Math 215
Spring 2010
Assignment 4
You are highly recommended to own a textbook, Boyce-DiPrima or another, to see a comprehensive treatment of the subjects and more examples.
§3.6: 8, 11, 13, 15, §3.7: 1, 7, 9, 20
• (§3.6: 8) Find the general solution of y 00 + 4y = 3 csc 2t,
0 < t < π/2.
Answer.
The solutions of the homogeneous equation is yc (t) = c1 cos 2t + c2 sin 2t.
The two functions y1 (t) = cos 2t and y2 (t) = sin 2t form a fundamental set of solutions,
with Wronskian W (y1 , y2 ) = 2. A particular solution is given by Y (t) = u1 (t)y1 (t) +
u2 (t)y2 (t), in which
Z
sin 2t(3 csc 2t)
3
u1 (t) = −
dt = − t,
W (t)
2
Z
cos 2t(3 csc 2t)
3
u2 (t) =
dt = ln | sin 2t|.
W (t)
4
Hence the particular solution is Y (t) = − 32 t cos 2t + 34 (sin 2t) ln | sin 2t|. The general
solution is
3
3
y(t) = c1 cos 2t + c2 sin 2t − t cos 2t + (sin 2t) ln | sin 2t|.
2
4
• (§3.6: 11) Find the general solution of y 00 − 5y 0 + 6y = g(t). Here g(t) is an arbitrary
continuous function.
Answer. Two linearly independent solutions of the homogeneous DE are y1 (t) = e3t
and y2 (t) = e2t , with Wronskian W (y1 , y2 ) = −e5t . By Theorem 3.6.1,
Z 2s
Z 3s
Z
e g(s)
e g(s)
3t
2t
Y (t) = −e
ds + e
ds = [e3(t−s) − e2(t−s) ]g(s)ds.
−e5s
−e5s
The general solution is y(t) = c1 e3t + c2 e2t + Y (t).
• (§3.6: 13) Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous
equation.
t2 y 00 − 2y = 3t2 − 1,
t > 0;
y1 (t) = t2 ,
y2 (t) = t−1 .
Answer.
t2 y100 − 2y1 = 2t2 − 2t2 = 0 and t2 y200 − 2y2 = 2t−1 − 2t−1 = 0. Their
Wronskian W (y1 , y2 ) = −3. The equation can be rewritten as
y 00 − 2t−2 y = g(t) = 3 − t−2 .
Using the method of variation of parameters, a particular solution is given by Y (t) =
u1 (t)y1 (t) + u2 (t)y2 (t), in which
Z −1
t (3 − t−2 )
1
u1 (t) = −
dt = t−2 + ln t,
W (t)
6
Z 2
t (3 − t−2 )
1
u2 (t) =
dt = − t3 + t/3.
W (t)
3
Hence the particular solution is Y (t) =
1
6
+ t2 ln t − t2 /3 + 1/3 = t2 ln t + 1/2 − t2 /3.
(The general solution is y(t) = c1 t2 + c2 t−1 + t2 ln t + 1/2 and hence we can take
Y (t) = t2 ln t + 1/2.)
• (§3.6: 15) Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous
equation.
ty 00 − (1 + t)y 0 + y = t2 e2t ,
t > 0;
y1 (t) = 1 + t,
y2 (t) = et .
Answer. ty100 − (1 + t)y10 + y1 = 0 − (1 + t) + (1 + t) = 0 and ty200 − (1 + t)y20 + y2 =
tet − (1 + t)et + et = 0. Their Wronskian W (y1 , y2 ) = tet . The equation can be
rewritten as
y 00 − (1 + t−1 )y 0 + t−1 y = g(t) = te2t .
Using the method of variation of parameters, a particular solution is given by Y (t) =
u1 (t)y1 (t) + u2 (t)y2 (t), in which
Z t 2t
e (te )
u1 (t) = −
dt = −e2t /2,
tet
Z
(1 + t)(te2t )
u2 (t) =
dt = tet .
tet
Hence the particular solution is Y (t) = −(1 + t)e2t /2 + et tet = 12 e2t (t − 1).
