Section 9.1 Modeling with Diﬀerential Equations

(1)

Section 9.1 Modeling with Differential Equations

804 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS

5. (a)  = sin  ⇒ ⁰= cos  ⇒ ⁰⁰= − sin .

LHS = ⁰⁰+  = − sin  + sin  = 0 6= sin , so  = sin  is not a solution of the differential equation.

(b)  = cos  ⇒ ⁰= − sin  ⇒ ⁰⁰= − cos .

LHS = ⁰⁰+  = − cos  + cos  = 0 6= sin , so  = cos  is not a solution of the differential equation.

(c)  = ¹₂ sin  ⇒ ⁰ = ¹₂( cos  + sin ) ⇒ ⁰⁰= ¹₂(− sin  + cos  + cos ).

LHS = ⁰⁰+  = ¹₂(− sin  + 2 cos ) +¹2 sin  = cos  6= sin , so  = ¹2 sin is not a solution of the differential equation.

(d)  = −¹2 cos  ⇒ ⁰ = −¹2(− sin  + cos ) ⇒ ⁰⁰= −¹2(− cos  − sin  − sin ).

LHS = ⁰⁰+  = −¹2(− cos  − 2 sin ) +

−¹2 cos 

= sin  = RHS, so  = −¹2 cos is a solution of the differential equation.

6. (a)  = ln  + 

 ⇒ ⁰= · (1) − (ln  + )

² =1 − ln  − 

² .

LHS = ²⁰+  = ²·1 − ln  − 

² +  ·ln  + 



= 1 − ln  −  + ln  +  = 1 = RHS, so  is a solution of the differential equation.

(b) A few notes about the graph of  = (ln  + ):

(1) There is a vertical asymptote of  = 0.

(2) There is a horizontal asymptote of  = 0.

(3)  = 0 ⇒ ln  +  = 0 ⇒  = ^−, so there is an -intercept at ^−.

(4) ⁰= 0 ⇒ ln  = 1 −  ⇒  = ¹^−, so there is a local maximum at  = ¹^−.

(c) (1) = 2 ⇒ 2 =ln 1 + 

1 ⇒ 2 = , so the solution is  = ln  + 2

 [shown in part (b)].

(d) (2) = 1 ⇒ 1 =ln 2 + 

2 ⇒ 2 + ln 2 +  ⇒  = 2 − ln 2, so the solution is  =ln  + 2 − ln 2

 [shown in part (b)].

7. (a) Since the derivative ⁰= −²is always negative (or 0 if  = 0), the function  must be decreasing (or equal to 0) on any interval on which it is deﬁned.

(b)  = 1

 +  ⇒ ⁰= − 1

( + )². LHS = ⁰= − 1

( + )² = −

 1

 + 

2

= −²= RHS

(c)  = 0 is a solution of ⁰= −²that is not a member of the family in part (b).

(d) If () = 1

 + , then (0) = 1 0 +  = 1

. Since (0) = 05, 1

 = 1

2 ⇒  = 2, so  = 1

 + 2.

13. (a) ⁰= 1 + ²+ ²≥ 1 and ⁰→ ∞ as  → ∞. The only curve satisfying these conditions is labeled III.

(b) ⁰= ^−²^−² 0if   0 and ⁰ 0if   0. The only curve with negative tangent slopes when   0 and positive tangent slopes when   0 is labeled I.

(c) ⁰= 1

1 + ^²^+²  0and ⁰→ 0 as  → ∞. The only curve satisfying these conditions is labeled IV.

(d) ⁰= sin() cos() = 0if  = 0, which is the solution graph labeled II.

14. (a) The coffee cools most quickly as soon as it is removed from the heat source. The rate of cooling decreases toward 0 since the coffee approaches room temperature.

(b) 

 = ( − ), where  is a proportionality constant,  is the temperature of the coffee, and  is the room temperature. The initial condition is (0) = 95^◦C. The answer and the model support each other because as  approaches ,  approaches 0, so the model seems appropriate.

(c)

15. (a)  increases most rapidly at the beginning, since there are usually many simple, easily-learned sub-skills associated with learning a skill. As  increases, we would expect  to remain positive, but decrease. This is because as time progresses, the only points left to learn are the more difﬁcult ones.

(b) 

 = ( −  ) is always positive, so the level of performance  is increasing. As  gets close to ,  gets close to 0; that is, the performance levels off, as explained in part (a).

(c)

16. (a) 

 = (∞− ). Assuming ∞ , we have   0 and

  0for all .

(b)

17. If () = 



1 − ^−^1−

= − ^^−^1−for   0, where   0,  0, 0    1, and  = (1 − ), then



 = 



0 − ^−^1−· 

(−¹^−)



= −^^−^1−· (−)(1 − )^−= (1 − )

^ ^−^1−= 

^(− ). The equation for  indicates that as  increases,  approaches . The differential equation indicates that as  increases, the rate of increase of  decreases steadily and approaches 0 as  approaches .