1. No category

Sample Solutions of Assignment 8 for MATH3270A for ODE
November 7,2013
a
Contact Mr. XIAO Yao(yxiaomath.cuhk.edu.hk)
directly if you have any questions on the
solutions
Section 7.6
In each of problems express the general solution of the given system of equations in term of
real-valued functions. Also draw a direction field, sketch a few of the trajectories, and describe
the behavior of the solutions as t → ∞.
−1 1
0
2. x =
x
−4 −1
9
2
0
5
4. x =
x
5

 − 2 −1
−3 0 2
8. x0 =  1 −1 0  x
−2 −1 0
Answer: 2. The eigen-system of matrix is
1
r1 = −1 + 2i,
1
2i
ξ =
1
r2 = −1 − 2i, ξ 2 =
−2i
Hence the general solution of the system is
cos 2t
sin 2t
−t
x = c1
e + c2
e−t
−2 sin 2t
2 cos 2t
When t −→ ∞, x −→ 0.
The graph are in the last page.
4. The eigen-system of matrix is
1 3
−3 − 3i
1
r1 = + i, ξ =
5
2 2
1 3
−3 + 3i
2
r2 = − i, ξ =
5
2 2
1
2
Hence the general solution of the system is
t
t
3(cos 32 t + sin 32 t)
3(− cos 32 t + sin 32 t)
2
x = c1
e + c2
e2
3
3
5 cos 2 t
5 sin 2 t
8. The eigen-system of matrix is


2
r1 = −2,
ξ 1 =  −2 
1
√


−
2i
√
1√ 
r2 = −1 − 2i, ξ 2 = 
−1 − 2i
 √

2i
√

r3 = −1 + 2i, ξ 3 =  −1
√
1 + 2i
Hence the general solution of the system is
√
√
√
√






2 sin√ 2t
2 cos√ 2t
2
 + c3 e−t 
x = c1  −2  e−2t + c2 e−t 
sin 2 √ 
√
√ − cos√ 2t √
√
1
2 cos 2t + sin 2t
cos 2t + 2 sin 2t
It is easy to see that all solutions converge to the equilibrium point (0, 0, 0). We omit the graph.
10. Find the solution of the given initial value problem. Describe the behavior of the solution as
t −→ ∞.
−3 2
1
x=
x, x(0) =
.
−1 −1
2
−3 2
Answer: Since A =
, so the eigenvalues of this matrix are λ = −2 ± i. For
−1 −1
1−i
1
λ = −2 + i, the corresponding eigenvector is ξ =
. For λ = −2 − i, the corresponding
1
−1 − i
eigenvector is ξ 2 =
.
1
The general solution is
cos t + 2 sin t
sin t − cos t
−2t
x(t) = c1
e + c2
e−2t .
cos t
sin t
3
1
From x(0) =
, we can get c1 = 1, c2 = 1.
2
Hence the solution is
cos t + 3 sin t
x(t) =
e−2t .
2 cos t + sin t
x −→ 0, as t −→ ∞.
12.
0
x =
− 45 2
−1 65
x
.
(a).Find the eigenvalues of two given system.
(b).Choose an initial point (other that the origin) and draw the corresponding trajectory in the
x1 x2 -plane.
(c). For your trajectory in part(b) draw the graphs of x1 versus t and of x2 versus t.
(d). For your trajectory
draw the graphs in three dimensional tx1 x2 -space.
4in part(b)
−5 2
Answer: (a) A =
, then we have λ = 15 ± i.
−1 56
−1
−1
1
1
1
2
(b). For λ = 5 + i, we have ξ =
. For For λ = 5 − i, we have ξ =
. Hence
1+i
1−i
the general solution is
t
t
−
sin
t
− cos t
e5 .
x(t) = c1
e 5 + c2
cos t + sin t
cos t − sin t
0
If we choose x(0) =
, then we have c1 0, c2 = 1.
1
(c) and (d) are omitted.
15.
0
x =
2
α
−5 −2
α 10
−1 −4
x.
19.
0
x =
(a).Determine the eigenvalues in terms of α.
x.
4
(b).Find the critical value or values of α where the qualitative nature of the phase portrait for
the system changes.
(c). Draw a phase portrait for a value of α slightly below, and for another value slightly above,
each critical value.
Answer: 15. The characteristic equation for the system is given by
r2 − 4 + 5α = 0.
The roots are
√
r1,2 = ± 4 − 5α.
We can easily find the critical value of α is 45 .
Answer: 19. The characteristic equation for the system is given by
r2 + (4 − α)r + 10 − 4α = 0.
The roots are
α √ 2
± α + 8α − 24.
2
√
√
First note thats the roots are complex when −4 − 2 10 < α < −4 + 2 10. We also find that
√
when −4 − 2 10 < α < 2, the equilibrium point is a stable spiral. For the case α = 2, the
√
equilibrium is a center. When 2 < α < −4 + 2 10, the equilibrium point is an unstable spiral.
r1,2 = −2 +
For all other cases, the roots are real. When α > 2.5, the roots have opposite signs, with the
√
equilibrium point being a saddle. For the case −4 + 2 10 < α < 2.5, the roots are both positive,
√
and the equilibrium point is an unstable node. Finally, when α < −4 − 2 10, both roots are
negative, with the equilibrium point being a stable node.
23.

