**Unformatted text preview: **1 Introduction to Differential Equations
Exercises 1.1 1. Second-order; linear.
2. Third-order; nonlinear because of (dy/dx)4 .
3. The diﬀerential equation is ﬁrst-order. Writing it in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear in
y because of y 2 . However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is linear in x.
4. The diﬀerential equation is ﬁrst-order. Writing it in the form u(dv/du) + (1 + u)v = ueu we see that it is linear
in v. However, writing it in the form (v + uv − ueu )(du/dv) + u = 0, we see that it is nonlinear in u.
5. Fourth-order; linear
6. Second-order; nonlinear because of cos(r + u)
7. Second-order; nonlinear because of 1 + (dy/dx)2 8. Second-order; nonlinear because of 1/R2
9. Third-order; linear
10. Second-order; nonlinear because of x2
˙
11. From y = e−x/2 we obtain y = − 1 e−x/2 . Then 2y + y = −e−x/2 + e−x/2 = 0.
2
12. From y = 6
5 − 6 e−20t we obtain dy/dt = 24e−20t , so that
5
dy
+ 20y = 24e−20t + 20
dt 6 6 −20t
− e
5 5 13. From y = e3x cos 2x we obtain y = 3e3x cos 2x − 2e3x sin 2x and y
y − 6y + 13y = 0. = 24.
= 5e3x cos 2x − 12e3x sin 2x, so that 14. From y = − cos x ln(sec x + tan x) we obtain y = −1 + sin x ln(sec x + tan x) and
y = tan x + cos x ln(sec x + tan x). Then y + y = tan x.
15. Writing ln(2X − 1) − ln(X − 1) = t and diﬀerentiating implicitly we obtain X 2
dX
1 dX
−
=1
2X − 1 dt
X − 1 dt
2
1
−
2X − 1 X − 1 4
2 dX
=1
dt
-4 2X − 2 − 2X + 1 dX
=1
(2X − 1)(X − 1) dt -2 2 4 t -2 dX
= −(2X − 1)(X − 1) = (X − 1)(1 − 2X).
dt -4 Exponentiating both sides of the implicit solution we obtain
2X − 1
et − 1
= et =⇒ 2X − 1 = Xet − et =⇒ (et − 1) = (et − 2)X =⇒ X = t
.
X −1
e −2
Solving et − 2 = 0 we get t = ln 2. Thus, the solution is deﬁned on (−∞, ln 2) or on (ln 2, ∞). The graph of the
solution deﬁned on (−∞, ln 2) is dashed, and the graph of the solution deﬁned on (ln 2, ∞) is solid. 1 Exercises 1.1 16. Implicitly diﬀerentiating the solution we obtain y dy
dy
−2x2
− 4xy + 2y
= 0 =⇒ −x2 dy − 2xy dx + y dy = 0
dx
dx
=⇒ 2xy dx + (x2 − y)dy = 0. 4
2 Using the quadratic formula to solve y 2 − 2x2 y − 1 = 0 for y, we get
√
√
y = 2x2 ± 4x4 + 4 /2 = x2 ± x4 + 1 . Thus, two explicit solutions are
√
√
y1 = x2 + x4 + 1 and y2 = x2 − x4 + 1 . Both solutions are deﬁned on -4 -2 2 4 x -2 (−∞, ∞). The graph of y1 (x) is solid and the graph of y2 is dashed. -4 17. Diﬀerentiating P = c1 et / (1 + c1 et ) we obtain
dP
(1 + c1 et ) c1 et − c1 et · c1 et
=
2
dt
(1 + c1 et )
=
x 18. Diﬀerentiating y = e−x 2 c1 et [(1 + c1 et ) − c1 et ]
= P (1 − P ).
1 + c1 et
1 + c1 et et dt + c1 e−x we obtain
2 2 0
x y = e−x ex − 2xe−x
2 2 2 x et dt − 2c1 xe−x = 1 − 2xe−x
2 2 2 0 et dt − 2c1 xe−x .
