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) + y2 = 1, we
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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) + y2 = 1, we see that it is nonlinear in
y because of y2. However, writing it in the form (y2 − 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
2 e−x/2. Then 2y" + y = −e−x/2 + e−x/2 = 0.
12. From y = 6
5 − 6
5 e−20t we obtain dy/dt = 24e−20t
, so that
dy
dt + 20y = 24e−20t + 20 "6
5 − 6
5
e−20t
#
= 24.
13. From y = e3x cos 2x we obtain y" = 3e3x cos 2x − 2e3x sin 2x and y"" = 5e3x cos 2x − 12e3x sin 2x, so that
y"" − 6y" + 13y = 0.
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
2
2X − 1
dX
dt − 1
X − 1
dX
dt = 1
" 2
2X − 1 − 1
X − 1
# dX
dt = 1
2X − 2 − 2X + 1
(2X − 1)(X − 1)
dX
dt = 1
dX
dt = −(2X − 1)(X − 1) = (X − 1)(1 − 2X).
Exponentiating both sides of the implicit solution we obtain
2X − 1
X − 1 = et =⇒ 2X − 1 = Xet − et =⇒ (et − 1) = (et − 2)X =⇒ X = et − 1
et − 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
-4 -2 2 4 x
-4
-2
2
4
y
Exercises 1.1
16. Implicitly differentiating the solution we obtain
−2x2 dy
dx − 4xy + 2y
dy
dx =0 =⇒ −x2 dy − 2xy dx + y dy = 0
=⇒ 2xy dx + (x2 − y)dy = 0.
Using the quadratic formula to solve y2 − 2x2y − 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
(−∞,∞). The graph of y1(x) is solid and the graph of y2 is dashed.
17. Differentiating P = c1et
/ (1 + c1et
) we obtain
dP
dt = (1 + c1et
) c1et − c1et · c1et
(1 + c1et)
2
= c1et
1 + c1et
[(1 + c1et
) − c1et
]
1 + c1et = P(1 − P).
18. Differentiating y = e−x2
& x
0
et2
dt + c1e−x2
we obtain
y" = e−x2
ex2
− 2xe−x2
& x
0
et2
dt − 2c1xe−x2
= 1 − 2xe−x2
& x
0
et2
dt − 2c1xe−x2
.
Substituting into the differential equation, we have
y" + 2xy = 1 − 2xe−x2
& x
0
et2
dt − 2c1xe−x2
+ 2xe−x2
& x
0
et2
dt + 2c1xe−x2
= 1.
19. From y = c1e2x + c2xe2x we obtain dy
dx = (2c1 + c2)e2x + 2c2xe2x and d2y
dx2 = (4c1 + 4c2)e2x + 4c2xe2x, so that
d2y
dx2 − 4
dy
dx + 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1)e2x + (4c2 − 8c2 + 4c2)xe2x = 0.
20. From y = c1x−1 + c2x + c3x ln x + 4x2 we obtain
dy
dx = −c1x−2 + c2 + c3 + c3 ln x + 8x,
d2y
dx2 = 2c1x−3 + c3x−1 + 8,
and
d3y
dx3 = −6c1x−4 − c3x−2,
so that
x3 d3y
dx3 + 2x2 d2y
dx2 − x
dy
dx + y
= (−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, x< 0
x2, x ≥ 0
we obtain y" =
' −2x, x < 0
2x, x ≥ 0
so that xy" − 2y = 0.
2
Exercises 1.1
22. The function y(x) is not continuous at x = 0 since lim x→0−
y(x) = 5 and limx→0+
y(x) = −5. Thus, y"
(x) does not
exist at x = 0.
23. From x = e−2t + 3e6t and y = −e−2t + 5e6t we obtain
dx
dt = −2e−2t + 18e6t and dy
dt = 2e−2t + 30e6t
.
Then
x + 3y = (e−2t + 3e6t
) + 3(−e−2t + 5e6t
)
= −2e−2t + 18e6t = dx
dt
and
5x + 3y = 5(e−2t + 3e6t
) + 3(−e−2t + 5e6t
)
= 2e−2t + 30e6t = dy
dt .
24. From x = cos 2t + sin 2t + 1
5 et and y = − cos 2t − sin 2t − 1
5 et we obtain
dx
dt = −2 sin 2t + 2 cos 2t +
1
5
et and dy
dt = 2 sin 2t − 2 cos 2t − 1
5
et
and
d2x
dt2 = −4 cos 2t − 4 sin 2t +
1
5
et and d2y
dt2 = 4 cos 2t + 4 sin 2t − 1
5
et
.
Then
4y + et = 4(− cos 2t − sin 2t − 1
5
et
) + et
= −4 cos 2t − 4 sin 2t +
1
5
et = d2x
dt2
and
4x − et = 4(cos 2t + sin 2t +
1
5
et
) − et
= 4 cos 2t + 4 sin 2t − 1
5
et = d2y
dt2 .
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 + y2 = 0 is y = 0, since if y &= 0, y2 > 0 and (y"
)2 + y2 ≥ y2 > 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 cost 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
c1y1(0) + c2y2(0) = 2
c1y"
1(0) + c2y"
2(0) = 0
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