Solutions of HW11
1. We have f (t) = 1−uπ/2 (t) for t ≥ 0. Then Laplace transform of the equation
is
π
1 − e− 2 s
2
s Y −1+Y =
,
s
where Y is the Laplacetransform of y.
−πs
−πs
2
We have then Y = 1−es 2 + 1 /(s2 + 1) = 1+s−e
s(s2 +1) .
1+s
Let us find the inverse Laplace transform of s(s1+s
2 +1) . We write s(s2 +1) =
Bs+C
A
2
2
s + s2 +1 . Then 1 + s = As + A + Bs + Cs, hence A = −B, A = 1, C = 1,
1
−s+1
and we get s(s1+s
2 +1) = s + s2 +1 and its inverse Laplace transform is 1−cos t+sin t.
Let us find the inverse Laplace transform of s(s21+1) = As + Bs+C
s2 +1 . Then
2
2
1 = As + A + Bs + Cs, hence A = −B, C = 0, and A = 1. We get s(s21+1) =
1
s
s − s2 +1 . It inverse Laplace transform is 1−cos t. It follows that inverse Laplace
πs
transform of s(se22+1) is uπ/2 (t)(1 − cos(t − π/2)) = uπ/2 (t)(1 − sin t).
We get then
y(t) = 1 − cos t + sin t − uπ/2 (t)(1 − sin t).
It is equal to 1 − cos t + sin t for 0 ≤ t ≤ π/2 and to 1 − cos t + sin t − 1 + sin t =
2 sin t − cos t for t ≥ π/2.
2. Applying Laplace transform, we get s2 Y − s + 2sY − 2 + 2Y = e−πs , hence
−πs
Y = s+2+e
s2 +2s+2 .
s+2
s+1+1
−t
Inverse Laplace transform of s2 +2s+2
= (s+1)
(cos t + sin t). Inverse
2 +1 is e
1
1
−t
Laplace transform of s2 +2s+2 = (s+1)2 +1 is e sin t, hence inverse Laplace
−πs
transform of s2e+2s+2 is uπ (t)e−(t−π) sin(t − π) = −uπ (t)e−t+π sin t.
It follows that y = e−t (cos t + sin t) − uπ (t)e−t+π sin t.
3. Applying the Laplace transform, we get
s2 Y + 2sY + 2Y =
π
s
+ e− 2 s .
s2 + 1
π
We get Y =
s
e− 2 s
(s2 +1)(s2 +2s+2) + s2 +2s+2 .
s
As+B
Cs+D
(s2 +1)(s2 +2s+2) = s2 +1 + (s+1)2 +1
3
2
3
2
2
+2Bs+2B+Cs
We have
= As +2As +2As+Bs
(s2 +1)((s+1)2 +1)
then (A + C)s + (2A + B + D)s + (2A + 2B + C)s + (2B + D) = s, hence

A+C = 0



2A + B + D = 0
2A + 2B + C = 1



2B + D = 0
3
Substituting the last equation D = −2B into the second we get 2A + B −
2B = 0, hence 2A = B. Substituting C = −A (from the first equation) and
1
+Cs+Ds2 +D
,
2A = B into the third equation, we get 2A + 4A − A = 1, hence 5A = 1,
A = 1/5, B = 2/5, C = −1/5, D = −4/5, and hence
s
(s2
+
1)(s2
+ 2s + 2)
=
1
2
5s + 5
2
s +1
+
1
− 15 s − 45
s+ 2
− 1 (s + 1) − 35
= 52 5 + 5
.
+ 2s + 2
s +1
(s + 1)2 + 1
s2
Consequently, the inverse Laplace transform of this function is 51 cos t + 25 sin t −
1 −t
cos t − 53 e−t sin t.
5e
1
1
−t
= (s+1)
sin t. Consequently,
Inverse Laplace transform of s2 +2s+2
2 +1 is e
−πs
2
the inverse Laplace transform of s2e+2s+2
is uπ/2 (t)e−(t−π/2) sin(t − π/2) =
−uπ/2 (t)e−(t−π/2) cos t.
Thus, solution of the equation is
y=
1
2
1
3
cos t + sin t − e−t cos t − e−t sin t − uπ/2 (t)e−(t−π/2) cos t.
5
5
5
5
4.
The matrix of the system is

1
 0
0
4
3
2

4
2 .
3
Its characteristic polynomial is (1 − λ)(λ2 − 6λ + 5) = (1 − λ)(λ − 1)(λ − 5).
The eigenvalues are λ = 1 (twice) and λ = 5.
For λ = 1 the eigenvectors are solutions of the homogeneous system of
algebraic equations with matrix


