3.1-3.3 Even-Numbered Homework Solutions
3.1
16. The general form of a linear, homogeneous, second-order equation with constant coefficients is
d2 y
dy
+p
+ qy = 0.
2
dt
dt
(a) Write the first-order system for this equation, and write this system in matrix
form. dy
dv
dY
0
1
Let
= v. Then
= −pv − qy. The matrix form of this sytem is
=
Y.
−q −p
dt
dt
dt
(b) Show that if q 6= 0, then the origin is the only equilibrium point of the system.
The determinant of the matrix in part (a) is given by (0)(−p) − (1)(−q) = q. The only equilibrium point of a
system is the origin if the determinant is not equal to zero.
(c) Show that if q 6= 0, then the only solution of the second-order equation with y constant is y(t) = 0 for all t.
dv
dy
= 0 and
= 0 for all t. Then in order for
Suppose that y(t) = C for all t, where C is some constant. Then
dt
dt
dv
dy
the equation
+p
+ qy = 0 to be true when q 6= 0, y(t) = 0 for all t.
dt
dt
d2 y
dy
= 0 into a first-order system using v =
as usual.
dt2
dt
dv
(a) Find the general solution for the
equation.
dt
dy
dv
If we let
= v, then
= 0. So v = C for some constant C.
dt
dt
dy
equation and find the general solution of the system.
(b) Substitute this solution into the
dt
dy
So
= v = C, so y = Ct + K, where K is another constant.
dt
18. Convert the second-order equation
26.
(a) We want
dY1
dY2
= AY1 and
= AY2 for matrix A given. Begin with Y1 :
dt
dt
dY1
= (−3e−3t , −3e−3t )
dt
= (−2e−3t − e−3t , 2e−3t − 5e−3t )
−2 −1 e−3t
=
2 −5 e−3t
= AY1 , as desired
dY2
= (−4e−4t , −8e−4t )
dt
= (−2e−4t − 2e−4t , 2e−4t − 5(2e−4t ))
−2 −1 e−4t
=
2 −5 2e−4t
= AY2 , as desired
(b) Since Y1 (0) = (1, 1) and Y2 (0) = (1, 2), neither is a multiple of the other, and the solutions are linearly
independent.
1
(c) Note that Y1 (0) + Y2 (0) = (1, 1) + (1, 2) = (2, 3). The function Y (t) = Y1 (t) + Y2 (t) =
e−3t + e−4t
e−3t + 2e−4t
is the
solution to the IVP.
28. Let A =
−2
3
−3
cos 3t
− sin 3t
2
, Y1 (t) = e−2t
, Y2 (t) = e−2t
, and Y (0) =
.
−2
sin 3t
cos 3t
3
(a) Check that the two functions are solutions of the system; if they are not solutions, then stop.
dY1
= AY1 :
Verify that
dt
dY1
−2e−2t cos 3t − 3e−2t sin 3t
=
3e−2t cos 3t − 2e−2t sin 3t
dt
−2t
−2 −3
e
cos 3t
=
, as desired.
3 −2
e−2t sin 3t
Verify that
dY2
= AY2 :
dt
2e−2t sin 3t − 3e−2t cos 3t
−3e−2t sin 3t − 2e−2t cos 3t
−2t
−2 −3
−e
sin 3t
=
, as desired.
3 −2
e−2t cos 3t
dY1
=
dt
(b) Check that the two solutions are linearly independent; if they are not linearly independent, then stop.
It is obvious that the two solutions are linearly independent.
(c) Find the solution to the linear system with the given initial value.
The general solution to the linear system is k1 Y1 (t) + k2 Y2 (t), where k1 and k2 are constants. Find k1 and k2 :
2
−2(0) k1 cos 3(0) − k2 sin 3(0)
=e
3
k1 sin 3(0) + k2 cos 3(0)
k1
=
.
k2
Hence, the solution to the system with the given initial value is Y (t) = 2Y1 (t) + 3Y2 (t).
3.2
−2
Y.
−3
−4 − λ
−2
(a) The matrix A − λI is given by
. So det A = λ2 + 7λ + 10 = (λ + 5)(λ + 2). The eigenvalues
−1
−3 − λ
are λ = −5 and λ = −2.
x1
−4 −2
(b) First, we need a vector v1 =
such that
v = −5v1 . We need to solve the system:
y1
−1 −3 1
2. We have the system
dY
=
dt
−4
−1
−4x1 − 2y1 = −5x1
−x1 − 3y1 = −5y1 , or
x1 − 2y1 = 0
−x1 + 2y1 = 0
.
x1
2
x2
−4
=
. We also need a vector v2 =
such that
y1
1
y2
−1
the system:
So, v1 =
2
−2
v = −2v2 . We need to solve
−3 2
−4x2 − 2y2 = −2x2
−x2 − 3y2 = −2y2 , or
−2x2 − 2y2 = 0
−x2 − y2 = 0
.
