5.1. Characteristic equation has one double root 1. Hence the general
solution of the corresponding homogeneous equation is C1 et + C2 tet . We look
for a particular solution of the non-homogeneous equation in the form y(t) =
t2 (At + B)et + C. We have y 0 = (3At2 + 2Bt)et + (At3 + Bt2 )et = (At3 + (3A +
B)t2 +2Bt)et and y 00 = (3At2 +2(3A+B)t+2B)et +(At3 +(3A+B)t2 +2Bt)et =
(At3 + (6A + B)t2 + (6A + 4B)t + 2B)et . Substituting this into the equation we
get
(At3 +(6A+B)t2 +(6A+4B)t+2B)et −2(At3 +(3A+B)t2 +2Bt)et +(At3 +Bt2 )et +C =
((6A + B − 6A − 2B + B)t2 + (6A + 4B − 4B)t + 2B)et + C = 6Atet + 2Bet + C =
tet + 4,
which implies that A = 1/6, B = 0, C = 4, so that the general solution is
1
y(t) = (C1 + C2 t)et + t3 et + 4.
6
Its derivative is y 0 (t) = C2 et + (C1 + C2 t)et + 12 t2 et + 16 t3 et . Substituting the
initial condition, we get
C1 + 4 = 1,
C2 + C1 = 1.
We get C1 = −3 and C2 = 4. Consequently, the solution of the initial value
problem is
1 3
y(t) =
t + 4t − 3 et + 4.
6
5.2. The unique repeated root of the characteristic polynomial is λ = −2.
It follows that a fundamental system of solutions of the homogeneous equation
is y1 (t) = e−2t and y2 (t) = te−2t . We are looking for a solution of the nonhomogeneous equation in the form y = u1 (t)y1 (t) + u2 (t)y2 (t) from the system
0
u1 y1 + u02 y2 = 0
u01 y10 + u02 y20 = t−2 e−2t
It our case it is
−2u01 e−2t
+
u01 e−2t +
0
u2 (−2te−2t
u02 te−2t = 0
+ e−2t ) = t−2 e−2t
It is equivalent to
u01 + u02 t
−u01 + u02 (−t + 1/2)
= 0
= t−2 /2
Adding the two equations, we get
u02 /2 = t−2 /2, or u02 = t−2 .
1
Then from the first equation we get u01 = −u02 t = −t−1 . Integrating u0i , we get
u2 (t) = −t−1 + C2 .
u1 (t) = − ln t + C1 ,
It follows that the general solution of the equation is
y(t) = − ln t · e−2t − e−2t + C1 e−2t + C2 te−2t .
κ
κ
5.3. Equation of the motion of a spring is u00 + m
u = 0, where m
can be
g
κ
found from equality m = L , where g is acceleration of free fall, and L is the
length by which the spring is stretched. We get therefore, equation (we use feet
and seconds)
32
u = 0.
u00 +
0.5
The initial condition is u(0) = 0.25 and u0 (0) = 0. The characteristic polynomial
is λ2 + 64. It follows that the general solution is u(t) = C1 sin 8t + C2 cos 8t.
Then u0 (t) = 8C1 cos 8t − 8C2 sin 8t. The initial condition gives C2 = 0.25 and
C1 = 0. Hence, after t seconds, the mass will be down from the equilibrium
position by 0.25 cos 8t feet, i.e., by 3 cos 8t inches. It follows that period of
motion is 2π/8 = π/4 seconds, amplitude is 6 inches, and frequency is 4/π sec−1
F
, where F is the external force, and m is
5.4. The equation is u00 + Lg u = m
the mass. We get therefore
u00 +
2 cos 3t
32
u=
,
0.125
4
or
1
cos 3t.
2
The initial condition is u(0) = 1/6, u0 (0) = 0.
5.5. We have to find the integral
Z ∞
t sin at e−st dt.
u00 + 162 u =
0
R
Let
at first the indefinite integrals I1 = sin at e−st dt and I2 =
R us find
cos at e−st dt, where a and s are parameters. Using integration by parts,
we get
Z
Z
1
1
1
a
−st
−st a
I1 = −
sin at de
= − sin at e +
cos at e−st dt = − sin at e−st + ·I2 ,
s
s
s
s
s
and
1
I2 = −
s
Z
−st
cos at de
1
a
= − cos at e−st −
s
s
Z
1
a
sin at e−st dt = − cos at e−st − ·I1 .
s
s
Substituting the second equation into the first, we get
1
1
a
1
a
a2
−st a
−st
I1 = − sin at e +
− cos at e
− · I1 = − sin at e−st − 2 cos at e−st − 2 ·I1 ,
s
s
s
s
s
s
s
2
hence
I1 =
1+
a2
s2
−1 1
a
− sin at e−st − 2 cos at e−st
s
s
=
−s sin at − a cos at −st
e
s2 + a2
Substituting this into the second equation, we get
1
a s sin at + a cos at −st
e
=
I2 = − cos at e−st + ·
s
s
s2 + a2
−(s2 + a2 ) cos at + as sin at + a2 cos at −st
a sin at − s cos at −st
e
e
=
2
2
s(s + a )
s2 + a2
Z
Let us compute now the integral
Z
Z
s
a
s sin at + a cos at −st
e
dt = tI1 + 2
I1 + 2
I2
t sin at e−st dt = t dI1 = tI1 +
s2 + a2
s + a2
s + a2
a
s
We have I1 (0) = − s2 +a
As t → ∞, I1 and I2 converge
2 , I2 (0) = − s2 +a2 .
exponentially to 0. It follows that
Z
0
∞
∞
a
s
=
I1 + 2
I2 t sin at e
dt = tI1 + 2
2
2
s +a
s +a
t=0
s
a
a
s
2as
0− − 2
· 2
− 2
· 2
.
= 2
2
2
2
2
s +a s +a
s +a s +a
(s + a2 )2
−st
3