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 equation1.2:

$L\frac{d^{2}I}{d{t}^2}+R\frac{dI}{dt}+\frac{I}{C}=\frac{dV}{dt}$

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$V=V_0\cos \mathrm{Ó\neg }t$, the equation 1.3 will become

$R\frac{dq}{dt}+\frac{q}{C}=V_0\cos \mathrm{Ó\neg }t$

Then the equation in the form will be

$\frac{dq}{dt}+\frac{q}{RC}=\frac{V_0}{R}\cos \mathrm{Ó\neg }t$

From the equation 3.4,

$$\begin{gathered} g=∫\frac{1}{RC}dt \\ =\frac{t}{RC} \\ {e}^g=e^{\frac{t}{RC}} \end{gathered}$$

From the equation 3.9 ,

$$\begin{gathered} {\text{qe}}^g\text{=∫}\left(\frac{V_{\mathrm{ο}}}{R}\text{cosÓ¬t}\right){\text{e}}^{\text{t/RC}}\text{dt} \\ =\frac{V_{∘}}{R}\frac{e^{-t/RC}}{{\mathrm{Ó\neg }}^2+\left(1/R{C}^2\right)}\left(\mathrm{Ó\neg }\sin \mathrm{Ó\neg }t+\frac{1}{RC}\cos \mathrm{Ó\neg }t\right)+\mathrm{Î\pm } \\ q=\frac{V_{∘}}{{\mathrm{Ó\neg }}^2{R}^2{C}^2}\left(\mathrm{Ó\neg }R{C}^2\sin \mathrm{Ó\neg }t+\cos \mathrm{Ó\neg }t\right)+\mathrm{Î\pm }{e}^{-t/RC} \end{gathered}$$

So, the general solution will be:

$q=\frac{V_{}∘}{{\mathrm{Ó\neg }}^2{R}^2{C}^2+1}\left(\mathrm{Ó\neg }R{C}^2\sin \mathrm{Ó\neg }t+C\cos \mathrm{Ó\neg }t\right)+\mathrm{Î\pm }{e}^{-t/RC}$