Practice Problems for Section 7.9: Solutions






cos 2t 
1
0 2 
] 1 Solve x0 = Ax + 
, x(0) =   , A = 
.
− sin 2t
0
−2 0
¯
¯
¯ −r
2
(i) Find eAt : |A − rI| = ¯¯
¯ −2 −r


¯
¯
¯
¯ = r 2 + 4 = 0 ⇒ r = ±2i.
¯
¯






−2i
2 
1
1
0
A − 2iI = 
⇒ eigenvector ξ =   =   + i   = a + ib.
−2 −2i
i
0
1




cos 2t 
sin 2t 
u(t) = a cos 2t − b sin 2t = 
, v(t) = a sin 2t + b cos 2t = 
.
− sin 2t
cos 2t


cos 2t sin 2t 
Hence Ψ(t) = [u(t), v(t)] = 
. Since Ψ(0) = I ⇒ Ψ(t) = eAt .
− sin 2t cos 2t





cos 2t sin 2t   1   cos 2t 
(ii) Homogeneous part of solution: eAt x0 = 
=
.
− sin 2t cos 2t
0
− sin 2t
(iii) Nonhomogeneous part of solution:







R
t
cos 2s − sin 2s   cos 2s   1 
⇒ 0t e−As f (s)ds =   ,
=
e−As f (s) = 
0
0
− sin 2s
sin 2s
cos 2s

R
⇒ eAt 0t e−As f (s)ds = 




cos 2t sin 2t   t   t cos 2t 
.
=
−t sin 2t
0
− sin 2t cos 2t
(iv) Full solution:
At
At
x(t) = e x0 + e
Z t
0

−As
e




cos 2t 
f (s)ds = (1 + t) 
.
− sin 2t



2t cos 2t 
1
0 2 
.
] 2 Solve x0 = Ax + 
, x(0) =   , A = 
−2 0
−2t sin 2t
0




cos 2t 
cos 2t sin 2t  At
, e x0 = 
.
A and x0 are as in ] 1 ⇒ eAt = 
− sin 2t cos 2t
− sin 2t
Compute nonhomogeneous part:







R
cos 2s − sin 2s   2s cos 2s   2s 
t2 
,
e−As f (s) = 
=
⇒ 0t e−As f (s)ds = 
0
sin 2s
cos 2s
−2s sin 2s
0

R
t
⇒ eAt 0 e−As f (s)ds = 




cos 2t sin 2t   t2   t2 cos 2t 
=
.
0
−t2 sin 2t
− sin 2t cos 2t
⇒ x(t) = eAt x0 + eAt
Z t
0


cos 2t 
e−As f (s)ds = (1 + t2 ) 
.
− sin 2t
1






36t2 
0
2 −4 
] 3 Solve x0 = Ax + 
, x(0) =   , A = 
.
6t
1
1 −2
¯
¯
¯ 2−r
At
(i) Find e : |A − rI| = ¯¯
¯
1


Verify A2 = 
−4
−2 − r
¯
¯
¯
¯ = r 2 = 0 ⇒ double eva r = 0.
¯
¯


0 0 
1 + 2t −4t 
⇒ eAt = I + At = 
.
0 0
t
1 − 2t

(ii) Homogeneous part of solution: eAt x0 = 
(iii) Nonhomogeneous part of solution:







1 + 2t −4t   0   −4t 
=
.
t
1 − 2t
1
1 − 2t


1 − 2s
4s   36s2   60s2 − 72s3

e−As f (s) = 
=
−s
1 + 2s
6s
6s + 12s2 − 36s3

R t −As
⇒ 0 e f (s)ds = 

20t3 − 18t4 
3t2 + 4t3 − 9t4




3
4
3
4
2
3
4
R
8t
+
6t
(1
+
2t)(20t
−
18t
)
−
4t(3t
+
4t
−
9t
)
t

=
⇒ eAt 0 e−As f (s)ds = 
3t2 − 2t3 + 3t4
t(20t3 − 18t4 ) + (1 − 2t)(3t2 + 4t3 − 9t4 )
(iv) Final solution:
x(t) = eAt x0 + eAt


Z t
0

e−As f (s)ds = 


3


(i) Find eAt : Matrix is upper triangular ⇒ triple eva r = 1.



0 2 3
0 0 4




2
3



Set B = A − I = 
 0 0 2  ⇒ B =  0 0 0  , B = 0,
0 0 0
0 0 0

hence eAt = et eBt

1 2t 3t + 2t2


.
= et (I + Bt + B 2 t2 /2) = et 
2t
 0 1

0 0
1
(ii) Compute solution:





