Therefore, from the above solution we can conclude that : x(t) = 10 He³ 3t e³t - 1.6 [៦] e -t -8.4e2t + -2e2t Use the method of variation of parameters to solve the initial value problem x' = Ax+f(t), x(a)=x using the following values. 4 -1 8 A= 16 |, f(t) = 7 1 + 4t x(0) = eAt -t = - 4 16t 1-4t x(t)= = ...
Therefore, from the above solution we can conclude that : x(t) = 10 He³ 3t e³t - 1.6 [៦] e -t -8.4e2t + -2e2t Use the method of variation of parameters to solve the initial value problem x' = Ax+f(t), x(a)=x using the following values. 4 -1 8 A= 16 |, f(t) = 7 1 + 4t x(0) = eAt -t = - 4 16t 1-4t x(t)= = ...
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
please provide the answer in the format of the example solution I attached.
![Therefore, from the above solution we can conclude that :
x(t) = 10 He³
3t
e³t - 1.6
[៦]
e
-t
-8.4e2t
+
-2e2t](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1ae1bb18-ef65-433d-a0dc-aafdea4542d4%2F3e45f74b-6477-4844-937b-78f1876b2b76%2Fww0jrfu_processed.png&w=3840&q=75)
Transcribed Image Text:Therefore, from the above solution we can conclude that :
x(t) = 10 He³
3t
e³t - 1.6
[៦]
e
-t
-8.4e2t
+
-2e2t
Transcribed Image Text:Use the method of variation of parameters to solve the initial value problem x' = Ax+f(t),
x(a)=x using the following values.
4
-1
8
A=
16
|, f(t) =
7
1 + 4t
x(0) =
eAt
-t
=
- 4
16t 1-4t
x(t)=
=
...
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