9. 10. |C/G|| y" + 3y' + 2y = 1 + d(t − 1), y" + 2y' + y = et +28(t− 2), y(0) = −1, y'(0) = 2 y(0)=1, y′(0) = −1
9. 10. |C/G|| y" + 3y' + 2y = 1 + d(t − 1), y" + 2y' + y = et +28(t− 2), y(0) = −1, y'(0) = 2 y(0)=1, y′(0) = −1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Numbers 10 and 15
![In Exercises 1-20 solve the initial value problem. Where indicated by C/G, graph the solution.
1.
y" + 3y' + 2y = 6e²t + 28(t − 1), y(0) = 2, y'(0) = −6
-t
2.
C/Gy" + y − 2y = −10e¯
-
+58(t-1),
y(0) = 7, y'(0)
3.
y" - 4y = 2e-t +58(t-1), y(0) = −1, y'(0) = 2
+
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
|C/G|y" + y = sin 3t+28(t — π/2),
y" + 4y = 4 + 8(t−3π),
y(0) = 0,
y" - y = 8+28(t — 2),
y(0) = -1,
y"+y' = et + 38(t — 6),
y(0) = -1,
y'(0) = 4
y" + 4y = 8e²t +8(t = π/2), y(0) = 8,_y'(0) = 0
y(0) = 1,
y'(0) = 1
y'(0)=1
y'(0) =
= −1,
C/Gy" + 3y' + 2y = 1 + 8(t − 1), y(0) = 1, y'(0) = −1
y" + 2y + y = et + 28 (t− 2), y(0)
y'(0) = 2
= −1
y(0) = 0,
y(0)
=
C/Gy" + 4y = sint + 8(t− π/2),
y' (0) = 2
y" + 2y + 2y = 8(t – π) — 38(t – 2π),
= -1, y'(0) = 2
y(0) = 1,
y'(0) =
y" + 4y' + 13y = 8(t − π/6) + 28(t — π/3),
2y" − 3y′ – 2y = 1 + d(t − 2), y(0) = -1,
4y" — 4y' + 5y = 4 sint - 4 cost + 8(t − π/2) — 8(t = π), y(0) = 1, y′(0) = 1
y' (0) = 2
= 2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f3c8c6e-d3d0-4d9a-a919-498816195678%2Fa8928729-6dcf-44d9-9c32-fd145a977ab3%2Frbms3fs_processed.png&w=3840&q=75)
Transcribed Image Text:In Exercises 1-20 solve the initial value problem. Where indicated by C/G, graph the solution.
1.
y" + 3y' + 2y = 6e²t + 28(t − 1), y(0) = 2, y'(0) = −6
-t
2.
C/Gy" + y − 2y = −10e¯
-
+58(t-1),
y(0) = 7, y'(0)
3.
y" - 4y = 2e-t +58(t-1), y(0) = −1, y'(0) = 2
+
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
|C/G|y" + y = sin 3t+28(t — π/2),
y" + 4y = 4 + 8(t−3π),
y(0) = 0,
y" - y = 8+28(t — 2),
y(0) = -1,
y"+y' = et + 38(t — 6),
y(0) = -1,
y'(0) = 4
y" + 4y = 8e²t +8(t = π/2), y(0) = 8,_y'(0) = 0
y(0) = 1,
y'(0) = 1
y'(0)=1
y'(0) =
= −1,
C/Gy" + 3y' + 2y = 1 + 8(t − 1), y(0) = 1, y'(0) = −1
y" + 2y + y = et + 28 (t− 2), y(0)
y'(0) = 2
= −1
y(0) = 0,
y(0)
=
C/Gy" + 4y = sint + 8(t− π/2),
y' (0) = 2
y" + 2y + 2y = 8(t – π) — 38(t – 2π),
= -1, y'(0) = 2
y(0) = 1,
y'(0) =
y" + 4y' + 13y = 8(t − π/6) + 28(t — π/3),
2y" − 3y′ – 2y = 1 + d(t − 2), y(0) = -1,
4y" — 4y' + 5y = 4 sint - 4 cost + 8(t − π/2) — 8(t = π), y(0) = 1, y′(0) = 1
y' (0) = 2
= 2
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