0 4 4
 0 2 2 .
0 2 2
The set of solutions of this system described by the condition x2 = −x3 . It
follows that (1, 0, 0) and (0, −1, 1) are eigenvectors.
For λ = 5 the corresponding system has matrix


−4 4
4
 0 −2 2  .
0
2 −2
A solution is (0, 1, 1). It follows that the general solution of the system is






1
0
0
~x = c1 et  0  + c2 et  −1  + c3  1  .
0
1
1
5. There was a typo in the problem. The second equation is x02 = −6x1 − 3x3 .
2

−4
Matrix of the system is  −6
0
2
0
8/3

−1
−3 . The characteristic polynomial
−2
is
−4 − λ
−6
0
2
−λ
8/3
−1
−3
−2 − λ
=
(−4 − λ)(−λ)(−2 − λ) + 16 − 12(2 + λ) − 8(4 + λ) =
− λ3 − 6λ2 − 8λ + 16 − 24 − 12λ − 32 − 8λ =
− λ3 − 6λ2 − 28λ − 40.
Substituting λ = −2 we get 8 − 6 · 4 + 28 · 2 − 40 = 8 − 24 + 56 − 40 = 0. Dividing
the characteristic polynomial by λ + 2, we get
−λ2 (λ + 2) − 4λ(λ + 2) − 20(λ + 2) = −(λ + 2)(λ2 + 4λ + 20).
√
Roots of the polynomial λ2 + 4λ + 20 are λ = −4± 216−80 = −2 ± 4i.
Let us find an eigenvector associated with the eigenvalue λ = −2. Subtracting it from the diagonal, we get




−2 2 −1
−2
2 −1
−2 2 −1
−2 0 −1
 −6
−4 0  →
2 −3  →  0
→
.
0 1 0
0 1 0
0 8/3 0
0 8/3 0

1
Hence an eigenvector is  0 .
−2
Let us find an eigenvector associated with the eigenvalue −2+4i. Subtracting
it from the diagonal, we get the matrix


−2 − 4i
2
−1

−6
2 − 4i −3  .
0
8/3
−4i

Divide the second row by −6 and put it on the first place:


1
−1/3 + 2/3i 1/2
 −2 − 4i
2
−1 
0
8/3
−4i
Multiply the first row by 2 + 4i and add it to the second row:


1 −1/3 + 2/3i 1/2
 0
−4/3
2i  .
0
8/3
−4i
3
The last two rows are proportional (the third is −2 times the second). Multiplying the second row by −3/4 we get
1 −1/3 + 2/3i
1/2
.
0
1
−3i/2
Let us multiply the second row by 1/3 − 2/3i and add to the first row:
1 0 −1/2 − i/2
.
0 1
−3i/2


1/2 + i/2
.
3i/2
It follows that an eigenvector is 
1
The corresponding complex solution of the differential equation is



 
 
1/2 + i/2
1/2
1/2
 = e−2t (cos 4t + i sin 4t)  0  +  3/2  i .
3i/2
e(−2+4i)t 
1
1
0
Its real and imaginary parts are







(cos 4t − sin 4t)/2
1/2
1/2
,
− 23 sin 4t
e−2t cos 4t  0  − sin 4t  3/2  = e−2t 
0
1
cos 4t
and





(cos 4t + sin 4t)/2
1/2
1/2
3
.
e−2t cos 4t  3/2  + sin 4t  0  = e−2t 
2 cos 4t
1
0
sin 4t


The general solution of the system is


 



(cos 4t − sin 4t)/2
(cos 4t + sin 4t)/2
1
3
 + c3 

− 23 sin 4t
e−2t c1  0  + c2 
2 cos 4t
−2
cos 4t
sin 4t
6. The characteristic polynomial of the system is λ2 − 6λ + 8, hence the eigenvalues are λ = 2 and λ = 4.
for the eigenvalue 2 are solutions of the system with matrix
The eigenvectors
3 −1
, e.g. (1, 3).
3 −1
1 −1
The eigenvectors for λ = 4 are solutions of the system with matrix
,
3 −3
e.g. (1, 1).
It follows that a fundamental system of solutions of the system of differential
equations is
2t 4t e
e
, ~x2 =
.
~x1 =
3e2t
e4t
4
Let
e2t e4t
,
3e2t e4t
1 1
1/2 −3/2
−1
, and X (0) =
, so that the fundaThen X(0) =
3 1
−1/2 1/2
mental matrix eAt of the system is
!
2t
−3e2t +e4t
e2t −e4t
e
e4t
1/2 −3/2
−1
2
2
.
X(t)X (0) =
=
3e2t −e4t
−9e2t +e4t
3e2t e4t
−1/2 1/2
X(t) =
2
5
2