− 41 1
0
x0 =  −1 − 14 0  x
0
0 − 41

.
(a).Find the eigenvalues of two given system.
(b).Choose an initial point (other that the origin) and draw the corresponding trajectory in the
x1 x2 -plane. Also draw the trajectories in the x1 x3 − and x2 x3 − planes.
(c). For the initial point in part(b) draw the corresponding trajectories in the x1 x2 x3 -space.
5

− 41
Answer: A =  −1
0
(b) and (c) are similar

1
0
− 14 0 , hence the eigenvalues are λ = − 41 ± 21 i.
0 − 14
to problem 12.
28. A mass m on a spring with constant k satisfies the differential equation (see section 3.8)
mu00 + ku = 0
where u(t) is the displacement at time t of the mass from its equilibrium position.
a). Let x1 = u and x2 = u0 , Show that the resulting system is
0 1
0
x =
x
k
−m
0
b). Find the eigenvalues of the matrix for the system in part (a).
c). Sketch several trajectories of the system. Choose one of your trajectories and sketch the
corresponding graphs of x1 versus t and of x2 versus t. Sketch both graphs on one set of axes.
d). What is the relation between the eigenvalues of the coefficients of the coefficient matrix and
the nature frequency of the spring-mass system?
Answer: a). By direct computation.
b). The eigenvalues of the matrix is
r
r1 = −i
k
,
m
r
r2 = i
k
m
c). Omitted here.
d). According Section 3.8, the nature frequency is
q
k
m
= |r|.
29. Consider the two-mass, three- spring system of example 3 in the text. Instead of converting
the problem into a system of four first order equations we indicate here how to proceed directly
from equations (22).
6
(a). Show that Eqs (22) can be written in the form
−2 32
00
x =
x = Ax.
4
−3
3
(b). Assume that x = ξert and show that (A − r2 I)ξ = 0.
(c). Find the eigenvalue and eigenvector of A.
(e). Write down expressions for x1 and x2 .
(f). By differentiating the results from part (d), write down expressions for x01 and x02 .
Answer: Since x001 = −2x1 + 3x2 , x002 = 34 x1 − 3x2 . So we have
−2 32
00
x =
x = Ax.
4
−3
3
(b). Let x = ξert , then we have x0 = rξert , x00 = r2 ξert . From the equation, we have r2 ξert =
2
Aξert =⇒ (A
− r I)ξ = 0.
3
−2 − λ
2
= 0, we have λ1 = −1, λ2 = −4. That is r2 = −1, r2 = −4. For
(c). From 4
−3
−
λ
3
3
3
2
1
.
, for λ2 = −4, we can get ξ =
λ1 = −1, we can get ξ =
−4
2
(d). We can x1 = 3c1 cos t + 3c2 sin t + 3c3 cos 2t + 3c4 sin 2t,x2 = 2c1 cos t + 2c2 sin t − 4c3 cos 2t −
4c4 sin 2t.
(e). From (d), we can easily get x01 = −3c1 sin t+3c2 cos t−6c3 sin 2t+6c4 cos 2t, x02 = −2c1 sin t+
2c2 cos t + 8c3 sin 2t − 8c4 cos 2t
Section 7.8
Find the general solution of the given systems of equations and describe how the solution behave
as t −→ ∞.
3
1
1. x0 =
x
−4 −1
3
− 2 − 14
3. x0 =
x
1