2 2 0 Substituting into the diﬀerential equation, we have
x y + 2xy = 1 − 2xe−x 2 2 2 2 0 19. From y = c1 e2x + c2 xe2x we obtain x et dt − 2c1 xe−x + 2xe−x et dt + 2c1 xe−x = 1.
2 2 0 dy
d2 y
= (2c1 + c2 )e2x + 2c2 xe2x and
= (4c1 + 4c2 )e2x + 4c2 xe2x , so that
dx
dx2 d2 y
dy
−4
+ 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1 )e2x + (4c2 − 8c2 + 4c2 )xe2x = 0.
2
dx
dx
20. From y = c1 x−1 + c2 x + c3 x ln x + 4x2 we obtain
dy
= −c1 x−2 + c2 + c3 + c3 ln x + 8x,
dx
d2 y
= 2c1 x−3 + c3 x−1 + 8,
dx2
and
d3 y
= −6c1 x−4 − c3 x−2 ,
dx3
so that
x3 d3 y
d2 y
dy
+ 2x2 2 − x
+y
3
dx
dx
dx
= (−6c1 + 4c1 + c1 + c1 )x−1 + (−c3 + 2c3 − c2 − c3 + c2 )x
+ (−c3 + c3 )x ln x + (16 − 8 + 4)x2
= 12x2 . 21. From y = −x2 ,
2 x , x<0
x≥0 we obtain y = −2x, x<0 2x, x≥0 2 so that xy − 2y = 0. Exercises 1.1
22. The function y(x) is not continuous at x = 0 since lim y(x) = 5 and lim y(x) = −5. Thus, y (x) does not
x→0− x→0+ exist at x = 0.
23. From x = e−2t + 3e6t and y = −e−2t + 5e6t we obtain
dx
= −2e−2t + 18e6t
dt and dy
= 2e−2t + 30e6t .
dt Then
x + 3y = (e2t + 3e6t ) + 3(−e−2t + 5e6t )
dx
= −2e−2t + 18e6t =
dt
and
5x + 3y = 5(e−2t + 3e6t ) + 3(−e−2t + 5e6t )
dy
= 2e−2t + 30e6t =
.
dt
24. From x = cos 2t + sin 2t + 1 et and y = − cos 2t − sin 2t − 1 et we obtain
5
5
dx
1
= −2 sin 2t + 2 cos 2t + et
dt
5
and and dy
1
= 2 sin 2t − 2 cos 2t − et
dt
5 d2 x
1
= −4 cos 2t − 4 sin 2t + et
dt2
5 and d2 y
1
= 4 cos 2t + 4 sin 2t − et .
dt2
5 Then
1
4y + et = 4(− cos 2t − sin 2t − et ) + et
5
1
d2 x
= −4 cos 2t − 4 sin 2t + et = 2
5
dt
and 1
4x − et = 4(cos 2t + sin 2t + et ) − et
5
1
d2 y
= 4 cos 2t + 4 sin 2t − et = 2 .
5
dt 25. An interval on which tan 5t is continuous is −π/2 < 5t < π/2, so 5 tan 5t will be a solution on (−π/10, π/10).
26. For (1 − sin t)−1/2 to be continuous we must have 1 − sin t > 0 or sin t < 1. Thus, (1 − sin t)−1/2 will be a
solution on (π/2, 5π/2).
27. (y )2 + 1 = 0 has no real solution.
28. The only solution of (y )2 + y 2 = 0 is y = 0, since if y = 0, y 2 > 0 and (y )2 + y 2 ≥ y 2 > 0.
29. The ﬁrst derivative of f (t) = et is et . The ﬁrst derivative of f (t) = ekt is kekt . The diﬀerential equations are
y = y and y = ky, respectively.
30. Any function of the form y = cet or y = ce−t is its own second derivative. The corresponding diﬀerential
equation is y − y = 0. Functions of the form y = c sin t or y = c cos t have second derivatives that are the
negatives of themselves. The diﬀerential equation is y + y = 0.
31. Since the nth derivative of φ(x) must exist if φ(x) is a solution of the nth order diﬀerential equation, all lowerorder derivatives of φ(x) must exist and be continuous. [Recall that a diﬀerentiable function is continuous.]
