⁰= − + 

 ⇒ ⁰+  = 

 ().

(b) The integrating factor is ^{ }= ^. Multiplying () by ^gives ^⁰+ ^ = ^

 ⇒ (^)⁰= 

 ^ ⇒

^ =

 

^ ⇒ ^ = 1

 ^+  ⇒  = 1

 + ^−. Since  =1

, we have

 = 1

1

 + ^−

⇒  = 

1 +  ^−, which agrees with Equation 9.4.7,  = 

1 + ^−, when  = .

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