 1 −2
1 1
1
5. x0 =  2 1 −1  x
0 −1 1
7
Answer: 1. The eigen-system of matrix is
1
r1 = r2 = 1, ξ =
1
−2
The generalized eigenvector is
η=
1
−1
Hence the fundamental set of solutions of the system is
1
t+1
1
t
2
x =
e, x =
et
−2
−t − 2
Hence the general solution of the system is
1
t+1
t
x = c1
e + c2
et
−2
−t − 2
3
− 2 − 14
x, so the eigenvalues are λ = −1. For λ = −1, we have
Answer: 3. A =
1
1
−
2
1
−1
1
1
ξ =
. For Aη = ξ , we can get that η =
. Hence the general solution is
−2
2
1
1
−1
−t
−t
−t
x(t) = c1 e
+ c2 te
+ c2 e
.
−2
−2
2
5. The eigen-system of matrix is



−3
0
r1 = −1, r2 = r3 = 2, ξ 1 =  4  , ξ 2 =  −1  ,
2
1

The generalized eigenvector is

−1
η= 1 
0

Hence the fundamental set of solutions of the system is






−3
0
−1
x1 =  4  e−t , x2 =  −1  e2t , x3 =  −t + 1  e2t
2
1
t
Hence the general solution of the system is






−3
0
−1
x = c1  4  e−t + c2  −1  e2t + c3  −t + 1  e2t
2
1
t
8
8. Find the solution of the given initial value problem. Draw the trajectory of the solution in
the x1 , x2 plane and also draw the graph of x1 versus t.
5 3 −2 2
3
0
x =
x, x(0) =
.
−3
− 23 12
Answer: . The characteristic equation is
r2 + 2r + 1 = 0,
so the roots is r1,2 = −1. The corresponding eigenvector is
1
1
ξ =
,
1
A second linearly independent solution is obtained by solving the system
3 3 η1
1
−2 2
.
=
3
3
η2
1
−2 2
Let η1 = k, we obtain that η2 = 23 + k. Hence the general solution is
0
1
1
−t
−t
x = c1
e + c2 [
te + 2 e−t ].
1
1
3
Imposing the initial conditions, we find that
c1 = 3, c2 = −9.
So the solution of IVP is
x=
3 − 9t
−3 − 9t
e−t .
11. Find the solution of the given initial value problem. Draw the trajectory of the solution in
the x1 , x2 , x3 space and also draw the

1
0

−4
x =
3
graph of x1 versus t.



0 0
−1
1 0  x, x(0) =  2  .
6 2
−30
Answer: . First, we 
can easily
 obtain the eigenvalues x1 = 1, r2 = 1, r3 = 2. For r1 = 1,
0
the eigenvalue is ξ 1 =  1 . Using the same method as in question 8, the second linearly
−6
9

  1 
 
0
−4
0
3





1
0 .
0
independent solution η = k
+
. For r3 = 2, the eigenvalue is ξ =
21
−6
1
−4
Hence the general solution is




 1 
 
0
0
−4
0
t
t
t







1
1
0
x = c1
e + c2 [
te +
e ] + c3 0  e2t .
21
−6
−6
1
−4
using the initial condition, we have
c1 = 2, c2 = 4, c3 = 3.
Hence the solution is
 