32. Solving the system
c1 y1 (0) + c2 y2 (0) = 2
c1 y1 (0) + c2 y2 (0) = 0 3 Exercises 1.1
for c1 and c2 we get
c1 = 2y2 (0)
y1 (0)y2 (0) − y1 (0)y2 (0) and c2 = − 2y1 (0)
.
y1 (0)y2 (0) − y1 (0)y2 (0) Thus, a particular solution is
y= 2y2 (0)
2y1 (0)
y1 −
y2 ,
y1 (0)y2 (0) − y1 (0)y2 (0)
y1 (0)y2 (0) − y1 (0)y2 (0) where we assume that y1 (0)y2 (0) − y1 (0)y2 (0) = 0.
33. For the ﬁrst-order diﬀerential equation integrate f (x). For the second-order diﬀerential equation integrate twice.
In the latter case we get y = ( f (t)dt)dt + c1 t + c2 .
34. Solving for y using the quadratic formula we obtain the two diﬀerential equations
y = 1
2 + 2 1 + 3t6
t and y = 1
2−2
t 1 + 3t6 , so the diﬀerential equation cannot be put in the form dy/dt = f (t, y).
35. The diﬀerential equation yy − ty = 0 has normal form dy/dt = t. These are not equivalent because y = 0 is a
solution of the ﬁrst diﬀerential equation but not a solution of the second.
36. Diﬀerentiating we get y = c1 + 3c2 t2 and y = 6c2 t. Then c2 = y /6t and c1 = y − ty /2, so
y= y − ty
2 y
6t t+ 1
t3 = ty − t2 y
3 and the diﬀerential equation is t2 y − 3ty + 3y = 0.
37. (a) From y = emt we obtain y = memt . Then y + 2y = 0 implies
memt + 2emt = (m + 2)emt = 0.
Since emt > 0 for all t, m = −2. Thus y = e−2t is a solution.
(b) From y = emt we obtain y = memt and y = m2 emt . Then y − 5y + 6y = 0 implies
m2 emt − 5memt + 6emt = (m − 2)(m − 3)emt = 0.
Since emt > 0 for all t, m = 2 and m = 3. Thus y = e2t and y = e3t are solutions.
(c) From y = tm we obtain y = mtm−1 and y = m(m − 1)tm−2 . Then ty + 2y = 0 implies
tm(m − 1)tm−2 + 2mtm−1 = [m(m − 1) + 2m]tm−1 = (m2 + m)tm−1
= m(m + 1)tm−1 = 0.
Since tm−1 > 0 for t > 0, m = 0 and m = −1. Thus y = 1 and y = t−1 are solutions.
(d) From y = tm we obtain y = mtm−1 and y = m(m − 1)tm−2 . Then t2 y − 7ty + 15y = 0 implies
t2 m(m − 1)tm−2 − 7tmtm−1 + 15tm = [m(m − 1) − 7m + 15]tm
= (m2 − 8m + 15)tm = (m − 3)(m − 5)tm = 0.
Since tm > 0 for t > 0, m = 3 and m = 5. Thus y = t3 and y = t5 are solutions.
38. When g(t) = 0, y = 0 is a solution of a linear equation.
39. (a) Solving (10 − 5y)/3x = 0 we see that y = 2 is a constant solution.
(b) Solving y 2 + 2y − 3 = (y + 3)(y − 1) = 0 we see that y = −3 and y = 1 are constant solutions.
(c) Since 1/(y − 1) = 0 has no solutions, the diﬀerential equation has no constant solutions. 4 Exercises 1.1
(d) Setting y = 0 we have y = 0 and 6y = 10. Thus y = 5/3 is a constant solution.
40. From y = (1 − y)/(x − 2) we see that a tangent line to the graph of y(x) is possibly vertical at x = 2. Intervals
of existence could be (−∞, 2) and (2, ∞).
41. One solution is given by the upper portion of the graph with domain approximately (0, 2.6). The other solution
is given by the lower portion of the graph, also with domain approximately (0, 2.6).
