W ARNING! This contains some sample problems. This should not... your only study guide for the test; these problems may...

W ARNING! This contains some sample problems. This should not... your only study guide for the test; these problems may...

SAMPLE PROBLEMS, MATERIAL AFTER EXAM 2
DR. NEAL BUSHAW
W ARNING! This contains some sample problems. This should not be used as
your only study guide for the test; these problems may well leave out important
subjects! Study your notes, homework, and webwork as well!
Problem 1. Find the solution of the given initial value problem.
(
1, 0 ≤ t < 3π
a) y 00 + y =
.
0, 3π < t < ∞
y(0) = 0, y 0 (0) =(1.
1, π ≤ t < 2π
b) y 00 + 2y 0 + 2y =
.
0, 0 ≤ t < π and t ≥ 2π
y(0) = 0, y 0 (0) = 1.
c) y 00 + 3y 0 + 2y = u2 (t).
y(0) = 0, y 0 (0) = 1.
d) y (4) − y = u1 (t) − u2 (t).
y(0) = y 0 (0) = y 00 (0) = y 000 (0) = 0.
Problem 2. Find an expression involving uc (t) for a function f that ramps up from
zero at t = t0 to the value h at t = t0 + k.
Problem 3. Find the solution of the given initial value problem.
a) y 00 + 2y 0 + 2y = δ(t − π); y(0) = 1, y 0 (0) = 0.
b) y 00 + 4y = δ(t − π) − δ(t − 2π); y(0) = 0, y 0 (0) = 0.
c) y 00 + 2y 0 + 3y = sin t + δ(t − 3π); y(0) = 0, y 0 (0) = 0.
Problem 4. Consider the initial value problem
y 00 + 2y 0 + 6y = 0,
a)
b)
c)
d)
y(0) = 2,
y 0 (0) = α ≥ 0.
Find the solution y(t) of this problem.
Find α such that y = 0 when t = 1.
Find, as a function of α, the smallest positive value of t for which y = 0.
Determine the limit of the expression found in part (c) as α → ∞.
1
2
DR. NEAL BUSHAW
Problem 5. Find a real valued solution to the following initial value problems.
a) y 00 − 6y 0 + 13y = 0; y(0) = 1, y 0 (0) = 1.
b) y 00 + 4y 0 + 4y = 0; y(0) = 1, y 0 (0) = −4.
c) y 00 + 3y 0 + 2y = 0; y(0) = 3, y 0 (0) = 0.
Problem 6. For which values of α (if any) are all solutions of y 00 − (2α − 1)y 0 +
α(α − 1)y = 0 unbounded as t → ∞?
Problem 7. The characteristic equation of a homogenous 9th order differential equation with constant coefficients has roots r = 0 with multiplicity 3, r = −2 with
multiplicity two, r = −3 ± 2i with multiplicity two. Write the general solution of
the differential equation. No, we haven’t done a problem like this. You can figure it
out!!
Problem 8.
a) The differential equation t2 y 00 + 3ty 0 + y = 0 has a solution y1 (t) = t−1 for
t > 0. Find the general solution.
b) The differential equation t2 y 00 − t(t + 2)y 0 + (t + 2)y = 0 has a solution y1 (t) = t
for t > 0. Find the general solution.
Problem 9. Find the general solution of the differential equation y 00 + 2y 0 + y = e−t .
Problem 10. Solve the initial value problem.
a) y 00 − y 0 − 2y = 6x + 6e−x ; y(0) = 1, y 0 (0) = 0.
b) y 00 − y 0 − 2y = 6te2t ; y(0) = 0, y 0 (0) = 1.
Problem 11. Determine the form of the particular solution Y (t) if the method of
undetermined coefficients is to be used. You do not need to determine the values of
the coefficients; It is good practice, but I haven’t checked that these will give you ‘nice
numbers’.
a) y 00 + 3y 0 = 2t2 + t2 e−3t + sin 3t.
b) y 00 + y = t(1 + sin t).
c) y 00 − 5y 0 + 6y = et cos 2t + e2t (3t + 4) sin t.
d) y 00 + 2y 0 + 2y = 3e−t + 2e−t cos t + 4e−t t2 sin t.
e) y 00 − 4y 0 + 4y = 2t2 + 4te2t + t sin 2t.
Problem 12. Consider the initial value problem y 00 + 4y = 0 with y(0) = −3,
y 0 (0) = 6. Write the solution in the form y = R cos (ω0 t − δ).
Problem 13. A mass of 2 kg stretches a spring 0.5m. If the mass is set in motion
from its equilibrium with a downward velocity of 10 cm/s, and there is no damping,
write an IVP for the position u in meters of the mass at any time t in seconds. Use
g = 9.8m/s2 for the acceleration due to gravity.
SAMPLE PROBLEMS, MATERIAL AFTER EXAM 2
3
Problem 14. A mass m = 1 is attached to a spring with constant k = 2 and
damping constant γ. Determine the value of γ so that the motion is critically damped.
Problem 15. Determine ω0 , R, δ so as to write the given expression in the form
u = R cos (ω0 t − δ).
a) u = 3 cos 2t +√
4 sin 2t.
b) u = − cos t + 3 sin t.
Problem 16. 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.
Problem 17. A mass weighing 16 lb stretches a spring 3 in. The mass is attached
to a viscous damper with a damping constant of 2 lb · s/f t. If the mass is set in
motion from its equilibrium position with a downward velocity of 3 in/s, find the
time when the mass first returns to its equilibrium position.
Problem 18. A mass weighing 4 lb stretches a spring 1.5in. The mass is given a
positive displacement of 2 in from its equiliburium position, and is released. Assuming there is no damping and that the mass is acted on by an external force of 2 cos 3t
lb, formulate and solve the initial value problem describing the motion of the mass.
t 1
t 1
Problem 19. Verify that x = e
+ 2te
is a solution of the system
0
1
2 −1
1
0
t
x =
x+e
.
3 −2
−1
 