0
−1
0
t
t





2
1
e + 3 0  e2t .
te +
x=4
1
−33
−6

15. Show that all solutions of the system
0
x =
a b
c d
x
approach zero as t → ∞ if and only if a + d < 0 and ad − bc > 0. Compare this result with that
of Problem 38 in Section 3.5.
Answer: The eigenfunction of matrix is
r2 − (a + d)r + (ad − bc) = 0
a). If r1 + r2 = a + d < 0 and r1 r2 = ad − bc > 0, the real part of two eigenvalue are negative.
Then the general solution of the system has the form
c1 er1 t ξ 1 + c2 er2 t ξ 2
or c1 er1 t ξ 1 + c2 er1 t (tξ 1 + ξ 2 )
It is clear that the solutions approach zero since r1 < 0 and r2 < 0.
b). If all the solutions of the system approach zero, the real part of the eigenvalues must be
negative. Hence a + d = r1 + r2 = Re(r1 ) + Re(r2 ) < 0. If there are two real eigenvalues, then
ad − bc = r1 r2 > 0. If there are two conjugate complex eigenvalues, then ad − bc = r1 r2 = |r1 |2 =
Re(r1 )2 + Im(r2 )2 > 0.
10
17.Consider again the system
0
x = Ax =
1 −1
1 3
x
(i)
that we discussed in Example 2.We found there that A has a double eigenvalue r1 = r2 = 2
with a single independent eigenvector ξ (1) = (1, −1)T ,or any nonzero multiple thereof.Thus one
solution of the system (i) is x(1) = ξ (1) e2t and a second independent solution has the form
x(2) = ξte2t + ηe2t ,
where ξ and η satisfy
(A − 2I)ξ = 0, (A − 2I)η = ξ.
(ii)
In the text we solved the first equation for ξ and then the second equation for η.Here we ask you
to proceed in the reverse order.
(a)Show that η satisfies (A − 2I)2 η = 0.
(b)Show that (A − 2I)2 = 0.Thus the genralized eigenvector η can be chosen arbitrarily,except
that it must be independent of ξ (1) .
(c)Let η = (0, −1)T .Then determine ξ from the second of Eqs.(ii) and observe that ξ = (1, −1)T =
ξ (1) . This choice of η reproduces the solution found in Example 2.
(d)Let η = (1, 0)T and determine the corresponding eigenvector ξ.
(e)Let η = (k1 , k2 )T ,where k1 and k2 are arbitrary numbers.Then determine ξ.How is it related
to the eigenvector ξ (1) ?
Answer:
(a)Multiply (A − 2I) on both sides of Eqs.(ii).
(b)
−1 −1
1
1
−1 −1
1
1
=
0 0
0 0
11
(c)
ξ = (A − 2I)η
−1 −1
0
=
1
1
−1
1
=
−1
(d)
ξ = (A − 2I)η
−1 −1
1
=
1
1
0
−1
=
1
(e)
ξ = (A − 2I)η
−1 −1
k1
=
1
1
k2
−1
= (k1 + k2 )
1
It’s a multiple of ξ (1)


1 1 1
18. Consider the system x0 = Ax =  2 1 −1  x
(i).
−3 2 4
(a). Show that r = 2 is an eigenvalue of algebraic multiplicity 3 of the coefficient matrix A and
12
that there is only one corresponding eigenvector, namely,


0
ξ1 =  1  .
−1
(b). Using the information in part (a), write down one solution x1 (t) of the system (i).
(c). To find a second solution assume that x = ξte2t + ηe2t . Show ξ and η satisfy the equations
(A − 2I)ξ = 0,
(A − 2I)η = ξ.
2
(d). To find a third solution assume that x = ξ( t2 )e2t + ηte2t + ζe2t . Show that ξ, η, ζ satisfy
(A − 2I)ξ = 0,
(A − 2I)η = ξ,
(A − 2I)ζ = η.
(e). Write down a fundamental matrix Ψ for system (i).
(f). Form a matrix T with the eigenvector ξ 1 in the first column and the generalized eigenvector
−1
−1
η and ζ in the second and third
 find T and form the product J = T AT .
 columns. Then
1 1 1
Answer: (a). Since A =  2 1 −1 , so the eigenvalues are λ = 2. For λ = 2, the
−3 2 4


0
ξ 1 =  1 .
−1


0
(b). x1 = e2t  1 .
−1
2t
(c).Let x = ξte + ηe2t , then we have x0 = ξe2t + 2ξte2t + 2ηe2t = Aξte2t + Aηe2t . Hence
(A − 2I)ξ = 0, (A − 2I)η = ξ.
 


 
1
0
1
and we can compute that η =  1 . Sox2 = te2t  1  + e2t  1 .
0
−1
0
t2 2t
2t
2t
(d). Let x = ξ( 2 )e + ηte + ζe , using the same idea we can show that
(A − 2I)ξ = 0,
(A − 2I)η = ξ,
(A − 2I)ζ = η.
13




 
 
2
0
1
2
t2 2t 
3
2t 
2t 




0 , and x = 2 e
1
1
0 . So
And we can compute that ζ =
+ te
+e
3
−1
0
3
the fundamental matrix is


0
1
t+2
2
Ψ(t) = e2t  1 t + 1 t2 + t  .
2
−1 −t − t2 + 3






0 1 2
−3 3
2
2 1 0
T =  1 1 0 , T −1 =  3 −2 −2 ,J =  0 2 1 .
−1 0 3
−1 1
1
0 0 2
19.Cosider the sytstem