42. One solution, with domain approximately (−∞, 1.6) is the portion of the graph in the second quadrant together
with the lower part of the graph in the ﬁrst quadrant. A second solution, with domain approximately (0, 1.6)
is the upper part of the graph in the ﬁrst quadrant. The third solution, with domain (0, ∞), is the part of the
graph in the fourth quadrant.
43. Diﬀerentiating (x3 + y 3 )/xy = 3c we obtain
xy(3x2 + 3y 2 y ) − (x3 + y 3 )(xy + y)
=0
x2 y 2
3x3 + 3xy 3 y − x4 y − x3 y − xy 3 y − y 4 = 0
(3xy 3 − x4 − xy 3 )y = −3x3 y + x3 y + y 4
y = y 4 − 2x3 y
y(y 3 − 2x3 )
.
=
3 − x4
2xy
x(2y 3 − x3 ) 44. A tangent line will be vertical where y is undeﬁned, or in this case, where x(2y 3 − x3 ) = 0. This gives x = 0
and 2y 3 = x3 . Substituting y 3 = x3 /2 into x3 + y 3 = 3xy we get
1
1
x3 + x3 = 3x
x
1/3
2
2
3 3
3
x = 1/3 x2
2
2
x3 = 22/3 x2
x2 (x − 22/3 ) = 0.
Thus, there are vertical tangent lines at x = 0 and x = 22/3 , or at (0, 0) and (22/3 , 21/3 ). Since 22/3 ≈ 1.59, the
estimates of the domains in Problem 42 were close.
45. Since φ (x) > 0 for all x in I, φ(x) is an increasing function on I. Hence, it can have no relative extrema on I.
46. (a) When y = 5, y = 0, so y = 5 is a solution of y = 5 − y.
(b) When y > 5, y < 0, and the solution must be decreasing. When y < 5, y > 0, and the solution must be
increasing. Thus, none of the curves in color can be solutions.
y (c) 10 5 −5 5 t 47. (a) y = 0 and y = a/b. 5 Exercises 1.1
(b) Since dy/dx = y(a − by) > 0 for 0 < y < a/b, y = φ(x) is increasing on this interval. Since dy/dx < 0 for
y < 0 or y > a/b, y = φ(x) is decreasing on these intervals.
(c) Using implicit diﬀerentiation we compute
d2 y
= y(−by ) + y (a − by) = y (a − 2by).
dx2
Solving d2 y/dx2 = 0 we obtain y = a/2b. Since d2 y/dx2 > 0 for 0 < y < a/2b and d2 y/dx2 < 0 for
a/2b < y < a/b, the graph of y = φ(x) has a point of inﬂection at y = a/2b.
(d) y y=a b y=0
x 48. (a) In Mathematica use
Clear[y]
y[x ]:= x Exp[5x] Cos[2x]
y[x]
y''''[x] − 20 y'''[x] + 158 y''[x] − 580 y'[x] + 841 y[x] // Simplify
(b) In Mathematica use
Clear[y]
y[x ]:= 20 Cos[5 Log[x]]/x − 3 Sin[5 Log[x]]/x
y[x]
xˆ3 y'''[x] + 2xˆ2 y''[x] + 20 x y'[x] − 78 y[x] // Simplify Exercises 1.2
1
1
1
=
we get c1 = −4. The solution is y =
.
3
1 + c1
1 − 4e−t
1
1
2
2. Solving 2 =
we get c1 = − e−1 . The solution is y =
.
1 + c1 e
2
2 − e−(t+1)
3. Using x = −c1 sin t + c2 cos t we obtain c1 = −1 and c2 = 8. The solution is x = − cos t + 8 sin t. 1. Solving − 4. Using x = −c1 sin t + c2 cos t we obtain c2 = 0 and −c1 = 1. The solution is x = − cos t.
5. Using x = −c1 sin t + c2 cos t we obtain √ 3
1
1
c1 + c2 =
2
2
2
√
3
1
− c1 +
c2 = 0.
