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materdei
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I'm stuck on a problem, and I'm in serious need of help.
I) Problem:
Find the solution to [itex] f (t) = 2 \int^t_0 f'(u) sin 3 (t-u) \ du + 2 cos (3t) [/itex].
Also find [itex] f (0) [/itex].II) Solution, so far:
[itex] F(s) = 2 (s F(s) - f(0)) * \frac{3}{(s^2 + 3^2)} + \frac{2}{(s^2 + 3^2)}[/itex]
[itex] F(s) (1 - \frac{(6 s)}{(s^2 + 3^2)} = \frac{(2 s - 6 f(0)}{(s^2 + 3^2)}[/itex]
[itex] F(s) = \frac{(2 s - 6 f(0)}{(s^2 + 3^2 - 6s)}[/itex]And this is basically where I'm stuck.
III) Proposed final solution:
Rewrites [itex] (s^2 + 3^2 - 6s)[/itex] to [itex](s - 3)^2[/itex]
[itex] F(s) = 2 s \frac{(1)}{(s - 3)^2)} - 6 f(0) \frac{(1)}{((s - 3)^2)}[/itex]
[itex] f(t) = 2 \mathcal {L^{-1}} (s) \mathcal {L^{-1}} \frac{(1)}{(s - 3)^2} - 6 f(0) \mathcal {L^{-1}} \frac{(1)}{(s - 3)^2} [/itex]
[itex] f(t) = 2 t * t e^{3t} - 6 f(0) t e^{3t} [/itex]
[itex] f(t) = 2 t^2 e^{3t} - 6 f(0) t e^{3t} [/itex]
Insert [itex] t = 0 [/itex]
[itex] f(0) = 0* 2 e^{3t} - 0 * 6 f(0) e^{3t} = 0 [/itex]
Hence, [itex] f(0) = 0 [/itex] and therefore [itex] - 6 f(0) t e^{3t} = 0 [/itex]
which gives [itex] f(t) = 2 t^2 e^{3t} [/itex]
However, I feel VERY unsure about this.
Any pointers and feedback are immensly welcome!
I'm stuck on a problem, and I'm in serious need of help.
I) Problem:
Find the solution to [itex] f (t) = 2 \int^t_0 f'(u) sin 3 (t-u) \ du + 2 cos (3t) [/itex].
Also find [itex] f (0) [/itex].II) Solution, so far:
[itex] F(s) = 2 (s F(s) - f(0)) * \frac{3}{(s^2 + 3^2)} + \frac{2}{(s^2 + 3^2)}[/itex]
[itex] F(s) (1 - \frac{(6 s)}{(s^2 + 3^2)} = \frac{(2 s - 6 f(0)}{(s^2 + 3^2)}[/itex]
[itex] F(s) = \frac{(2 s - 6 f(0)}{(s^2 + 3^2 - 6s)}[/itex]And this is basically where I'm stuck.
III) Proposed final solution:
Rewrites [itex] (s^2 + 3^2 - 6s)[/itex] to [itex](s - 3)^2[/itex]
[itex] F(s) = 2 s \frac{(1)}{(s - 3)^2)} - 6 f(0) \frac{(1)}{((s - 3)^2)}[/itex]
[itex] f(t) = 2 \mathcal {L^{-1}} (s) \mathcal {L^{-1}} \frac{(1)}{(s - 3)^2} - 6 f(0) \mathcal {L^{-1}} \frac{(1)}{(s - 3)^2} [/itex]
[itex] f(t) = 2 t * t e^{3t} - 6 f(0) t e^{3t} [/itex]
[itex] f(t) = 2 t^2 e^{3t} - 6 f(0) t e^{3t} [/itex]
Insert [itex] t = 0 [/itex]
[itex] f(0) = 0* 2 e^{3t} - 0 * 6 f(0) e^{3t} = 0 [/itex]
Hence, [itex] f(0) = 0 [/itex] and therefore [itex] - 6 f(0) t e^{3t} = 0 [/itex]
which gives [itex] f(t) = 2 t^2 e^{3t} [/itex]
However, I feel VERY unsure about this.
Any pointers and feedback are immensly welcome!
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