Question 13 dx Given: = 4x - 2y + 2 U(t-1) dt dy = 3x -y + U(t-1) dt x(0) = 0, y(0) = 1/ The answer for x is Ox=et-e²t + [-et-² + e²(t-2)] (t - 2) x = et-e²t + [-et-¹ + e²(t-1)] U(t − 1) 2t - - Ox=et+e²t + [-et-1 + e²(t-²)] U(t − 2) Ox=et + e²t + [-et-¹ + e²(t-¹)] U(t − 1)
Question 13 dx Given: = 4x - 2y + 2 U(t-1) dt dy = 3x -y + U(t-1) dt x(0) = 0, y(0) = 1/ The answer for x is Ox=et-e²t + [-et-² + e²(t-2)] (t - 2) x = et-e²t + [-et-¹ + e²(t-1)] U(t − 1) 2t - - Ox=et+e²t + [-et-1 + e²(t-²)] U(t − 2) Ox=et + e²t + [-et-¹ + e²(t-¹)] U(t − 1)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Question 13
Given: = 4x - 2y + 2 U(t-1)
dt
dy
dt
= 3x -y + U(t-1)
x(0) = 0, y (0) = /
The answer for x is
Ox=et-e²t + [-et-² + e²(t-2)] (t− 2)
O
x = et-e²t + [-et-¹ + e²(t-¹)] U(t − 1)
Ox=et +
x = et +
e²t + [-et-¹ + e²(t−2)] Ú(t − 2)
e²t + [-et-¹ + e²(t-1)] U(t − 1)
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0a27103f-8949-4cc0-b106-57b2f9bb7122%2F5c43981f-b354-4142-86c0-0e6fced1c516%2Fld3zvec_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 13
Given: = 4x - 2y + 2 U(t-1)
dt
dy
dt
= 3x -y + U(t-1)
x(0) = 0, y (0) = /
The answer for x is
Ox=et-e²t + [-et-² + e²(t-2)] (t− 2)
O
x = et-e²t + [-et-¹ + e²(t-¹)] U(t − 1)
Ox=et +
x = et +
e²t + [-et-¹ + e²(t−2)] Ú(t − 2)
e²t + [-et-¹ + e²(t-1)] U(t − 1)
2
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