Question

Solve the given initial value problem. Give the largest interval lover which the general solution is defined.

$\frac{\mathbf{di}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{R}}{\mathbf{L}}\mathbf{i}\mathbf{=}\frac{\mathbf{E}}{\mathbf{L}}\mathbf{,}\mathbf{i}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{i}}_{\mathbf{0}}$

L, R, E, and i₀constant.

Short answer

The solution for the given initial value is$i=\frac{E}{R}+\left(i_0-\frac{E}{R}\right)e^{-\frac{Rt}{L}}$

Step-by-step solution

The given equation in the standard form and determine the integrated factor

Dividing the given differential equation by; we get the standard form

$\frac{di}{dt}+\frac{R}{L}i=\frac{E}{L}$

$$\begin{gathered} \text{Here}P\left(t\right)=\frac{R}{L}\text{and}f\left(t\right)=\frac{E}{L}\text{; which are continuous on}(-\infty ,\infty ) \\ \text{Integrating factor}=e^{\int \frac{R}{L}dt} \\ =e^{\frac{Rt}{L}} \end{gathered}$$

Determine the general solution for the given differential equation

Multiplying the standard form with the integrating factor, we get

which is same as$\frac{d}{dt}\left[i{e}^{\frac{Rt}{L}}\right]=\frac{E}{L}{e}^{\frac{R}{L}t}$

On integration, we obtain

$$\begin{gathered} e^{\frac{Rt}{L}}=\frac{E}{L}\left(\frac{L}{R}{e}^{\frac{Rt}{L}}\right)+c \\ i=\frac{E}{R}+c{e}^{-\frac{Rt}{L}};\text{for}-\infty <t<\infty \end{gathered}$$

If$\left(0\right)=i_0$; we get$i_0=\frac{E}{R}+c{e}^{-0}$Or

$c=i_0-\frac{E}{R}$

Thus, $i=\frac{E}{R}+\left(i_0-\frac{E}{R}\right)e^{-\frac{Rt}{L}}$; is the solution of the given initial value problem.