
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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I understand the entire solution for this question except for the choice of setting A as -1/3 why do we need it to be -1/3? Could you explain how to solve this problem and detail the reason why we do each step?

Transcribed Image Text:Problem 3 (restated for convenience): A causal system with input x(t) and output y(t) obeys the
differential equation
d²y dy
dt² dt
- 2y(t) =
dx
dt
+3x(t)
=
3(b) 10 points Assume that the output obeys the initial condition y(0) = 3, and the input is
x(t) eu(t) as in 5(a). Can you find an initial condition for the derivative of the output, y'(0-),
such that y(t) does not blow up as t gets large? If so, specify the value of the initial condition. If not,
say why not.
Zero input response
5(64(5) -4(0))-(0)-(sy (5) - 4(0))-2Y(=0
-ylo
5²-s-2) Y(St) = sy(o-) -y (σ`¯)+y' (0¯ )
Y(5) 2
35 - 3+ y' (0°)
35-3+4(0)
(5-2) (5+1)
A
=
B
+
5-2
5+1
A =
6-3+y' (0-)
=
1+5607/3
3
라
Need A = -1/3 to cancel out e²t term
in zero state response
| + 4 (0¯ ) = -1/2 =)) y ( 0 ) == 4
3
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