2
2
√
√
3
3
1
1
Solving we ﬁnd c1 =
and c2 = . The solution is x =
cos t + sin t.
4
4
4
4 6 Exercises 1.2
6. Using x = −c1 sin t + c2 cos t we obtain √ √
√
2
2
c1 +
c2 = 2
2
2
√
√
√
2
2
−
c1 +
c2 = 2 2 .
2
2 Solving we ﬁnd c1 = −1 and c2 = 3. The solution is x = − cos t + 3 sin t.
7. From the initial conditions we obtain the system
c1 + c2 = 1
c1 − c2 = 2.
Solving we get c1 = 3
2 and c2 = − 1 . A solution of the initial-value problem is y = 3 ex − 1 e−x .
2
2
2 8. From the initial conditions we obtain the system
c1 e + c2 e−1 = 0
c1 e − c2 e−1 = e.
Solving we get c1 = 1
2 and c2 = − 1 e2 . A solution of the initial-value problem is y = 1 ex 1 e2−x .
2
2
2 9. From the initial conditions we obtain c1 e−1 + c2 e = 5
c1 e−1 − c2 e = −5. Solving we get c1 = 0 and c2 = 5e−1 . A solution of the initial-value problem is y = 5e−x−1 .
10. From the initial conditions we obtain
c1 + c2 = 0
c1 − c2 = 0.
Solving we get c1 = c2 = 0. A solution of the initial-value problem is y = 0.
11. Two solutions are y = 0 and y = x3 .
12. Two solutions are y = 0 and y = x2 . (Also, any constant multiple of x2 is a solution.)
∂f
2
= y −1/3 . Thus the diﬀerential equation will have a unique solution in any
∂y
3
rectangular region of the plane where y = 0. 13. For f (x, y) = y 2/3 we have 14. For f (x, y) = √ xy we have ∂f
1
=
∂y
2 x
. Thus the diﬀerential equation will have a unique solution in any
y region where x > 0 and y > 0 or where x < 0 and y < 0.
15. For f (x, y) = y
∂f
1
we have
= . Thus the diﬀerential equation will have a unique solution in any region
x
∂y
x where x = 0.
16. For f (x, y) = x + y we have ∂f
= 1. Thus the diﬀerential equation will have a unique solution in the entire
∂y plane.
17. For f (x, y) = x2
2x2 y
∂f
=
we have
2 . Thus the diﬀerential equation will have a unique solution in
4 − y2
∂y
(4 − y 2 ) any region where y < −2, −2 < y < 2, or y > 2. 7 Exercises 1.2 18. For f (x, y) = x2
∂f
−3x2 y 2
we have
=
2 . Thus the diﬀerential equation will have a unique solution in
3
1+y
∂y
(1 + y 3 ) any region where y = −1.
y2
∂f
2x2 y
we have
=
2 . Thus the diﬀerential equation will have a unique solution in
2
+y
∂y
(x2 + y 2 )
any region not containing (0, 0). 19. For f (x, y) = 20. For f (x, y) = x2 y+x
∂f
−2x
. Thus the diﬀerential equation will have a unique solution in any
we have
=
y−x
∂y
(y − x)2 region where y < x or where y > x.
21. The diﬀerential equation has a unique solution at (1, 4).
22. The diﬀerential equation is not guaranteed to have a unique solution at (5, 3).
23. The diﬀerential equation is not guaranteed to have a unique solution at (2, −3).
24. The diﬀerential equation is not guaranteed to have a unique solution at (−1, 1).
25. (a) A one-parameter family of solutions is y = cx. Since y = c, xy = xc = y and y(0) = c · 0 = 0.
(b) Writing the equation in the form y = y/x we see that R cannot contain any point on the y-axis. Thus,
any rectangular region disjoint from the y-axis and containing (x0 , y0 ) will determine an interval around x0
and a unique solution through (x0 , y0 ). Since x0 = 0 in part (a) we are not guaranteed a unique solution
through (0, 0).