5 −3 −2
x0 = Ax =  8 −5 −4  x.
−4 3
3

(i)
(a)Show that r = 1 is a triple eigenvalue of the coefficient matrix A and that there are only two
linearly independent eigenvectors,which we may take as

 

1
0
ξ (1) =  0  ξ (2) =  2  . (ii)
2
−3
Write down two linearly independent solutions x(1) (t) and x(2) (t) of Eqs.(i).
(b)To find a third solution,assume that x = ξtet + ηet ;then show that ξ and η must satisfy
(A − I)ξ = 0,
(A − I)η = ξ.
(iii)
(iv)
(c)Equation (iii) is satisfied if ξ is an eigenvector,so one way to proceed is to choose ξ to be
a suitable linear combination of ξ (1) and ξ (2) so that Eq.(iv) is solvable,and then to solve that
equation for η.However,let us proceed in a different way and follow the pattern of Problem 17.
First,show that η satisfies
(A − I)2 η = 0.
Further,show that (A − I)2 = 0.Thus η can be chosen arbitrarily,except that it must be independent of ξ (1) and ξ (2) .
14
(d)A convenient choice for η is η =
0, 0, 1
T
.Find the corresponding ξ from Eq.(iv).Verify that
ξ is an eigenvector.
(e)Write down a fundamental matrix Ψ(t) for the system (i).
(f)Form a matrix T with the eigenvector ξ (1) in the first column and with the eigenvector ξ
from part (d) and the generalized eigenvector η in the other two columns.Find T−1 and form the
product J = T−1 AT. The matrix T is the Jordan form of A.
Answer:
(a)

5−λ
−3
−2
−5 − λ −4 
|A − λI| = det  8
−4
3
3−λ

= (1 − λ)3
So λ = 1 is a triple eigenvalue of the coefficient matrix.Suppose the eigenvector is (x1 , x2 , x3 )T ,then

 


x1
x1
5 −3 −2
 8 −5 −4   x2  =  x2 
x3
x3
−4 3
3

 4x1 − 3x2 − 2x3 = 0
8x1 − 6x2 − 4x3 = 0

−4x1 + 3x2 + 2x3 = 0
The coefficient matrix of this system admits a order of 1,thus there are only two independent
eigenvectors,by caculating:

1
ξ (1) =  0  ,
2
 
1
x(1) (t) = et  0 
2


0
ξ (2) =  2 
−3


0
x(2) (t) = et  2 
−3

(b)ξ is the eigenvector corresponding to eigenvalue 1,thus Eqs.(iii) is obvious.Substitute the
expression of η into the equation and we get (iv). Take ξ = x1 , x2 , x3 ,suppose η = (y1 , y2 , y3 )T
thus we have


 

x1
x1
5 −3 −2
 8 −5 −4   x2  =  x2 
−4 3
3
x3
x3
15


 

4 −3 −2
y1
x1
 8 −6 −4   y2  =  x2 
−4 3
2
y3
x3

 4x1 − 3x2 − 2x3 = 0
8x1 − 6x2 − 4x3 = 0

−4x1 + 3x2 + 2x3 = 0

 4y1 − 3y2 − 2y3 = x1
8y1 − 6y2 − 4y3 = x2
 −4y + 3y + 2y = x
1
2
3
3
Finally,ξ = (−2, −4, 2)T and η = (0, 0, 1)T so the third solution is:


 
−2
0
(3)
t



−4 te +
0  et
x (t) =
2
1
(c)Multiply (A − I) on both sides of Eqs.(iii),substitute it into (iv) and we get that. Secondly

 

0 0 0
4 −3 −2
4 −3 −2
 8 −6 −4   8 −6 −4  =  0 0 0 
0 0 0
−4 3
2
−4 3
2

Thus η can be chosen arbitrarily.
(d)

  
x1
0
4 −3 −2
 8 −6 −4   0  =  x2 
−4 3
2
1
x3


−2
⇒
ξ =  −4 
2

(e)


1 0
−2t
−4t 
Ψ(t) = et  0 2
2 −3 2t + 1
or


1 −2 −2t
Ψ(t) = et  0 −4 −4t 
2 2 2t + 1
16
(f)


