*This preview shows
page 1. Sign up
to
view the full content.*

**Unformatted text preview: **coeﬃcient matrix has a repeated eigenvalue if and only if
LC = 4R2 C 2 , or L = 4R2 C .
(b) We next consider the case when R = 1, C = 1, and L = 4. In this case,
L = 4R2 C , so the coeﬃcient matrix has a repeated eigenvalue (λ = − 1 ). In
2
order to ﬁnd I (t) and V (t), we need to solve the diﬀerential equation
λ=− 1
d I (t)
0
4
=
−1 −1
dt V (t) with the initial condition I (t)
V (t) I (0)
1
=
.
V (0)
2 Using the solution formula discussed in lecture, we obtain
I (t)
= eλt (1 − λt)
V (t)
t
t
= e− 2 (1 + )
2
t
1+t
= e− 2
.
2 − 2t I (0)
I (0)
+ tA
V (0)
V (0)
1
1
+t 2
−3
2 t t Therefore, I (t) = e− 2 (1 + t) and V (t) = e− 2 (2 − 2t).
Section 4.7.
Problem 3: We are going to look for a solution of the form
x(t) = u1 (t) x1 (t) + u2 (t) x2 (t) = u1 (t) e−t −1
1
.
+ u2 (t) et
1
0 In order for the function x(t) to be a solution of the inhomogeneous ODE
x′ (t) = −1 −2
e−t
,
x(t) +
t
0
1 it is necessary and suﬃcient that
u′1 (t) e−t e−t
−1
1
.
=
+ u′2 (t) et
t
1
0
2 This gives
u′1 (t) = tet + 1
and
u′2 (t) = te−t .
Therefore,...

View
Full
Document