Section 9.3 Separable Equations

Section 9.3 Separable Equations

42. In an elementary chemical reaction, single molecules of two reactants A and B form a molecule of the product C: A + B → C. The law of mass action states that the rate of reaction is proportional to the product of the concentrations of A and B:

d[C]

dt = k[A][B]

(See Example 3.7.4.) Thus, if the initial concentrations are [A]= a moles/L and [B]= b moles/L and we write x =[C], then we have

dx

dt = k(a − x)(b − x)

(a) Assuming that a 6= b, find x as a function of t. Use the fact that the initial concentration of C is 0.

(b) Find x(t) assuming that a = b. How does this expression for x(t) simplify if it is known that [C]= ¹₂a after 20 seconds?

Solution:

SECTION 9.3 SEPARABLE EQUATIONS ¤ 821 38.From Exercise 9.2.28, 

 = −50¹( − 20) ⇔

 

 − 20= 

−50¹

 ⇔ ln| − 20| = −50¹ +  ⇔

 − 20 = ^−50 ⇔ () = ^−50+ 20. (0) = 95 ⇔ 95 =  + 20 ⇔  = 75 ⇔

() = 75^−50+ 20.

39. 

 = ( −  ) ⇔

 

 −  =



(−)  ⇔ ln| − | = − +  ⇔ | − | = ^{−+} ⇔

 −  = ^− [ = ±^] ⇔  =  + ^−. If we assume that performance is at level 0 when  = 0, then

 (0) = 0 ⇔ 0 =  +  ⇔  = − ⇔  () =  − ^−. lim

→∞ () =  −  · 0 = .

40. (a) 

 = ( − )( − ),  6= . Using partial fractions, 1

( − )( − ) =1( − )

 −  −1( − )

 −  , so

 

( − )( − ) =



  ⇒ 1

 − (− ln | − | + ln | − |) =  +  ⇒ ln



 − 

 − 



 = ( − )( + ).

The concentrations [A] =  −  and [B] =  −  cannot be negative, so  − 

 − ≥ 0 and



 − 

 − 



 =  − 

 − . We now have ln

 − 

 − 



= ( − )( + ). Since (0) = 0, we get ln







= ( − ). Hence,

ln

 − 

 − 



= ( − ) + ln







⇒  − 

 − = 

^{(−)} ⇒  = [^{(−)}− 1]

^{(−)} − 1= [^{(−)}− 1]

^{(−)}−  moles

L .

(b) If  = , then

 = ( − )², so

 

( − )² =



 and 1

 − =  + . Since (0) = 0, we get  = 1

. Thus,  −  = 1

 + 1and  =  − 

 + 1 = ²

 + 1 moles

L . Suppose  = [C] = 2 when  = 20. Then

(20) = 2 ⇒ 

2 = 20²

20 + 1 ⇒ 40² = 20² +  ⇒ 20² =  ⇒  = 1 20, so

 = ²(20)

1 + (20) = 20

1 + 20 = 

 + 20 moles

L .

41. (a) If  = , then

 = ( − )( − )^12becomes 

 = ( − )^32 ⇒ ( − )^−32 =   ⇒

( − )^−32 =

  ⇒ 2( − )^−12=  +  [by substitution] ⇒ 2

 +  =√

 −  ⇒

 2

 + 

2

=  −  ⇒ () =  − 4

( + )². The initial concentration of HBr is 0, so (0) = 0 ⇒ 0 =  − 4

² ⇒ 4

² =  ⇒ ²= 4

 ⇒  = 2√

 [ is positive since  +  = 2( − )^−12 0].

Thus, () =  − 4 ( + 2√ )².

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44. A sphere with radius 1 m has temperature 15^◦C. It lies inside a concentric sphere with radius 2 m and temperature 25^◦C. The temperature T (r) at a distance r from the common center of the spheres satisfies the differential equation

d²T dr² +2

r dT dr = 0

If we let S = dT /dr, then S satisfies a first-order differential equation. Solve it to find an expression for the temperature T (r) between the spheres.

Solution:

1

822 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS (b)

 = ( − )( − )^12 ⇒ 

( − )√

 − =   ⇒

 

( − )√

 − =



  ().

From the hint,  =√

 −  ⇒ ²=  −  ⇒ 2  = −, so

 

( − )√

 −  =

 −2 

[ − ( − ²)]= −2

 

 −  + ² = −2

 

√ − 2

+ ²

= −217

 1

√ − tan⁻¹ 

√ − 



So () becomes −2

√ − tan⁻¹

√ − 

√ −  =  + . Now (0) = 0 ⇒  = −2

√ − tan⁻¹

√

√ −  and we have

√−2

 − tan⁻¹

√ − 

√ −  =  − 2

√ − tan⁻¹

√

√ −  ⇒ 2

√ − 

 tan⁻¹

 

 − − tan⁻¹

 − 

 − 



=  ⇒

() = 2

√

 − 

 tan⁻¹

 

 − − tan⁻¹

 − 

 − 

 .

