Download MTH 264 SECTION 3.3 20 DELTA COLLEGE The slope field for the

MTH 264
DELTA COLLEGE
SECTION 3.3 20
The slope field for the system
dx
dt
dy
dt
=
2x + 6y
=
2x − 2y
is shown on the right.
(a) Determine the type of the equilibrium point at the origin.
(b) Calculate all straight line solutions.
(c) Plot the x(t)- and y(t)-graphs (t ≥ 0) for the initial
conditions A(1, −1), B(3, 1), C(0, −1), D(−1, 2)
Solution:
(a) The matrix that represents this system is A =
2
2
6
−2
.
The characteristic equation of this matrix is: λ2 − 16 = 0. This equation factors into (λ − 4)(λ + 4) = 0, and
so it has two distinct, real roots: λ1 = 4 and λ2 = −4. Since one eigenvalue is positive and one eigenvalue is
negative, the equilibrium point at the origin is a saddle.
(b) For eigenvalue λ1 = 4, we can select:
−2
6
x
0
=
2 −6
y
0
For eigenvalue λ2 = −4, we can select:
6 6
x
0
=
2 2
y
0
eigenvector
eigenvector
~v1 =
1
−1
3
1
~v2 =
~A (t) =
The two straight line solutions of this system (through A, B) are Y
3
1
4t
e
~B (t) =
and Y
1
−1
e−4t .
(c) To construct solutions with initial conditions of C(0, −1) and D(−1, 2), we need to find linear combinations
~A and Y
~B that equal (0, −1) and (−1, 2), respectively, that
of the initial conditions of the straight line solutions Y
is solve the systems:
0
3
1
−1
3
1
= k1
+ k2
and
= l1
+ l2
−1
1
−1
2
1
−1
The reader is left to solve these systems and verify that the solutions are k1 = −1/4 and k2 = 3/4, and
l1 = 1/4 and l2 = −7/4, yielding solutions:
3
1 3
7
1 3
1
1
4t
−4t
4t
~
~
e +
e
and
YD (t) =
e −
e−4t
YC (t) = −
4 1
4 −1
4 1
4 −1
Below we sketch the four solutions in the phase plane, the four x(t)-graphs and then the four y(t)-graphs. It
is imperative that you be able to identify which graphs come from which solutions. They are not labeled here
and you should be sure that you can decide which is which.