Determine the solution of a differential equation governing an electric circuit satisfying given conditions

The current in an electric circuit is governed by the second-order differential equation

$I''(t) + I(t) = G(t)$

where $G(t)$ is a step function defined as

$G(t) = \begin{cases} 1 & \text{if } 0 \leq t \leq 2 \pi, \\ 0 & \text{otherwise}. \end{cases}$

Determine the solution which satisfies $I(0)= 0$ and $I'(0) = 1$ .

Since $G(t) = 0$ for $\geq 2 \pi$ we have the homogeneous equation

$I''(t) + I(t) = 0 \qquad \text{for} \quad t \geq 2 \pi.$

This has solutions

$I(t) = A \sin t + B \cos t \quad \implies \quad I'(t) = A \cos t - B \sin t.$

So, $I(0) = 0$ implies $B = 0$ and $I'(0) = 1$ implies $A = 1$ . Therefore,

$I(t) = \sin t \qquad \text{for} \quad t \geq 2 \pi.$

For $0 \leq t \leq 2 \pi$ , we have

$I''(t) + I(t) = 1.$

This has a particular solution given by $I_1(t) = 1$ . Hence, the general solution is

$I(t) = A \sin t + B \cos t + 1.$

Then $I(0) = 0$ implies $B = -1$ and $I'(0) = 1$ implies $A = 1$ . Therefore,

$I(t) = \sin t - \cos t + 1 \qquad \text{for} \quad 0 \leq t \leq 2 \pi.$

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