(c) The piecewise-deﬁned function which satisﬁes y(0) = 0 is not a solution since it is not diﬀerentiable at
x = 0.
d
26. (a) Since
tan(x + c) = sec2 (x + c) = 1 + tan2 (x + c), we see that y = tan(x + c) satisﬁes the diﬀerential
dx
equation.
(b) Solving y(0) = tan c = 0 we obtain c = 0 and y = tan x. Since tan x is discontinuous at x = ±π/2, the
solution is not deﬁned on (−2, 2) because it contains ±π/2.
(c) The largest interval on which the solution can exist is (−π/2, π/2).
27. (a) Since d
1
1
1
= y 2 , we see that y = −
−
=
is a solution of the diﬀerential equation.
dt
t+c
(t + c)2
t+c (b) Solving y(0) = −1/c = 1 we obtain c = −1 and y = 1/(1 − t). Solving y(0) = −1/c = −1 we obtain c = 1
and y = −1/(1 + t). Being sure to include t = 0, we see that the interval of existence of y = 1/(1 − t) is
(−∞, 1), while the interval of existence of y = −1/(1 + t) is (−1, ∞).
(c) Solving y(0) = −1/c = y0 we obtain c = −1/y0 and
y=− 1
y0
=
,
−1/y0 + t
1 − y0 t y0 = 0. Since we must have −1/y0 + t = 0, the largest interval of existence (which must contain 0) is either
(−∞, 1/y0 ) when y0 > 0 or (1/y0 , ∞) when y0 < 0.
(d) By inspection we see that y = 0 is a solution on (−∞, ∞).
28. (a) Diﬀerentiating 3x2 − y 2 = c we get 6x − 2yy = 0 or yy = 3x. 8 Exercises 1.2
y (b) Solving 3x2 − y 2 = 3 for y we get
y = φ1 (x) = 3(x2 − 1) , y = φ2 (x) = − 3(x2 − 1) ,
y = φ3 (x) = 3(x2 − 1) , y = φ4 (x) = − 3(x2 − 1) , 4 1 < x < ∞, 2 1 < x < ∞,
−∞ < x < −1, -4 2 −∞ < x < −1. 4 x 2 -2 4 x -2 Only y = φ3 (x) satisﬁes y(−2) = 3. -4 (c) Setting x = 2 and y = −4 in 3x2 − y 2 = c we get 12 − 16 = −4 = c, so the
explicit solution is y
4 y = − 3x2 + 4 , −∞ < x < ∞. 2
-4 -2
-2
-4 (d) Setting c = 0 we have y =
29. When x = 0 and y = 1
2 √ √
3x and y = − 3x, both deﬁned on (−∞, ∞). , y = −1, so the only plausible solution curve is the one with negative slope at (0, 1 ),
2 or the black curve.
30. If the solution is tangent to the x-axis at (x0 , 0), then y = 0 when x = x0 and y = 0. Substituting these values
into y + 2y = 3x − 6 we get 0 + 0 = 3x0 − 6 or x0 = 2.
31. The theorem guarantees a unique (meaning single) solution through any point. Thus, there cannot be two
distinct solutions through any point.
32. Diﬀerentiating y = x4 /16 we obtain y = x3 /4 = x(x4 /16)1/2 , so the ﬁrst
function is a solution of the diﬀerential equation. Since y = 0 satisﬁes the y
5
4
3
2
1 diﬀerential equation and limx→0+ x3 /4 = 0, the second function is also a solution
of the diﬀerential equation. Both functions satisfy the condition y(2) = 1.
Theorem 1.1 simply guarantees the existence of a unique solution on some
interval containing (2, 1). In this, such an interval could be (0, 4), where the two
functions are identical. -4 2 -2 4x 2 4x y
5
4
3
2
1
-4 -2 33. The antiderivative of y = 8e2x + 6x is y = 4e2x + 3x2 + c. Setting x = 0 and y = 9 we get 9 = 4 + c, so c = 5
and y = 4e2x + 3x2 + 5.
34. The antiderivative of y = 12x − 2 is y = 6x2 − 2x + c1 . From the equation of the tangent line we see that
when x = 1, y = 4 and y = −1 (the slope of the tangent line). Solving −1 = 6(1)2 − 2(1) + c1 we get c1 = −5.
