**Unformatted text preview: **Introduction to Diﬀerential Equations – Math 286 X1
Fall 2009
Homework 8 Solutions
1. Solve
−1 1
−4 −1 x′ = 1
1 x(0) = x, . Solution: We ﬁrst compute the eigenvalues and eigenvectors. The characteristic polynomial is
λ2 + 2λ + 5 = 0,
which has roots
λ= −2 ± √
4 − 20
= −1 ± 2i.
2 Choosing λ = −1 + 2i, we have
−2i
1
−4 −2i A − (−1 + 2i)I =
and solving
−2i
1
−4 −2i gives x
y 0
0 = −2ix + y = 0,
−4x − 2iy = 0.
One solution to this is x = 1, y = 2i, so our eigenvector can be chosen as
1
2i v= . Our two solutions will be the real and imaginary parts of eλt v, which we compute as
e(−1+2i)t 1
2i 1
2i = e−t (cos(2t) + i sin(2t)) = e−t cos(2t) + i sin(2t)
2i cos(2t) − 2 sin(2t) Taking real and imaginary parts gives two solutions:
x1 = e−t , x2 = e−t sin(2t)
2 cos(2t) . cos(2t)
−2 sin(2t) + C2 e−t sin(2t)
2 cos(2t) . cos(2t)
−2 sin(2t) Thus the general solution is
x(t) = C1 e−t
Plugging in t = 0 gives
x(0) = C1
2C2 , so C1 = 1, C2 = 1/2, so the solution is
x(t) = e−t cos(2t)
−2 sin(2t) 1 1
+ e−t
2 sin(2t)
2 cos(2t) . . 2. Solve 3
0 x′ = 1
3 1
1 x(0) = x, . Solution: The characteristic polynomial is
λ2 − 6λ + 9 = 0,
which is factored as (λ − 3)2 = 0, so it has a double root of 3. We have that
0 1
0 0 , = A − 3I = 0
0 and to get the eigenvector we get
0
0 1
0 x
y
1
0 or the equation y = 0. A solution to this is v1 = , . This gives us one solution to the ODE, namely
1
0 e3t . We need to now ﬁnd the generalized eigenvector, which we ﬁnd by solving the system
(A − 3I)v2 = v1 ,
or 0
0 1
0 x
y
0
1 or the equation y = 1. So we can choose v2 =
is , . Then we know another solution to this problem
1
0 x2 = e3t (tv1 + v2 ) = e3t t 1
0 = 0
1 + t
1 = e3t Thus our general solution is
x(t) = C1 e3t 1
0 + C2 e3t t
1 . Plugging in t = 0 gives
x(0) = C1
C2 , so C1 = C2 = 1, and our solution is
x(t) = e3t 1
0 + e3t 2 t
1 = e3t t+1
1 . . 3. Solve 3
0 x′ = 1
2 1
1 x(0) = x, . Solution: The characteristic polynomial is
λ2 − 5λ + 6 = 0,
which has roots λ = 2, 3. We have
A − 2I =
1
−1 so the eigenvector for λ = 2 is 1
0 , , and
0
0 A − 3I =
so the eigenvector for λ = 3 is 1 1
0 0 1
−1 , . Thus the general solution is x(t) = C1 e2t 1
−1 x(0) = C1 1
−1 1
0 + C2 e3t + C2 . Plugging in t = 0 gives
1
0 , which gives the system
C1 + C2 = 1,
−C1 = 1,
which has solution C1 = −1, C2 = 2. Thus our solution is
1
−1 x(t) = −e2t 1
0 + 2e3t . 4. For each of these matrices A, compute etA :
A= 2
0 0
2 , 2
0 A= 0
−1 , A= 7 −4
8 −5 . Solution: In each case, we need to compute the eigenvalues and eigenvectors of the matrix ﬁrst. So
we proceed:
(a) The characteristic polynomial is
λ2 − 4λ + 4 = 0,
which has a double root of 2. We have
A − 2I =
3 0
0 0
0 , so that all vectors are eigenvectors and we can choose what we like. Choose
1
0 v1 = , 0
1 v2 = , and this gives
1 0
0 1 S= . Since S is the identity matrix, S −1 = S, so we have
e2t
0 etA = S 0
e2t e2t
0 S −1 = I 0
e2t I= (b) the characteristic polynomial is
λ2 − λ − 2 = 0, which has roots λ = 2, −1. For λ = 2, we have 0
0 and 0
0 0
1 x
y 0
1 , = A − 2I = 0
0 0
0 , gives
y = 0,
so we have v1 = 1
0 . For λ = −1, we have
A+I = and we have v1 = 0
1 3
0 . Thus we have
S= 1 0
0 1 . Similarly to above, we have
S e2t
0 0
e −t S= e2t
0 0
e−t since S is the identity.
(c) The characteristic polynomial is
λ2 − 2λ − 3 = 0, which has roots λ = 3, −1. We have A − 3I =
and our eigenvector can be 1
1 4
8 −4
−8 , and
A+I =
4 8
8 −4
−4 , e2t
0 0
e2t . so our eigenvector can be 1
2 . Thus S=
and
S −1 = 1
det S 1 1
1 2 , 2 −1
−1 1 = 2 −1
−1 1 2 −1
−1 1 = −e−t + 2e3t
−2e−t + 2e3t . Thus
etA = 1
1 1
2 e3t
0 0
e −t 5 e−t − e3t
2e−t − e3t . ...

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