Answers and Hints to Review Questions for Test 3
1. (a) Find the general solution to the linear system of differential equations
dY
−2 −3
=
Y.
3 −2
dt
(b) Find the specific solution that satisfies Y (0) =
2
.
−1
(c) What is the natural period of the solutions?
−2 − λ
−3
Answer. (a) The eigenvalues are found by solving 3 −2 − λ
2
λ + 4λ + 13 = 0 and so
√
−2 ± 16 − 52
= −2 ± 3i.
λ=
2
Using λ = −2 + 3i we solve
−3i −3
0
v1
=
3 −3i
v2
0
= 0. Therefore,
One such eigenvector is when v1 = i and v2 = 1. Thus the complex solution is
i
Y (t) =
e−2t [cos(3t) + i sin(3t)]
1
− sin(3t)
cos(3t)
−2t
=
e +i
e−2t
cos(3t)
sin(3t)
Therefore, the general (real) solution is
− sin(3t)
cos(3t)
−2t
Y (t) = c1
e + c2
e−2t
cos(3t)
sin(3t)
(b) For the specific solution, one can see that c1 = −1 and c2 = 2, and so
sin(3t)
cos(3t)
−2t
Y (t) =
e +2
e−2t
− cos(3t)
sin(3t)
(c)
2π
.
3
2. (a) Find the general solution to the linear system of differential equations
dY
2 1
=
Y.
−1 4
dt
(b) Find the specific solution that satisfies Y (0) =
2
.
−1
2−λ
1
Answer. (a) The eigenvalues are found by solving −1 4 − λ
λ = 3 is a repeated eigenvalue.
= λ2 − 6λ + 9 = 0. Thus
The general solution is then
Y (t) =
x0
y0
3t
e +
−1 1
−1 1
x0
y0
te3t
or, simplifying, the general solution becomes
x0
−x0 + y0
3t
Y (t) =
e +
te3t
y0
−x0 + y0
(b) For the given initial condition, x0 = 2 and y0 = −1, thus the specific solution is
2
−3
3t
Y (t) =
e +
te3t
−1
−3
3. Sketch the phase portraits for the systems in Questions 1 and 2, and sketch the specific
solutions on those phase portraits.
Answer. Check your answers using HPG System solver.
4. Sketch the phase portraits for the following systems of differential equations with the help
of the given information about their eigenvalues and/or eigenvectors. Also sketch the solution
curve with the initial condition Y (0) = (1, 0).
dY
−2 −1
1
=
Y . Eigenvalues: λ = −3 repeated; eigenvector:
.
(a)
1 −4
1
dt
dY
2
4 2
−1
(b)
=
Y . Eigenvalues: λ = 0, 5 eigenvectors:
,
.
2
1
2
1
dt
dY
2
−4 −2
−1
(c)
=
Y . Eigenvalues: λ = 0, −5 eigenvectors:
,
.
−2 −1
2
1
dt
Answer. (a) See text Section 3.5#3. (b) See text Section 3.5#19. (c) Looks like (b) with
arrows reversed.
5. Write the general solution to the system in 4(b) in both component form, and in vector
form.
Answer. In vector form:
Y (t) = c1
−1
2
+ c2
2
1
In component form, x(t) = −c1 + 2c2 e5t , y(t) = 2c1 + c2 e5t .
e5t
6. Given the following systems of differential equations, and the corresponding eigenvalues,
for each system
(a) determine if the origin is a spiral sink, spiral source, or a center,
(b) determine the natural period of the oscillations,
(c) determine whether the solutions go in a clockwise or counterclockwise direction around the
origin,
(d) sketch the xy-phase portrait for the system, and the x(t)- and y(t)- graphs for solutions
with the indicated initial conditions.
dY
0 2
(i)
=
Y ; initial condition Y0 = (1, 0); the eigenvalues are λ = ±2i.
−2 0
dt
√
dY
−1
2
(ii)
=
Y ; initial condition Y0 = (0, 1); the eigenvalues are λ = −1 ± 2i.
−1 −1
dt
√
dY
2 −6
(iii)
=
Y ; initial condition Y0 = (2, 1); the eigenvalues are λ = (3 ± i 47)/2.
2
1
dt
Answer. (i) See answer to 3.4#3. (ii) See answer to 3.4#5. (iii) See answer to 3.4#7.