1 −2 0
T =  0 −4 0  ,
2 2 1


1 −1/2 0
T−1 =  0 −1/4 0  ,
−2 3/2 1


1 0 0
J =  0 1 0 .
0 0 1
λ 1
20. Let J =
, where λ is an arbitrary real number.
0 λ
(a). Find J 2 , J 3 and J 4 .
n
λ nλn−1
n
.
(b). Use an inductive argument to show that J =
0
λn
(c). Determine exp(Jt).
(d). Use exp Jt to solve the initial value problem x0 = Jx, x(0) = x0 .
4
3
λ 4λ3
λ 3λ2
λ2 2λ
4
3
.
,J =
,J =
Answer: . (a). J =
0 λ4
0 λ3
0 λ2
(b). n = 1, it holds. Suppose n = k holds, then for n = k + 1, we have
k+1
λ
(k + 1)λk
k+1
k
J
=J J =
.
0
λk+1
2
(c). exp(Jt) =
P
J n tn
n=0 n!
(d). x(t) = exp(Jt)x0 .
P
=
λn tn
n=0 n!
0
λn−1 tn
n!
λn tn
n=0 n!
P
Pn=0
=
eλt teλt
0 eλt
.
Section 7.9
In each of problem find the general solution of the given the system of equations.
17
t 2 −1
e
1. x =
x+
3 −2
t
2
1
− cos t
0
3. x =
x+
−5 −2
sin t
−3 4
8
t
5. x0 =
x+
−t−2
−2 −4
2
1 4
0
t
7. x =
x+e
−1
1 1
2
1
csc t
0
12. x =
x+
,
−5 −2
sec t
0
π
2
<t<π
Answer: 1. The eigen-system of matrix is
1
−3
1
−1
1
r1 = 1, ξ =
2
r2 = −1, ξ =
Hence general solution of the homogeneous system is
−3
1
t
−t
x = c1 e
+ c2 e
1
−1
1
0
t
. Since g(t) =
e+
t, we can assume that the particular solution is y = atet +bt+c+
0
1
1 3 −3
0
−4
t
2
,b =
,c =
,d =
.
de . using the equation we can get that a =
1
1
2
−1
−2
4
Hence the solution is
3 1
1
−3
0
1
1
t
−t
t
2
x = c1
e + c2
te +
t−
et .
e +
−
1
3
2
−1
− 12
4 −1
3. The eigen-system of matrix is
1
5
2−i
5
2+i
r1 = i, ξ =
2
r2 = −i, ξ =
Hence general solution of the homogeneous system is
−2 cos t + sin t
−2 sin t − cos t
+ c2
.
x = c1
cos t
sin t
18
Since g(t) =
−1
0
cos t +
0
1
sin t, we can assume that the particular solution is y =
1 −1
−2
(at + b) cos t + (ct + d) sin t. using the equation we can get that a =
,b =
,c =
1
2
1
−1
,d =
. Hence the solution is
3
− 52
1 1
−2 cos t + sin t
−2 sin t − cos t
−1
−1
−2
cos t+
x = c1
+c2
+
t cos t+
t sin t+
cos t
sin t
1
3
2
− 25
5. The eigen-system of matrix is
r1 = r2 = 0,
Hence general solution of the homogeneous system is
1 −2
−2
−2
u = c1
+ c2 t
+
1
1
0
Then a particular solution is
Z t
Ψ−1 (s)g(s)ds
x = Ψ(t)c + Ψ(t)
1 1 −2
8
2
−2
−2
−2
−1
= c1
+ c2 t
−
ln t +
t +
t−2
+
1
−4
0
1
0
0
7. The eigen-system of matrix is
−2
r1 = −1, ξ =
1
2
r2 = 3, ξ 2 =
.
1
1
t
We
assume
that the particular solution is y = ae . Using the equation we can get that a =
2
1
,. Hence the solution is
2
−1
1
2
−2
2
3t
−t
x = c1
e + c2
e +
et .
1
1
2 −1
12. The eigen-system of matrix is
1
−1
2−i
−1
2+i
r1 = i, ξ =
2
r2 = −i, ξ =
19
Hence general solution of the homogeneous system is
− cos t
− sin t
x = c1
+ c2
.
2 cos t + sin t
− cos t + 2 sin t
csc t
Since g(t) =
, we can assume that the solution is y = u(t)Ψ(t). And Ψ−1 (t) =
sec t
