Solutions for Chapter 4
Section 4.1
(18) Sol: Note that the frequencey in cos t is ! = 1:Set
cos t
sin t = A cos (t
)
Dierentiate both sides,
sin t
cos t =
A sin (t
)
Let t = 0; we nd
1 = A cos
1 = A sin
So A =
p
1+1=
p
2;
= 5 =4;and
cos t
sin t =
p
2 cos
Solutions for Chapter 6
Section 6.1
(4) Sol: x01 = x2 ; x02 = x3 ; x03 = 2x1 + x2 + 3x3 + sin t
e2t
e3t
~
(6) Sol: X (t) = c1 3t + c2
2e2t
e
(18) Sol: W [~ 1 ; ~ 2 ] =
x x
2e t
=
3e t
e3t
e3t
5e2t 6= 0
cos 3t sin 3t
= sin2 3t + cos2 3t = 1 6= 0
sin 3t cos
Solutions for Chapter 1
Section 1.2
0
(2) LHS = y0 = (3t + t2 ) = 3 + 2t
1
1
RHS = y + t = (3t + t2 ) + t = 3 + t + t = 3 + 2t
t
t
* LHS
RHS
) y = 3t + t2 is a solution
(10) y 0 =
p
t
; the general solution y = c
y
t2 :
(14) Set y (y + 1) = 0;we nd y = 0;
Solutions for Chapter 3
Section 3.1
(8) (DC)T =
1 6
1 7
1 1
; B =
1 0
(24) Counterexample: A =
(A + B) (A
2
0
1
: Then A2
1
B 2 6=
B) ; because
(A + B) (A
B) =
A2
B2 =
3 6
;
0 1
2 1
1 1
4
0
3
=
1
2 4
1 0
(42) (AB)T = B T AT . This identity is not so obvio