7. Find a system with the same eigenvalues as in 6(i), but whose solutions travel in the opposite
direction.
dY
0 −2
Answer. Just consider the negative of the equations in 6(i):
=
Y.
2
0
dt
8. Find equations involving a, b, c and d for the matrix A =
a b
c d
so that:
(a) A has 0 as an eigenvalue;
(b) A has a repeated eigenvalue.
Answer. First, the eigenvalues of A satisfy the equation λ2 − (a + d)λ + (ad − bc) = 0.
(a) If ad − bc = 0, then λ = 0 is a solution to the previous equation.
(b) If the discriminant in the quadratic equation is 0, then there is a repeated real root, i.e., if
(a + d)2 − 4(ad − bc) = 0.
9. Find the general solutions to the spring mass system where b,m and k are as given below.
Classify each system as undamped, underdamped, critically damped, or overdamped, and find
the period of the oscillations (if applicable).
(a) m = 2, b = 0 and k = 18
(b) m = 3, b = 6 and k = 15
(c) m = 1, b = 10 and k = 25
(d) m = 1, b = 7 and k = 10.
Answer. (a) y = c1 cos(3t) + c2 sin(3t), and the system is undamped with period
2π
.
3
(b) y = c1 e−t cos(2t) + c2 e−t sin(2t), and the system is underdamped with period π.
(c) y = c1 e−5t + c2 te−5t , and the system is critically damped, nonperiodic.
(d) y = c1 e−2t + c2 e−5t and the system is overdamped, nonperiodic.
10. Solve the equation in 9(d) subject to the initial conditions (a) y(0) = 0 and y 0 (0) = 0, and
(b) y(0) = 2, y 0 (0) = 0.
Answer.
(a) y(t) = 0.
(b) y(t) =
10 −2t
e
3
− 43 e−5t .
11. Do Section 3.7, Exercise # 7.
12. Find the general solutions to the following homogeneous differential equations. Use MathCAD or some other form of technology to help you find the roots of the characteristic equations.
(a) y (4) + 2y (3) − 7y 00 − 20y 0 − 12y = 0.
(b) y (4) + 8y (3) + 42y 00 + 104y 0 + 169y = 0.
(c) y (5) − 5y (4) + 2y (3) − 10y 00 + y 0 − 5y = 0.
(d) y (7) + 7y (6) + 17y (5) + 15y (4) − 5y (3) − 19y 00 − 13y 0 − 3y = 0.
Answer. The zeros of the corresponding characteristic equation were found using MathCAD
and are listed with each answer.
(a) The zeros are −1, 3, −2, −2, thus the general solution is
y = c1 e−2t + c2 te−2t + c3 e−t + c4 e3t .
(b) The zeros are −2 ± 3i, −2 ± 3i, thus the general solution is
y = c1 e−2t cos(3t) + c2 e−2t sin(3t) + c3 te−2t cos(3t) + c4 te−2t sin(3t).
(c) The zeros are ±i, ±i, 5, thus the general solution is
y = c1 cos(t) + c2 sin(t) + c3 t cos(t) + c4 t sin(t) + c5 e5t .
(d) The zeros are −1 (multiplicity 5), 1 and 3, thus the general solution is
y = c1 e−t + c2 te−t + c3 t2 e−t + c4 t3 e−t + c5 t4 e−t + c6 et + c7 e3t .
13. Find the form of the particular solution necessary for solving the following nonhomogeneous
differential equations. Do not find the particular solution.
(a) y 00 − 6y 0 + 9y = e2t
(b) y 00 − 5y 0 + 6y = e2t
(c) y 00 − 4y 0 + 4y = e2t
(d) y 00 − 4y 0 + 4y = t2
(e) y 00 − 6y 0 + 9y = t2 + e2t
(f) y 00 − 4y 0 + 4y = sin(6t)
(g) y 00 − 4y 0 + 4y = t cos(6t)
(h) y 00 + 36y = 10 sin(6t)
Answer. (a) yp = Ae2t
(b) yp = Ate2t
(c) yp = At2 e2t
(d) yp = At2 + Bt + C
(e) yp = At2 + Bt + C + De2t
(f) yp = A sin(6t) + B sin(6t)
(g) yp = A sin(6t) + B sin(6t) + At sin(6t) + Bt sin(6t)
(h) yp = A sin(6t) + B sin(6t) + At sin(6t) + Bt sin(6t)
