Answer
$i=\frac{12}{5}e^{-3t}+\frac{3}{5}[3\sin (4t)-4 \cos (4t)]$
Work Step by Step
Using RC Circuit we get:
$$\frac{di}{dt}+\frac{R}{L}i=\frac{E}{L}$$
We are given:
$E=10\sin 4t$
$L=\frac{2}{3}h$
$R=2$
Substituting:
$$\frac{di}{dt}+3i=15\sin 4t$$
The solution can be found by:
$$Q=e^{-\int3 dt}(c+\int 15\sin (4t)e^{3t}dt)$$
$$Q=e^{-3t}(c+15\sin (4t)e^{3t}dt)$$
$$Q=e^{-3t}[c+\frac{3}{5}e^{3t}(3\sin (4t)-4\cos(4t)]$$
$$Q=ce^{-3t}+\frac{3}{5}[3\sin (4t)-4\cos(4t)]$$
Since $t=0,i=0$, we get:
$$0=c+\frac{3}{5}(3\sin 0-4\cos0)\rightarrow c=\frac{12}{5}$$
Thus the current in the circuit for $t \geq 0$ is:
$$i=\frac{12}{5}e^{-3t}+\frac{3}{5}[3\sin (4t)-4 \cos (4t)]$$
