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Assignment 2 (MATH 215, Q1)
1. Solve the boundary-value problem if possible.
(a) y 00 + 9y = 0, y(0) = 1, y(π/2) = 0
Solution. The characteristic equation r2 + 9 = 0 has two roots r1 = 3i and r2 = −3i.
The general solution of the differential equation is y = C1 cos(3x) + C2 sin(3x). Since
y(0) = 1, we have C1 = 1. Moreover, y(π/2) = 0 gives C2 = 0. Therefore, the desired
solution is
y = cos(3x).
(b) y 00 + 2y 0 + 5y = 0, y(0) = 1, y(π) = 2
Solution. The characteristic equation r2 + 2r + 5 = 0 has two roots r1 = 1 + 2i and
r2 = 1 − 2i. The general solution of the differential equation is
y = C1 ex cos(2x) + C2 ex sin(2x).
Since y(0) = 1, we have C1 = 1. Moreover, y(π) = 0 gives eπ C1 = 2. Therefore, the
boundary-value problem has no solutions.
2. Solve the following nonhomogeneous differential equations.
(a) y 00 + y = sec x, 0 < x < π/2
Solution. The general solution of the equation y 00 + y = 0 is y = C1 cos x + C2 sin x.
Let yp (x) = u1 (x) cos x + u2 (x) sin x, where u1 and u2 satisfy
u01 cos x + u02 sin x = 0
−u01 sin x + u02 cos x = sec x.
Solving for u01 and u02 , we obtain
u01 (x) = −
sin x
cos x
and u02 (x) = 1.
Integrating gives
Z
u1 (x) = −
sin x
dx = ln(cos x)
cos x
Z
and u2 (x) =
dx = x.
Hence, a particular solution is yp (x) = ln(cos x) cos x + x sin x. We conclude that the
general solution of the equation y 00 + y = sec x (0 < x < π/2) is
y = ln(cos x) cos x + x sin x + C1 cos x + C2 sin x.
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(b) 2y 00 + y 0 = cos (2x)
Solution. The general solution of the equation 2y 00 + y 0 = 0 is yc = C1 e−x/2 + C2 . Let
yp (x) = u1 (x)e−x/2 + u2 (x), where u1 and u2 satisfy
u01 e−x/2 + u02 = 0
2 u01 (−1/2)e−x/2 = cos(2x).
Solving for u01 and u02 , we obtain
u01 (x) = −ex/2 cos(2x)
Integrating gives u2 (x) =
Z
u1 (x) =
R
and u02 (x) = cos(2x).
cos(2x) dx = sin(2x)/2 and
−ex/2 cos(2x)dx = −ex/2
2 cos(2x) + 8 sin(2x)
.
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Hence, a particular solution is
yp (x) = u1 (x)e−x/2 + u2 (x) =
−4 cos(2x) + sin(2x)
.
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We conclude that the general solution of the equation 2y 00 + y 0 = cos (2x) is
y=
−4 cos(2x) + sin(2x)
+ C1 e−x/2 + C2 .
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3. Solve the initial-value problem.
√
(a) y 0 + y = x e−x , y(0) = 3
Solution. An integrating factor is ex . Multiplying both sides of the equation by ex
gives
√
√
d x
or
(e y) = x.
ex y 0 + ex y = x
dx
It follows that
Z
√
2
x
e y=
x dx = x3/2 + C.
3
Hence, the general solution is y = 23 x3/2 e−x + Ce−x . Since y(0) = 3, we obtain C = 3.
Therefore, the desired solution is
y=
2 3/2 −x
x e + 3e−x .
3
(b) y 00 + 2y 0 − 3y = ex , y(0) = 1, y 0 (0) = 0
2
Solution. The characteristic equation r2 + 2r − 3 = 0 has two roots r1 = −3 and
r2 = 1. Let yp (x) = u1 (x)e−3x + u2 (x)ex , where u1 and u2 satisfy
u01 e−3x + u02 ex = 0
u01 (−3e−3x ) + u02 ex = ex .
Solving for u01 and u02 , we obtain u01 (x) = −e4x /4 and u02 (x) = 1/4. Hence,
Z
u1 (x) = −
1 4x
1
e dx = − e4x
4
16
and u2 (x) =
x
.
4
The general solution of the differential equation is
y=
1 x
xe + C1 e−3x + C2 ex .
4
The initial-value conditions y(0) = 1 and y 0 (0) = 0 give C1 =
Therefore, the desired solution is
y=
5
16
and C2 =
11
16 .
5 −3x 11 x
1 x
xe +
e
+
e .
4
16
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4. A spring with a 3-kg mass is held stretched 0.6 m beyond its natural length by a force
of 20 N. If the spring begins at its equilibrium position but a push gives it an initial
velocity of 1.2 m/s, find the position of the mass after t seconds.
Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the
differential equation
d2 x
m 2 + kx = 0.
dt
Here, m = 3(kg), k = 20/0.6 = 100/3(N/m). Thus, ω 2 = k/m = 100/9, so ω = 10/3.
The general solution is
x(t) = C1 cos
It follows that
x0 (t) = −
10 10 t + C2 sin
t .
3
3
10 10
10 10
C1 sin
t + C2 cos
t .
3
3
3
3
Now x(0) = 0 gives C1 = 0, and x0 (0) = 1.2 gives C2 = 0.36. Thus, the position of
the mass after t seconds is
10 t .
x(t) = 0.36 sin
3
3
5. In an LR circuit a generator supplies a voltage of E(t) = 40 sin (60t) volts, the inductance is 1 H, the resistance is 20 Ω, and I(0) = 1 A.
(a) Find I(t).
(b) Find the current after 0.1 s.
Solution. Since L = 1, R = 20, E(t) = 40 sin(60t), the current I satisfies the differential equation
dI
+ 20I = 40 sin(60t).
dt
This is a first order linear equation. We observe that e20t is an integration factor.
Thus,
e20t
dI
+ 20Ie20t = 40e20t sin(60t)
dt
or
d 20t e I = 40e20t sin(60t).
dt
It follows that
20t
e
Z
I=
40e20t sin(60t) dt =
e20t
sin(60t) − 3 cos(60t) + C.
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Since I(0) = 1, we have C = 8/5. Thus, the desired solution is
I(t) =
8
1
sin(60t) − 3 cos(60t) + e−20t .
5
5
When t = 0.1,
I(0.1) ≈ −0.415(A).
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