− cos t + 2 sin t − sin t − 2 cos t
. From u0 (t) = Ψ−1 (t)g(t), we can get
sin t
− cos t
(−2t + ln(cos t) + ln(sin t) + c1 ) cos t + (2t + 2c1 + c2 + 2 ln(sin t)) sin t
x=
(6t − 2 ln(cos t) + c2 ) cos t − (2t + 5c1 + 2c2 + ln(cos t) + t ln(sin t)) sin t
1.Find the general solution of the following system
0
2 −5
1 −2
(a)x =
0
(b)x =
x+
2 −5
1 −2
− cos t
sin t
x+
csc t
sec t
t e
2 −1
x+
(c)x =
t
3 −2
(change the second entry of the rightest term to be t)
0
4 −2
(d)x =
8 −4
(change the 2-2 entry of the matrix to be −4)
0
x+
t−3
−t−2
Answer:
(a)The eigen-system of matrix is
1
5
2−i
5
2+i
r1 = i, ξ =
2
r2 = −i, ξ =
20
Hence general solution of the homogeneous system is
5 cos t
5 sin t
x = c1
+ c2
.
2 cos t + sin t
− cos t + 2 sin t
− cos t
Since g(t) =
, we can assume that the solution is y = Ψ(t)u(t). And
sin t
1 cos t − 2 sin t 5 sin t
−1
Ψ (t) =
5 2 cos t + sin t −5 cos t
From u0 (t) = Ψ−1 (t)g(t), we can get
Z
t
Ψ−1 (s)g(s)ds.
x = cΨ(t) + Ψ(t)
Or in the other way around ,set the particular solution to be xp = a cos t + b sin t and substitute
it into the equation we get:
1
a=
3
4
1
1
,b =
3
0
2
Thus the general solution is:
1 0
1 4
5 cos t
5 sin t
X(t) = c1
+ c2
cos t +
sin t
+
2 cos t + sin t
− cos t + 2 sin t
3 1
3 2
(b)The eigen-system of matrix is
r1 = i, ξ 1 =
2
r2 = −i, ξ =
5
2−i
5
2+i
Hence general solution of the homogeneous system is
5 cos t
5 sin t
x = c1
+ c2
.
2 cos t + sin t
− cos t + 2 sin t
csc t
Since g(t) =
, we can assume that the solution is y = Ψ(t)u(t). And Ψ−1 (t) =
sec t
cos t − 2 sin t 5 sin t
1
. From u0 (t) = Ψ−1 (t)g(t), we can get
5
2 cos t + sin t −5 cos t
1
2
2
4
5 cos t
5 sin t
x = [ ln(sin t)−ln(− cos t)− t+c1 ]
+[ ln(sin t)− t+c2 ]
.
2 cos t + sin t
− cos t + 2 sin t
5
5
5
5
21
(c) The eigen-system of matrix is
1
1
1
1
3
r1 = 1, ξ =
2
r2 = −1, ξ =
Hence general solution of the homogeneous system is
1
1
−t
t
+ c2 e
x = c1 e
3
1
1
0
t
. Since g(t) =
e+
t, we can assume that the particular solution is y = atet +bt+c+
0
1
1
1
0
1
3
1
t
de . using the equation we can get that a = 2
,b =
,c = −
, d = −4
.
1
2
1
3
Hence the solution is
1 1
3 1
1
0
1
1
t
t
−t
te +
t−
−
et .
x = c1
e + c2
e +
1
2
1
3
1
3
2
4
(d)The eigen-system of matrix is
r1 = r2 = 0, ξ 1 =
1
2
The generalized eigenvector is
η=
0
− 12
Hence general solution of the homogeneous system is
0
1
1
u = c1
+ c2 t
+
2
2
− 21
The fundamental matrix is
Ψ(t) =
1
t
2 2t −
1
2
Hence the fundamental matrix such that Φ(0) = I is
1 + 4t −2t
Φ(t) = Ψ(t)Ψ(0) =
8t
1 − 4t
22
Then a specular solution is
Z
t
Ψ−1 (s)g(s)ds
−3 Z t
1 − 4s 2s
s
ds
= Ψ(t)c + Ψ(t)
−s−2
4
−2
Z t −3
s − 4s−2 − 2s−1
ds
= Ψ(t)c + Ψ(t)
4s−3 + 2s−2
1 0
1
1
2
2
−1
= c1
+ c2 t
+
−
ln t +
t − 2 t−2
2
2
4
5
− 21
0
x = Ψ(t)c + Ψ(t)