91Ó°ÊÓ

The linear differential equation is defined by$\mathbf{\text{x'}}\mathbf{+}\mathbf{\text{Px}}\mathbf{=}\mathbf{\text{Q}}$, whereP and Qare numeric constants or functions in x. It is made up of a yand a yderivative. The differential equation is called the first-order linear differential equation because it is a first-order differentiation.

Given equation 1.2:

$L\frac{dI}{dt}+RI+\frac{q}{C}=V$

Also given RI circuit$\frac{1}{C}=0$ with$V=V_0\cos \mathrm{Ó\neg }t\left(\mathrm{Ó\neg }=cons\tan t\right)$

From $V=V_0\cos \mathrm{Ó\neg }t\left(\mathrm{Ó\neg }=cons\tan t\right)$, the equation 1.2 will become

$L\frac{dI}{dt}+RI=V_0\cos \mathrm{Ó\neg }t$

Then the equation in the form will be

$\frac{dI}{dt}+\frac{R}{L}I=\frac{V_0}{L}\cos \mathrm{Ó\neg }t$

From the equation 3.4,

$$\begin{gathered} f=∫\frac{R}{L}dt \\ =\frac{R}{L}t \\ {e}^f=e^{\frac{Rt}{L}} \end{gathered}$$

From the equation 3.9,

$$\begin{gathered} {\text{le}}^f\text{=∫}\left(\frac{V_{∘}}{L}\text{cosÓ¬t}\right){\text{e}}^{\text{Rt/L}}\text{dt} \\ =\frac{V_{∘}}{L}\frac{e^{-Rt/L}}{{\mathrm{Ó\neg }}^2+\left(R/L^2\right)}\left(\mathrm{Ó\neg }\sin \mathrm{Ó\neg }t+\frac{R}{L}\cos \mathrm{Ó\neg }t\right)+\mathrm{Î\pm } \\ I=\frac{V_{∘}}{{\mathrm{Ó\neg }}^2{L}^2+R^2}\left(\mathrm{Ó\neg }L\sin \mathrm{Ó\neg }t+R\cos \mathrm{Ó\neg }t\right)+\mathrm{Î\pm }{e}^{-Rt/L} \end{gathered}$$

So, the general solution will be:

$I=\frac{V_{∘}}{{\mathrm{Ó\neg }}^2{L}^2{R}^2}\left(\mathrm{Ó\neg }L\sin \mathrm{Ó\neg }t+R\cos \mathrm{Ó\neg }t\right)+\mathrm{Î\pm }{e}^{-Rt/L}$