Solution Found!
The current I(t) in an LC series circuit is governed by
Chapter 7, Problem 20E(choose chapter or problem)
The current in an LC series circuit is governed by the initial value problem
\(I^{\prime \prime}(t)+4 I(t)=g(t)\);
\(I(0)=1, I^{\prime}(0)=3\),
where
\(g(t):=\left\{\begin{array}{l}
3 \sin t \\
0
\end{array}\right.
\) \(\begin{array}{l}
0 \leq t \leq 2 \pi \\
2 \pi<t
\end{array}
\)
Determine the current as a function of time t.
Equation transcription:
Text transcription:
I^{\prime \prime}(t)+4 I(t)=g(t)
I(0)=1, I^{\prime}(0)=3
g(t):=\left\{\begin{array}{l}
3 \sin t \\
0
\end{array}\right.
\begin{array}{l}
0 \leq t \leq 2 \pi \\
2 \pi<t
\end{array}
Questions & Answers
QUESTION:
The current in an LC series circuit is governed by the initial value problem
\(I^{\prime \prime}(t)+4 I(t)=g(t)\);
\(I(0)=1, I^{\prime}(0)=3\),
where
\(g(t):=\left\{\begin{array}{l}
3 \sin t \\
0
\end{array}\right.
\) \(\begin{array}{l}
0 \leq t \leq 2 \pi \\
2 \pi<t
\end{array}
\)
Determine the current as a function of time t.
Equation transcription:
Text transcription:
I^{\prime \prime}(t)+4 I(t)=g(t)
I(0)=1, I^{\prime}(0)=3
g(t):=\left\{\begin{array}{l}
3 \sin t \\
0
\end{array}\right.
\begin{array}{l}
0 \leq t \leq 2 \pi \\
2 \pi<t
\end{array}
ANSWER:
Solution. Our aim is to find the general solution of I ( t ) using laplace differential operator.
............................... (1)
Step 1. To solve this type of differential equation by using Laplace differential operator we use the following references.
Reference 1.
Reference 2.
Reference 3. Let
Reference 4.using the definition of laplace transforms
Reference 5. Using the formula
