91Ó°ÊÓ

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

Given equation 1.2:

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

And equation

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

From$V=V_0{e}^{i\mathrm{Ó\neg }t}$ , the equation 1.2 will become

$L\frac{dI}{dt}+RI={\text{V}}_{\text{0}}{e}^{i\mathrm{Ó\neg }t}$

Then the equation in the form will be

$\frac{dI}{dt}+\frac{R}{L}I=\frac{{\text{V}}_{\text{0}}{e}^{i\mathrm{Ó\neg }t}}{L}$

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 equation3.9,

$$\begin{gathered} {\text{Ie}}^f\text{=∫}\left(\frac{V_0{e}^{i\mathrm{Ó\neg }t}}{L}\right){\text{e}}^{\text{Rt/L}}\text{dt} \\ =\frac{V_{∘}}{L}\frac{e^{Rt/L}}{i\mathrm{Ó\neg }+\left(R/L\right)}{e}^{i\mathrm{Ó\neg }t}+\mathrm{Î\pm } \\ I=\frac{V_{∘}}{i\mathrm{Ó\neg }L+R}{e}^{i\mathrm{Ó\neg }t}+\mathrm{Î\pm }{e}^{Rt/L} \end{gathered}$$

So, the real part of this solution will be:

$\text{Re(I)=}\frac{V∘R}{{\mathrm{Ó\neg }}^2{L}^2+R^2}{\text{cosÓ¬t+αe}}^{Rt/L}$

From $V=V_0{e}^{i\mathrm{Ó\neg }t}$, the equation 1.3 will become

Then the equation in the form will be

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,

$$\begin{gathered} {\text{qe}}^g\text{=∫}\left(\frac{V_{∘}}{R}{\text{e}}^{i\mathrm{Ó\neg }t}\right){\text{e}}^{\text{t/RC}}\text{dt} \\ =\frac{V_{∘}}{R}\frac{e^{t/RC}}{i\mathrm{Ó\neg }+\left(1/RC\right)}{e}^{i\mathrm{Ó\neg }t}+\mathrm{β} \\ q=\frac{V_{∘}C}{i\mathrm{Ó\neg }RC+1}{e}^{i\mathrm{Ó\neg }t}+\mathrm{β}{e}^{-t/RC} \end{gathered}$$

So, the part of this solution will be:

$\text{Re(q)=}\frac{V_{∘}C}{{\mathrm{Ó\neg }}^2{R}^2{C}^2+1}{\text{cosÓ¬t+βe}}^{-t/RC}$