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Math 3D - Homework 10 Solutions Paul Macklin March 20, 2005 Page 344, #2 (5 points): Find the general solution to the following system of equations:
x˙ =
(^) x.
Solution: First, we find the determinant of A − λI: ∣∣ ∣∣ ∣∣
1 − λ − 5 0 1 −(3 + λ) 0 0 0 1 − λ
∣∣ =^ −(1^ −^ λ)^2 (3 +^ λ) + 5(1^ −^ λ) = (1 − λ)(2 + 2λ + λ^2 ) = (1 − λ)(λ + 1 − i)(λ + 1 + i). Let us first find an eigenvector v 1 = (x, y, z)T^ for the eigenvalue 1. Plugging λ = 1 into (A − λI)v 1 = 0, we have
x y z
whereby y = 0 and x = 4y = 0. Therefore, an eigenvector is
v 1 =
We now find an eigenvector for λ = − 1 − i.
2 + i − 5 0 1 −(2 + i) 0 0 0 2 + i
x y z
By the last equation, z = 0, and by the second, x = (2 + i)y. Therefore, and eigenvector is
2 + i 1 0
and a solution is
e−(1+i)t
2 + i 1 0
(^) = e−t^ (cos(t) − i sin(t))
2 + i 1 0
= e−t
2 cos t − sin t cos t 0
(^) − ie−t
2 sin t + cos t sin t 0
The real and imaginary parts are linearly independent solutions, and so the general solution is
x(t) = c 1 e−t
2 cos t − sin t cos t 0
(^) + c 2 e−t
2 sin t + cos t sin t 0
(^) + c 3 et
Page 344, #6 (5 points): Solve the given initial-value problem:
x˙ =
x, x(0) =
Solution: First, we find the determinant of A − λI:
∣∣ ∣∣^3 − 4 λ^ −(1 +−^2 λ)
∣∣ = (λ + 1)(λ − 3) + 8 = λ^2 − 2 λ + 5.
The roots to this polynomial (by the quadratic formula) are 1± 2 i. Let us find an eigenvector for 1 − 2 i. [2(1 + i) − 2 4 −2(−1 + i)
] [x y
]
By the first equation, x(1 + i) = y, and so an eigenvector is (^) [ 1 1 + i
]
and so a solution is
e(1−^2 it)
[ 1
1 + i
]
= et^ (cos 2t − i sin 2t)
[ 1
1 + i
]
Thus, both the real and imaginary parts are linearly independent solutions, and so the general solution is
x(t) = c 1 et
[ (^) cos 2t cos 2t + sin 2t
]
[ (^) − sin 2t cos 2t − sin 2t
]
By the initial condition, (^) [ 1 5
]
= c 1
[ 1
]
[ 0
]
whence c 1 = 1 and c 2 = 4. Thus, the solution is
x(t) = et
[ (^) cos 2t cos 2t + sin 2t
]
[ (^) − sin 2t cos 2t − sin 2t
]
= et
[ (^) cos 2t − 4 sin 2t 5 cos 2t − 3 sin 2t
]
To find a second solution, we seek v = (x, y, z)T^ such that (A − 2 I)^2 v = 0. First, note that
(A − 2 I)^2 =
Now, let’s find that vector:
R^2 + −→R^3 →R^3
whereby x = y + z, and so a good choice is v = (1, 0 , 1)T^. (We can pick y = 0, z = 1. It’s good to pick vectors with as many zeros as possible without making them the zero vector, as that simplifies your multiplications, etc.) Therefore, we can form a second solution by
x 2 = e^2 t^ (I + (A − 2 I)t)
= e^2 t
(^) t
= e^2 t
t 1 − t
Since the eigenvalue λ = 2 is repeated three times, we seek a third solution for λ = 2. We do this by looking for a vector v = (x, y, z)T^ such that (A − 2 I)^3 v = 0. First,
(A − 2 I)^3 =
and so any v will satisfy (A − 2 I)^3 v = 0. A good choice is v = (0, 0 , 1)T^. (Again, under the operating assumption that it’s good to have the vector have as many zeros as possible without being the zero vector.) Hence, a third solution is
x 3 (t) = e^2 t
I + (A − 2 I)t + (A − 2 I)^2 2!^1 t^2
= e^2 t
(^) t +
(^12) t^2
= e^2 t
t −t + 12 t^2 2 t − 12 t^2
Thus, the general solution is
x(t) = c 1 x 1 (t) + c 2 x 1 (t) + c 3 x 1 (t) = c 1 e^2 t
(^) + c 2 e^2 t
t 1 − t
(^) + c 3 e^2 t
t −t + 12 t^2 2 t − 12 t^2