• (§3.7: 1) Determine ω0 , R and δ so as to write the expression u = 3 cos 2t + 4 sin 2t in
the form u = R cos(ω0 t − δ).
√
Answer.
R cos δ = 3 and R sin δ = 4, so R = 32 + 42 = 5 and δ = arctan(4/3) ≈
0.9273. Hence
u = 5 cos(2t − 0.9273).
2
• (§3.7: 7) A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward,
contracting the spring a distance of 1 in., and then set in motion with a downward
velocity of 2 ft/s, and if there is no damping, find the position u of the mass at any
time t. Determine the frequency, period, amplitude, and phase of the motion.
Answer.
We use ft-lb-s for the units and note 1 ft = 12 in. The spring constant
is k = 3/(1/4) = 12 lb/ft. Mass m = 3/32 lb-s2 /ft. Since there is no damping, the
3 00
equation of motion is 32
u + 12u = 0, that is,
u00 + 128u = 0.
The initial conditions are
u(0) = −1/12, u0 (0) = 2.
√
√
The general solution is u(t) = A cos 8 2t+B sin 8 2t. Invoking the initial conditions,
we have
√
√
√
1
2
u(t) = − cos 8 2t +
sin 8 2t.
12
8
p
√
√
√
Thus R = 11/288 ft, δ = π−arctan(3/ 2) rad, ω0 = 8 2 rad/s, and T = π/(4 2)s.
• (§3.7: 9) A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also
attached to a viscous damper with a damping constant of 400 dyn·s/cm. If the mass
is pulled down an additional 2 cm and then released, find its position u at any time t.
Plot u versus t. Determine the quasi frequency and the quasi period. Determine the
ratio of the quasi period to the period of the corresponding undamped motion. Also
find the time τ such that |u(t)| < 0.05 cm for all t > τ .
Answer. We use cm-g-s as the units. The spring constant is k = (20)(980)/5 = 3920
dyne/cm. Mass m = 20 g. The equation of motion is 20u00 + 400u0 + 3920u = 0 or
u00 + 20u0 + 196u = 0.
The initial conditions are
u0 (0) = 0.
√
√
−10t sin 4 6t. Invoking the initial
The general solution is u(t) = Ae−10t
√ cos 4 6t + Be
conditions, we have A = 2, B = 5/ 6,
√
√
5
u(t) = e−10t [2 cos 4 6t + √ sin 4 6t] (cm).
6
√
π
The quasi-frequency is µ = 4 6. The quasi period is Td = 2π/µ = 2√
. The
6
u(0) = 2,
undamped motion has period T = 2π/14 = π/7. The ratio Td /T =
7
√
.
2 6
To find an upper bound for τ , write u in the form
p
√
u(t) = 4 + 25/6 e−10t cos(4 6t − δ).
p
To guarantee u(t) ≤ .05, it suffices that 4 + 25/6 e−10t ≤ .05, which yields τ = .4046.
3
• (§3.7: 20) Assume that the system described by the equation mu00 + γu0 + ku = 0
is critically damped and that the initial conditions are u(0) = u0 and u0 (0) = v0 . If
v0 = 0, show that u → 0 as t → ∞ but that u is never zero. If u0 is positive, determine
a condition on v0 that will ensure that the mass passes through its equilibrium position
after it is released.
Answer. The general solution for the critical damping case is u(t) = (A+Bt)e−γt/2m .
The I.C. u(0) = u0 implies A = u0 , and u0 (0) = v0 implies A(−γ/2m)+B = v0 . Hence
u(t) = [u0 + (v0 + γu0 /2m)t]e−γt/2m .
If v0 = 0, then u = u0 (1 + γt/2m)e−γt/2m , which is never zero since γ and m are
positive. By L’Hospital’s Rule u → 0 as t → ∞.
If u0 > 0 and u(t1 ) = 0 for some t1 > 0, we need (v0 + γu0 /2m) < 0, that is
v0 < −γu0 /2m.
4