So, v2 =
x2
y2
=
1
.
1
(e) The general solution for this system is Y (t) = k1 e
−5t
2
−2t 1
+ k2 e
.
1
1
1
Y.
4
2−λ
1
(a) The matrix A − λI is given by
. So det A = λ2 − 6λ + 9 = (λ − 3)2 . The eigenvalue is λ = 3.
−1
4−λ
x
2 1 x1
x
(b) We need a vector v = 1 such that
= 3 1 . So, we have the system
y1
−1 4 y1
y1
dY
=
4. We have the system
dt
2
−1
2x1 + y1 = 3x1
−x1 + 4y1 = 3y1 , which is equal to the system
−x1 + y1 = 0
−x1 + y1 = 0
These equations specify the line x1 = y1 , so the vector v =
1
is an eigenvector for λ = 3.
1
(e) System does not have two distinct eigenvalues, so part (e) doesn’t have to be completed.
dY
=
6. We have the system
dt
5
9
4
Y.
0
(a) Use the techniques in exercises 3.2.2 and 3.2.4 above to check that the eigenvalues are λ = 9 and λ = −4.
1
(b) Check that the corresponding eigenvector for λ = 9 is v1 =
and the corresponding eigenvector for λ = −4 is
1
−4
v2 =
.
9
9t 1
−4t −4
(c) The general solution for this system is Y (t) = k1 e
+ k2 e
.
1
9
dY
=
8. We have the system
dt
2
−1
−1
Y.
1
(a) Use the techniques in exercises
3.2.2 and 3.2.4
√
√ above, as well as the quadratic formula, to check that the
3+ 5
3− 5
eigenvalues are λ =
and λ =
.
2
2
√
√ 3+ 5
−1 − 5
(b) Check that the corresponding eigenvector for λ =
is v1 =
and the corresponding eigenvector
2
2
√
√ 3− 5
−1 + 5
for λ =
is v2 =
.
2
2
√
√
3+ 5 3− 5 √ √ t −1 −
t −1 +
5
5
+ k2 e 2
.
(c) The general solution for this system is Y (t) = k1 e 2
2
2
3
12. Solve the following initial value problem for the given initial conditions:
dx
= 3x
dt
dy
= x − 2y
dt
x(0)
1
(a)
=
y(0)
0
dY
3 0
=
Convert the system to into matrix form:
Y . The eigenvalues are λ1 = 3 and λ2 = −2 with
dt 1 −2
5
0
5
0
corresponding eigenvectors v1 =
and v2 =
. So the general solution is Y (t) = k1 e3t
+ k2 e−2t
.
1
1
1
1
1
1
Then 1 = x(0) = 5k1 e3(0) = 5k1 ⇒ k1 = and 0 = y(0) = k1 e3(0) + k2 e−2(0) = k1 + k2 ⇒ k2 = − . So the
5
5
1 3t 5
1 −2t 0
solution for the given initial condition is Y (t) = e
− e
.
1
1
5
5
x(0)
0
(b)
=
y(0)
1
We have 0 = x(0) = 5k1 , which implies k1 = 0. Then k2 = 0. So the solution is Y (t) = 0.
x(0)
2
(c)
=
y(0)
2
2
2
2
2
5
0
− e−2t
.
We have 2 = x(0) = 5k1 , which implies k1 = . Then k2 = − . So the solution is Y (t) = e3t
1
1
5
5
5
5
dY
4 −2
14. Solve the initial-value problem
=
Y , Y (0) = Y0 for the following values of Y0 .
1 1
dt
1
(a) Y0 =
0
Using the techniques in exercise 3.2.12above,
check that
the eigenvalues for this system are λ1 = 3 and λ2 = 2
2
1
with corresponding eigenvectors v1 =
and v2 =
. Hence, the general solution is
1
1
2
1
Y (t) = k1 e3t
+ k2 e2t
. Check that k1 = 1 and k2 = −1. The solution of the initial-value problem is thus
1
1
2
1
Y (t) = e3t
− e2t
.
1
1
2
(b) Y0 =
1
2
Check that k1 = 1 and k2 = 0. The solution of the initial-value problem is thus Y (t) = e3t
.
1
−1
(c) Y0 =
−2
2
1
Check that k1 = 1 and k2 = −3. The solution of the initial-value problem is thus Y (t) = e3t
− 3e2t
.
1
1
4
3.3
2.
6.
5
8.
10.
(a)
6
(b)
(c)
7
12.
(a)
(b)
8
(c)
dY
2 6
=
.
2 −2
dt
(a) The eigenvalues are λ1 = −4 and λ2 = 4, so the equilibrium point at the origin is a saddle.
−4t −x1
(b) For λ1 = −4, solutions of the form e
are straight-line solutions. For λ2 = 4, solutions of the form
x1
3x1
e4t
are straight-line solutions.
x1
20. You are given the slope field for the system
9