42. If  =

, then

 = ²

². The differential equation²

² +2





 = 0can be written as

 +2

 = 0. Thus,



 = −2

 ⇒ 

 = −2

 ⇒

 1

 =



−2

 ⇒ ln|| = −2 ln|| + . Assuming  =   0 and   0, we have  = ^{−2 ln +}= ^{ln −2}^ = ⁻² [ = ^] ⇒  = 1

²  ⇒ 

 = 1

² ⇒

 = 1

²  ⇒



 =

 1

²   ⇒  () = −

 + .

 (1) = 15 ⇒ 15 = − +  (1) and  (2) = 25 ⇒ 25 = −¹2 + (2).

Now solve for  and : −2(2) + (1) ⇒ −35 = −, so  = 35 and  = 20, and  () = −20 + 35.

43. (a) 

 =  −  ⇒ 

 = −( − ) ⇒

 

 −  =



− ⇒ (1) ln| − | = − + ¹ ⇒

ln| − | = − + ² ⇒ | − | = ^{−+2} ⇒  −  = ³^− ⇒  = ³^−+  ⇒

() = 4^−+ . (0) = 0 ⇒ ⁰= 4+  ⇒ ⁴ = 0−  ⇒

() = (0− )^−+ .

(b) If 0 , then 0−   0 and the formula for () shows that () increases and lim

→∞() = .

As  increases, the formula for () shows how the role of 0steadily diminishes as that of  increases.

44. (a) Use 1 billion dollars as the -unit and 1 day as the -unit. Initially, there is $10 billion of old currency in circulation, so all of the $50 million returned to the banks is old. At time , the amount of new currency is () billion dollars, so 10 − () billion dollars of currency is old. The fraction of circulating money that is old is [10 − ()]10, and the amount of old currency being returned to the banks each day is10 − ()

10 005billion dollars. This amount of new currency per day is introduced into circulation, so

 =10 − 

10 · 005 = 0005(10 − ) billion dollars per day.

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48. The air in a room with volume 180 m³contains 0.15% carbon dioxide initially. Fresher air with only 0.05% carbon dioxide flows into the room at a rate of 2 m³/min and the mixed air flows out at the same rate. Find the percentage of carbon dioxide in the room as a function of time. What happens in the long run?

Solution:

SECTION 9.3 SEPARABLE EQUATIONS ¤ 823 (b) 

10 − = 0005  ⇒ −

10 −  = −0005  ⇒ ln(10 − ) = −0005 +  ⇒ 10 −  = ^{−0005}, where  = ^ ⇒ () = 10 − ^{−0005}. From (0) = 0, we get  = 10, so () = 10(1 − ^{−0005}).

(c) The new bills make up 90% of the circulating currency when () = 09 · 10 = 9 billion dollars.

9 = 10(1 − ^{−0005}) ⇒ 09 = 1 − ^{−0005} ⇒ ^{−0005} = 01 ⇒ −0005 = − ln 10 ⇒

 = 200 ln 10 ≈ 460517 days ≈ 126 years.

45. (a) Let () be the amount of salt (in kg) after  minutes. Then (0) = 15. The amount of liquid in the tank is 1000 L at all times, so the concentration at time  (in minutes) is ()1000 kgL and

 = −

() 1000

kg L



10 L min



= −() 100

kg min.

 

 = − 1 100



 ⇒ ln  = − 

100+ , and (0) = 15 ⇒ ln 15 = , so ln  = ln 15 −  100. It follows that ln 

15



= −  100and 

15= ^−100, so  = 15^−100kg.

(b) After 20 minutes,  = 15^−20100= 15^−02≈ 123 kg.