14. Find the general solutions to the following differential equations.
(a) y 00 + 4y 0 + 3y = e−2t
(b) y 00 + 4y 0 + 3y = t + 1
(c) y 00 + 5y 0 + 6y = e−t + 4
(d) y 00 + 4y 0 + 13y = 3 cos(2t)
Answer. (a) y = c1 e−3t + c2 e−t − e−2t
(b) y = c1 e−3t + c2 e−t + 31 t −
1
9
(c) y = c1 e−3t + c2 e−2t + 12 e−t +
2
3
(d) y = c1 e−2t cos(3t) + c2 e−2t sin(3t) +
27
145
cos(2t) +
24
145
sin(2t)
15. Find the solutions to 14(c) subject to the initial condition y(0) = 0 and y 0 (0) = 0.
Answer. y(t) =
11 −3t
e
6
− 3e−2t + 21 e−t + 23 . As t → ∞, y(t) → 23 .
16. For what values of k will pure resonance occur in the following undamped spring systems?
(a) y 00 + ky = −2 sin(3t).
(b) y 00 + ky = 5 cos(6t).
Answer. (a) k = 9 (b) k = 36.
17. Compute the solution to y 00 + 9y = 2 cos(3t) subject to y(0) = 2 and y 0 (0) = −9.
Answer. See Section 4.3 #13.
18. Suppose the suspension system of the average car can be fairly well modeled by an
underdamped harmonic oscillator with a natural period of 2 seconds. How far apart should
speed bumps be placed so that a car traveling at 10 miles per hour over several bumps will
bounce more and more violently with each bump?
Answer. Place the speed bumps so that the car will hit them every two seconds (the natural
period of the car). In 2 seconds, the car travels 176/6 feet, or 29 feet, 4 inches.
19. (a) (A Superposition Principle) Suppose y1 is a solution to ay 00 + by 0 + cy = f (t) and y2 is a
solution to ay 00 + by 0 + cy = g(t) where the a, b and c are the same constants in each equation,
but f and g may be different. Show that y = αy1 + βy2 is a solution to
ay 00 + by 0 + cy = αf (t) + βg(t)
where α and β are constants.
(b) Given that y1 (t) = − 51 sin(3t) is a solution to
y 00 + 4y = sin(3t)
and that y2 (t) = 34 t sin(2t) is a solution to
y 00 + 4y = 3 cos(2t),
find the general solution to
y 00 + 4y = −5 sin(3t) + 9 cos(2t)
Answer. (a) Use the linearity property of derivatives and plug in the results:
ay 00 + by 0 + cy = a(αy100 + βy200 ) + b(αy10 + βy20 ) + c(αy1 + βy2 )
= α(ay100 + by10 + cy1 ) + β(ay200 + by20 + cy2 )
= αf (t) + βg(t).
(b) y = c1 cos(2t) + c2 sin(2t) + sin(3t) + 49 t sin(2t).
20. (a) Show that y(t) = k1 cos(ωt)+k2 sin(ωt) can be written in the form y(t) = A cos(ωt−φ).
Find formulas for A and φ.
√
(b) Convert y(t) = 3 cos(4t) − sin(4t) to the form y(t) = A cos(ωt − φ).
√
(c) Convert y(t) = − cos(4t) + 3 sin(4t) to the form y(t) = A cos(ωt − φ).
Answer. Using the identity cos(α − β) = cos α cos β + sin α sin β one obtains
A cos(ωt − φ) = A[cos(ωt) cos(φ) + sin(ωt) sin(φ)]
= k1 cos(ωt) + k2 sin(ωt).
Therefore, k1 = A cos φ and k2 = A sin φ, consequently,
q
k1
k2
(1)
A = k12 + k22 , cos φ = , and sin φ = .
A
A
√
√
(b) First A = 3 + 1 = 2. Thus sin φ = − 21 and cos φ = 23 . Therefore, one choice for φ is
φ = −π/6. Thus y(t) = 2 cos(4t + π/6).
√
√
(c) First A = 1 + 3 = 2. Thus sin φ = 23 and cos φ = − 12 . Therefore, one choice for φ is
φ = 32 π. Thus y(t) = 2 cos(4t − 23 π).
As a double check in each of (b) and (c), you can graph the original function along with your
answer and see if the graphs are the same.