1 −2s −3s + 2s2
0
−18s + 12s2






 0  = 

e−As f (s) = e−s 
1
−2s
−12s
 0




s
0 0
1
6
6e
2

−4t + 8t + 6t

1 − 2t + 3t2 − 2t3 + 3t4
0
0
1 2 3






0





] 4 Solve x = Ax +  0  , x(0) =  0  , A =  0 1 2  .
6et
0
0 0 1

4


−9t2 + 4t3


R t −As
, hence
⇒ 0 e f (s)ds = 
−6t2


6t

x(t) = eAt
Z t
0


1 2t 3t + 2t2
−9t2 + 4t3






e−As f (s)ds = et 
2t
−6t2
 0 1


0 0
1
6t




(−9t2 + 4t3 ) − 12t3 + 6t(3t + 2t2 )
9t2 + 4t3





 = et 
= et 
6t2
−6t2 + 12t2




6t
6t






100t 
0
6 −7 
] 5 Solve x0 = Ax + 
, x(0) =   , A = 
.
−50t
0
1 −2
>> syms t s;A=sym([6 -7;1 -2]);f=[100*s;-50*s];
>> simplify(int(expm(-A*s)*f,0,t));
>> x=simplify(expm(A*t)*ans)
x =
[ -exp(5*t)*(110*t*exp(-5*t)-68*exp(-5*t)-7+75*exp(-6*t))]
[ -exp(5*t)*(80*t*exp(-5*t)-74*exp(-5*t)+75*exp(-6*t)-1)]






2 −5 
0
4t
.
] 6 Solve x0 = Ax +   , x(0) =   , A = 
1
0
1 −2
>> syms t s;A=sym([2 -5;1 -2]);f=[4*s;1];
>> simplify(int(expm(-A*s)*f,0,t));
>> x=simplify(expm(A*t)*ans)
x =
[
cos(t)-8*sin(t)+8*t-1]
[ -3*sin(t)+2*cos(t)+4*t-2]






4 cos t 
3
2 −5 
] 7 Solve x0 = Ax + 
, x(0) =   , A = 
.
6 sin t
5
1 −2
3
>> syms t s;A=sym([2 -5;1 -2]);f=[4*cos(s);6*sin(s)];
>> simplify(int(expm(-A*s)*f,0,t));
>> x=simplify(expm(A*t)*([3;5]+ans))
x =
[ 3*cos(t)+17*t*cos(t)-32*sin(t)+4*sin(t)*t]
[ -13*sin(t)+5*sin(t)*t+5*cos(t)+6*t*cos(t)]






e−t cos 2t
0
−1 0
0






0
−t





] 8 Solve x = Ax +  e sin 2t  , x(0) =  0  , A =  3 1 −2  .
0
0
3 2
2
>> syms t s;A=sym([-1 0 0;3 1 -2;3 2 1]),f=[exp(-s)*cos(2*s);exp(-s)*sin(2*s);0];
>> x=simplify(factor(simple(expm(A*t)*int(expm(-A*s)*f,0,t))))
x =
[
1/2*sin(2*t)*exp(-t)]
[ -1/20*(18*exp(-2*t)*sin(2*t)+11*exp(-2*t)*cos(2*t)-11*cos(2*t)-7*sin(2*t))*exp(t)]
[
-1/20*exp(t)*(4*exp(-2*t)*sin(2*t)-7*exp(-2*t)*cos(2*t)-11*sin(2*t)+7*cos(2*t))]
]9




0
2 −1 
(a) Find a particular solution to x0 = Ax + t   , where A = 
,
1
3 −2
via the method of undetermined coefficients.
Eigenvalues of A: (2 − r)(−2 − r) + 3 = r2 − 1 = 0 ⇒ r = ±1. Since 0 is not an eigenvalue, the
form of the particular solution is xp (t) = at + b. Sub this in ODE system:




0
0
x0p = a = Axp + t   = Aat + Ab + t   .
1
1
From coefficients of t0 ⇒ a = Ab. 








0
0
2 −1   0   1 
From coefficients of t1 ⇒ 0 = Aa +   ⇒ a = −A−1   = − 
=
.
1
1
3 −2
1
2
4





2 −1   1   0 
Then b = A−1 a = 
=
, hence
3 −2
2
−1


t
.
xp (t) = at + b = 
2t − 1




1
2 −1 
(b) Find a particular solution to x0 = Ax + et   , where A = 
,
0
3 −2
via the method of undetermined coefficients.
Matrix is the same as in (a). Since 1 is an eigenvalue of A, the form of the particular solution is
xp (t) = et (at + b). Sub this in ODE system:




1
1
x0p = et (at + a + b) = Axp + et   = et (Aat + Ab +  ).
0
0
From coefficients of t1 et ⇒ a = Aa (∗)




1
1
From coefficients of t0 et ⇒ a + b = Ab +   ⇒ (A − I)b = a −   (∗∗)
0
0


1
(∗) ⇒ a is an eigenvector of A for r = 1. It is easy to see that   is such an eigenvector,
1


1
hence a = ξ   with ξ to be determined. An equation for ξ follows from the requirement that
0


1
(∗∗) has a solution. This is equivalent to yT (a −  ) = 0 for any solution of (A − I)T y = 0.
0


3 
An eigenvector of AT for the eigenvalue 1 is y = 
, thus the condition for (∗∗) to have a
−1
solution is










ξ
1
3/2 
1
1/2 
[3, −1](  −  ) = 2ξ − 3 = 0 ⇒ ξ = 3/2 ⇒ a = 
and a −   = 
.
ξ
0
3/2
0
3/2
Since it is now
that (∗∗) has a solution, we only need to evaluate the first component.



 guaranteed
1/2
b
1
, this yields b1 − b2 = 1/2. We choose b2 = 0 ⇒ b1 = 1/2, hence b = 
.
Writing b = 
0
b2
The particular solution then becomes


et 3t + 1 
xp (t) = et (at + b) = 
.
2
3t
5