46.Let () be the amount of carbon dioxide in the room after  minutes. Then (0) = 00015(180) = 027 m³. The amount of air in the room is 180 m³at all times, so the percentage at time  (in mimutes) is ()180 × 100, and the change in the amount of carbon dioxide with respect to time is



 = (00005)

 2 m³

min



−() 180

 2 m³

min



= 0001 − 

90= 9 − 100

9000 m³ min

Hence,

 

9 − 100 =

 

9000and − 1

100ln |9 − 100| = 1

9000 + . Because (0) = 027, we have

−100¹ ln 18 = , so −100¹ ln |9 − 100| = 9000¹  − 100¹ ln 18 ⇒ ln|9 − 100| = −90¹ + ln 18 ⇒

ln |9 − 100| = ln ^−90+ ln 18 ⇒ ln |9 − 100| = ln(18^−90), and |9 − 100| = 18^−90. Since  is continuous,

(0) = 027, and the right-hand side is never zero, we deduce that 9 − 100 is always negative. Thus, |9 − 100| = 100 − 9 and we have 100 − 9 = 18^−90 ⇒ 100 = 9 + 18^−90 ⇒  = 009 + 018^−90. The percentage of carbon dioxide in the room is

() = 

180× 100 = 009 + 018^−90

180 × 100 = (00005 + 0001^−90) × 100 = 005 + 01^−90 In the long run, we have lim

→∞() = 005 + 01(0) = 005;that is, the amount of carbon dioxide approaches 005% as time goes on.

47.Let () be the amount of alcohol in the vat after  minutes. Then (0) = 004(2000) = 80 L. The amount of beer in the vat is 2000L at all times, so the percentage at time  (in minutes) is ()2000 × 100, and the change in the amount of alcohol with respect to time  is



 =rate in − rate out = 006

 20 L

min



− () 2000

 20 L

min



= 12 − 

100=120 −  100

L min Hence, 

120 −  =

 

100and − ln |120 − | = 100¹  + . Because (0) = 80, we have − ln 40 = , so

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54. A model for tumor growth is given by the Gompertz equation dV

dt = a(ln b − ln V )V

where a and b are positive constants and V is the volume of the tumor measured in mm³. (a) Find a family of solutions for tumor volume as a function of time.

(b) Find the solution that has an initial tumor volume of V (0) = 1 mm³. Solution:

2

SECTION 9.3 SEPARABLE EQUATIONS ¤ 825 (b) 

 = −² ⇒ 

² = −

 ⇒ −1

 = −

 +  ⇒ 1

 =

 − . Since (0) = ⁰,

 = −1

0

and 1

 = 

+ 1

0

. Therefore, () = 1

 + 10

= 0

0 + . 

 = 0

0 +  ⇒

() = 



 0

0 +  = 

 ln|⁰ + | + ⁰. Since (0) = 0, we get 0 =

 ln  + ⁰ ⇒

⁰= 0−

 ln  ⇒ () = ⁰+

(ln|⁰ + | − ln ) = ⁰+

 ln





0 + 





.

We can rewrite the formulas for () and () as () = 0

1 + (0)and () = 0+

 ln



1 +0

 



.

Remarks: This model of horizontal motion through a resistive medium was designed to handle the case in which 0 0.

Then the term −²representing the resisting force causes the object to decelerate. The absolute value in the expression for () is unnecessary (since , 0, and  are all positive), and lim

→∞() = ∞. In other words, the object travels infinitely far. However, lim

→∞() = 0. When 0 0, the term −²increases the magnitude of the object’s negative velocity. According to the formula for (), the position of the object approaches −∞ as  approaches (−⁰):

→−(lim 0)() = −∞. Again the object travels infinitely far, but this time the feat is accomplished in a finite amount of time. Notice also that lim

→−(0)() = −∞ when ⁰ 0, showing that the speed of the object increases without limit.

51. (a) 1

1

1

 =  1

2

2

 ⇒ 

(ln 1) = 

( ln 2) ⇒

 

(ln 1)  =

 

(ln ^2)  ⇒ ln 1= ln ^2+  ⇒ 1= ^{ln 2+ } = ^{ln 2}^ ⇒ 1= ^2, where  = ^.

(b) From part (a) with 1= , 2= , and  = 00794, we have  = ^00794.

52. (a) 

 = (ln  − ln  ) ⇒ 

 = − (ln  − ln ) ⇒ 

 ln( ) = −  ⇒

 

 ln( ) =



−  ⇒

 1

 =



− 

 = ln( ),

 = (1 ) 



⇒ ln || = − +  ⇒

|| = ^−^ ⇒  = ^− [where  = ±^] ⇒ ln() = ^− ⇒ 

 = ^{−} ⇒

 = ^{−}with  6= 0.

(b)  (0) = 1 ⇒ 1 = ^{−(0)} ⇒ 1 = ^ ⇒  = ^−, so  = ^−^{−}= ^{−−}= ^{(−−1)}.

53. (a) The rate of growth of the area is jointly proportional to

()and  − (); that is, the rate is proportional to the product of those two quantities. So for some constant ,  = √

 ( − ). We are interested in the maximum of the function  (when the tissue grows the fastest), so we differentiate, using the Chain Rule and then substituting for

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