 
 
1
0
1
Problem 20. Are the vectors x1 = −1, x2 = 1, x3 = 1 linearly inde1
1
1
pendent?
−2 −6
0
Problem 21. Consider the system x =
x. Two solutions of the system
0
1
−2 t
1 2t
are x1 =
e and x2 =
e .
1
0
a) Use the Wronskian to verify that the two solutions are linearly independent.
b) Write the general solution of the system.
4
DR. NEAL BUSHAW
Problem 22. Suppose that the system x0 = Ax has the general solution


 
 
 
x1 (t)
−2
1
0
t
−2t







x(t) = x2 (t) = c1 1 e + c2 0 e + c3 1 e−t .
x3 (t)
0
1
1
 
1
Given the initial condition x(0) =  1 , find x1 (t), x2 (t), and x3 (t).
−1
1 −3
0
Problem 23. Solve the initial value problem x = Ax with A =
and
0 −2
1
x(0) =
.
3
Problem 24. Solve the initial value problem
x0 = x + 2y
y 0 = 4x + 3y
with x(0) = 3, y(0) = 0.
Solutions: Don’t Look! Not yet! Are you sure?!
Problem 1.
a) y = 1 − cos t + sin t − u3π (t)(1 + cos t).
b) y = e−t sin t+ 21 uπ (t)(1+e−(t−π) cos t+e−(t−π) sin t)− 12 u2π (t)(1−e−(t−2π) cos t−
e−(t−2π) sin t).
c) y = e−t − e−2t + u2 (t)( 12 − e−(t−2) + 12 e−2(t−2) ).
d) y = u1 (t)h(t − 1) − u2 (t)h(t − 2), where h(t) = −1 + 12 (cos t + cosh t).
Problem 2. f (t) = (ut0 (t − t0 ) − ut0 +k (t)(t − (t0 + k))) hk .
Problem 3.
a) y = e−t cos t + e−t sin t + uπ (t)e−(t−π) sin(t − π).
b) y = 21 uπ (t) sin 2(t − π) − 12 u2π (t) sin 2(t − 2π).
√
√
c) y = 41 sin t − 14 cos t + 14 e−t cos 2t + √12 u3π e−(t−3π) sin 2(t − 3π).
Problem 4.
√
a) y = 2e−t cos 5t +
b) α = 1.50878.√
c) t =
d) √π5 .
π−arctan
√
5
2 5
2+α
.
α+2
√ e−t
5
sin
√
5t.
SAMPLE PROBLEMS, MATERIAL AFTER EXAM 2
5
Problem 5.
a) y = −e3t sin 2t + e3t cos 2t.
b) y = (1 − 2t)e−2t .
c) y = −3e−2t + 6e−t .
Problem 6. α > 1.
Problem 7. c1 +c2 t+c3 t2 +c4 e−2t +c5 te−2t +e−3t (c6 cos 2t+c7 sin 2t)+te−3t (c8 cos 2t+
c9 sin 2t).
Problem 8.
a) y = c1 t−1 + c2 t−1 ln t.
b) y = c1 t + c2tet .
Problem 9. y = c1 e−t + c2 te−t + 12 t2 e−t .
Problem 10.
a) y = 32 e2t − 2e−t + 23 − 3t −2te−t .
b) y = 59 e2t − 59 e−t + t2 − 32 t e2t .
Problem 11.
a) Y (t) = t(A0 t2 + A1 t + A2 ) + t(B0 t2 + B1 t + B2 )e−3t + E sin 3t + F cos 3t.
b) Y (t) = A0 t + A1 + t(B0 t + B + 1) sin t + t(C0 t + C1 ) cos t.
c) Y (t) = A0 et cos 2t + A1 et sin 2t + e2t (B0 t + B1 ) sin t + e2t (C0 t + C1 ) cos t.
d) Y (t) = A0 e−t + te−t (B0 t2 + B1 t + B2 ) sin t + te−t (C0 t2 + C1 t + C2 ) cos t.
e) Y (t) = A0 t2 + A1 t + A2 + t2 (B0 t + B1 )e2t + (C0 t + C1 ) sin 2t + (D0 t + D1 ) cos 2t.
√
Problem 12. y = −3 cos 2t + 3 sin 2t = 3 2 cos(2t − 3π
).
4
Problem 13. 2u00 + 39.2u = 0, u(0) = 0, u0 (0) = 0.1.
√
Problem 14. γ = 2 2.
Problem 15.
a) u = 5 cos (2t − δ), δ = arctan 4/3 = 0.9273
b) u = 2 cos (t − 2π/3).
√
√
√
1
Problem 16. u(t) = 4√1 2 sin 8 2t − 12
cos 8 2t, ω = 8 2, T =
δ = 2.0113.
Problem 17. t =
π
√
.
2 31
Problem 18. u00 + 256u = 16 cos 3t. u(0) = 1/6, u0 (0) = 0.
Problem 20. Yes.
π
√
,
4 2
R=
p
11/288,
6
DR. NEAL BUSHAW
Problem 21.
−2et e2t = −e3 t 6= 0, so they are linearly independent.
a) W (x1 , x2 )(t) = t
e
0
−2 t
1 2t
b) x(t) = c1
e + c2
e .
1
0
Problem 22.
x1 (t) = 6et − 5e−2t
x2 (t) = −3et + 4e−t
x3 (t) = −5e−2t + 4e−t
−2t 1
t 1
.
+ 3e
Problem 23. x(t) = −2e
1
0
Problem 24.
x(t) = 2e−t + e5t
y(t) = −2e−t + 2e5t