The antiderivative of y = 6x2 − 2x − 5 is y = 2x3 − x2 − 5x + c2 . Setting x = 1 and y = 4 we get 4 = −4 + c2 ,
so c2 = 8 and y = 2x3 − x2 − 5x + 8. 9 Exercises 1.3 Exercises 1.3
dP
dP
= kP + r;
= kP − r
dt
dt
2. Let b be the rate of births and d the rate of deaths. Then b = k1 P and d = k2 P . Since dP/dt = b − d, the
diﬀerential equation is dP/dt = k1 P − k2 P . 1. 3. Let b be the rate of births and d the rate of deaths. Then b = k1 P and d = k2 P 2 . Since dP/dt = b − d, the
diﬀerential equation is dP/dt = k1 P − k2 P 2 .
4. Let P (t) be the number of owls present at time t. Then dP/dt = k(P − 200 + 10t).
5. From the graph we estimate T0 = 180◦ and Tm = 75◦ . We observe that when T = 85, dT /dt ≈ −1. From the
diﬀerential equation we then have
dT /dt
−1
k=
= −0.1.
=
T − Tm
85 − 75
6. By inspecting the graph we take Tm to be Tm (t) = 80 − 30 cos πt/12. Then the temperature of the body at time
t is determined by the diﬀerential equation
dT
π
= k T − 80 − 30 cos t
dt
12 , t > 0. 7. The number of students with the ﬂu is x and the number not infected is 1000 − x, so dx/dt = kx(1000 − x).
8. By analogy with diﬀerential equation modeling the spread of a disease we assume that the rate at which the
technological innovation is adopted is proportional to the number of people who have adopted the innovation
and also to the number of people, y(t), who have not yet adopted it. If one person who has adopted the
innovation is introduced into the population then x + y = n + 1 and
dx
= kx(n + 1 − x),
dt x(0) = 1. 9. The rate at which salt is leaving the tank is
(3 gal/min) · A
lb/gal
300 = A
lb/min.
100 Thus dA/dt = A/100.
10. The rate at which salt is entering the tank is
R1 = (3 gal/min) · (2 lb/gal) = 6 lb/min.
Since the solution is pumped out at a slower rate, it is accumulating at the rate of (3 − 2)gal/min = 1 gal/min.
After t minutes there are 300 + t gallons of brine in the tank. The rate at which salt is leaving is
R2 = (2 gal/min) · A
lb/gal
300 + t = 2A
lb/min.
300 + t The diﬀerential equation is
dA
2A
=6−
.
dt
300 + t
11. The volume of water in the tank at time t is V = Aw h. The diﬀerential equation is then
dh
1 dV
1
−cA0
=
=
dt
Aw dt
Aw 10 2gh =− cA0
Aw 2gh . Exercises 1.3 Using A0 = π 2
12 2 = π
, Aw = 102 = 100, and g = 32, this becomes
36
dh
cπ √
cπ/36 √
64h = −
h.
=−
dt
100
450 12. The volume of water in the tank at time t is V = 1 πr2 h = 1 Aw h. Using the formula from Problem 11 for the
3
3
volume of water leaving the tank we see that the diﬀerential equation is
dh
3 dV
3
(−cAh
=
=
dt
Aw dt
Aw 2gh ) = − 3cAh
Aw 2gh . Using Ah = π(2/12)2 = π/36, g = 32, and c = 0.6, this becomes
dh
0.4π 1/2
3(0.6)π/36 √
64h = −
h .
=−
dt
Aw
Aw
To ﬁnd Aw we let r be the radius of the top of the water. Then r/h = 8/20, so r = 2h/5 and Aw = π(2h/5)2 =
4πh2 /25. Thus
dh
0.4π
=−
h1/2 = −2.5h−3/2 .
dt
4πh2 /25
dq
d2 q
di
dq
and L 2 + R
= E(t) we obtain L + Ri = E(t).
dt
dt
dt
dt
dq
1
14. By Kirchoﬀ’s second law we obta...

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