Question

Solve for i(t) for the circuite, given that V(t) = 10 sin5t V, R=4 W and L =2 H. (Ri + L dt/di = v)

Answer 1

Answer on Question #49747 – Math – Differential Calculus | Equations

Solve for $i(t)$ for the circuit, given that $V(t) = 10 \sin 5t \, V$, $R = 4 \, W$ and $L = 2 \, H$. ($Ri + L \, di/dt = v$)

Solution.

$$Ri + L \, \frac{di}{dt} = v \rightarrow 2 \, \frac{di}{dt} + 4i = 10 \sin 5t \rightarrow \frac{di}{dt} + 2i = 5 \sin 5t$$

Multiply both sides by $e^{2t}$ and obtain

$$e^{2t} \, \frac{di}{dt} + 2e^{2t}i = 5e^{2t} \sin 5t \rightarrow \frac{d}{dt}(e^{2t}i) = 5e^{2t} \sin 5t \rightarrow$$

Integrate both sides with respect to $t$

$$\rightarrow e^{2t}i = \int 5e^{2t} \sin 5t \, dt$$

To take integral $\int e^{2t} \sin 5t \, dt$, use 2 times integration by parts:

$$f = \sin 5t, \, g = \frac{1}{2}e^{2t}, \, df = 5 \cos 5t \, dt, \, dg = e^{2t} \, dt$$

So, $\int e^{2t} \sin 5t \, dt = \frac{1}{2}e^{2t} \sin 5t - \frac{5}{2} \int e^{2t} \cos 5t \, dt$

Then $f = \cos 5t$, $g = \frac{1}{2}e^{2t}$, $df = -5 \sin 5t \, dt$, $dg = e^{2t} \, dt$

So $\int e^{2t} \sin 5t \, dt = \frac{1}{2}e^{2t} \sin 5t - \frac{5}{2} \left( \frac{1}{2}e^{2t} \cos 5t + \frac{5}{2} \int e^{2t} \sin 5t \, dt \right) =$

$$\begin{array}{l} = \frac{1}{2}e^{2t} \sin 5t - \frac{5}{4}e^{2t} \cos 5t - \frac{25}{4} \int e^{2t} \sin 5t \, dt \rightarrow \\ \left(1 + \frac{25}{4}\right) \int e^{2t} \sin 5t \, dt = \frac{1}{2}e^{2t} \sin 5t - \frac{5}{4}e^{2t} \cos 5t \rightarrow \\ \rightarrow \int e^{2t} \sin 5t \, dt = \frac{4}{29} \left( \frac{1}{2}e^{2t} \sin 5t - \frac{5}{4}e^{2t} \cos 5t \right) \end{array}$$

Thus, $e^{2t}i = \frac{5}{29} (-5\cos 5t + 2\sin 5t)e^{2t} + c \rightarrow$

$\rightarrow$ $i(t) = ce^{-2t} - \frac{25}{29}\cos 5t + \frac{10}{29}\sin 5t$ , where $c$ is an arbitrary real constant.

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