x' (t) + 4x' (t) = −2t – sin(t – 4), with x(4) = 0, and x'(4) = 1.

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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Solve the following initial value problems by using Laplace transforms:
I did attached the question and the answer based on a calculator that I used. Please write neatly.

x' (t) + 4x' (t) = −2t – sin(t – 4), with x(4) = 0, and x'(4) = 1.
Transcribed Image Text:x' (t) + 4x' (t) = −2t – sin(t – 4), with x(4) = 0, and x'(4) = 1.
laplace transform.x + 4x = -2t-sin(t-4), x(4) = 0, x' (4) = 1
Solution
15
32
-4(t-4)
X =
15(t-4)
8
4
4
+cos (1−4) + sin(t− 4)
17
383
544
∙e
Transcribed Image Text:laplace transform.x + 4x = -2t-sin(t-4), x(4) = 0, x' (4) = 1 Solution 15 32 -4(t-4) X = 15(t-4) 8 4 4 +cos (1−4) + sin(t− 4) 17 383 544 ∙e
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