Question

The current I(t) in an LC series circuit is governed by the initial value problem....

Answer 1

I''(t)+4I(t)=g(t) I(0)=1, I'(0)=3

\[g(t)=3\sin(t), 0<t<2\pi \]
\[g(t)=0, 2\pi<t\]

Answer 2

\[I''(t)+4I(t)=\begin{cases}3\sin t&\text{for }0<t<2\pi\\0&\text{for }t>2\pi\end{cases}\]
with \(I(0)=1\) and \(I'(0)=3\).

Take the Laplace transform of both sides:
\[\begin{align*}
\mathcal{L}\left\{I''(t)+4I(t)\right\}&=\mathcal{L}\left\{g(t)\right\}\\\\
\mathcal{L}\left\{I''(t)\right\}+4\mathcal{L}\left\{I(t)\right\}&=\mathcal{L}\left\{g(t)\right\}\\\\
s^2\mathcal{L}\left\{I(t)\right\}-sI(0)-I'(0)+4\mathcal{L}\left\{I(t)\right\}&=\mathcal{L}\left\{g(t)\right\}
\end{align*}\]
If you're like me and don't commit your transforms to memory, you can always derive them using the definition of the transform.
\[\begin{align*}\mathcal{L}\{g(t)\}&=\int_0^\infty g(t)e^{-st}~dt\\\\
&=3\int_0^{2\pi}\sin t~e^{-st}~dt+\int_{2\pi}^\infty 0e^{-st}~dt\\\\
{\bf L}&=3\int_0^{2\pi}\sin t~e^{-st}~dt\end{align*}\]
Once you find \(\bf L\), you have
\[\begin{align*}
s^2\mathcal{L}\left\{I(t)\right\}-sI(0)-I'(0)+4\mathcal{L}\left\{I(t)\right\}&={\bf L}\\\\
s^2\mathcal{L}\left\{I(t)\right\}-s-3+4\mathcal{L}\left\{I(t)\right\}&={\bf L}\\\\
\left(s^2+4\right)\mathcal{L}\left\{I(t)\right\}&={\bf L}+s+3\\\\
\mathcal{L}\left\{I(t)\right\}&=\frac{{\bf L}+s+3}{s^2+4}\\\\
I(t)&=\mathcal{L}^{-1}\left\{\frac{{\bf L}+s+3}{s^2+4}\right\}
\end{align*}\]

Answer 3

After working the problem this is what I got. Not sure if I arrived at the correct solution.

Answer 4

Your work for finding \({\bf L}\) above is correct. I figured I'd go around defining \(g(t)\) in terms of the step function, we get the same result in either case.

Your final answer should include the step function somewhere. I think you overlooked something.

Answer 5

\[\begin{align*}
I(t)&=\mathcal{L}^{-1}\left\{\frac{\dfrac{3\left(1-e^{-2\pi s}\right)}{s^2+1}+s+3}{s^2+4}\right\}
\end{align*}\]
Rewriting the argument on the rhs:
\[\frac{3\left(1-e^{-2\pi s}\right)+(s+3)(s^2+1)}{(s^2+1)(s^2+4)}\]
\[-\frac{3e^{-2\pi s}}{(s^2+1)(s^2+4)}+\frac{s^3+3s^2+s+6}{(s^2+1)(s^2+4)}\]
Your work for decomposing the second fraction and finding the inverse transform is correct, but you seem to have neglected the first fraction.

Formula 27 gives you just what you need to figure out the inverse transform:
http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
(definitely a useful resource to keep handy)

Answer 6

To summarize: What you've done is correct so far, but you're missing part of the answer.

Answer 7

Ok I see what I missed. I wasn't sure how to include the

\[-3e ^{-2\pi s}\]
into the decomposition. Now I see it and the step function that is included in the final answer.

I use that same table and Pauls online math notes as a resource!

Thanks for your help!!!!!!!