Question

In Problems 51 and 52 solve equation (10) subject to $\mathit{i}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$with L, R, C, and E(t) as given. Use a graphing utility to graph the solution for$\mathbf{0}\mathbf{⩽}\mathit{t}\mathbf{⩽}\mathbf{3}$.

$$\begin{gathered} \mathit{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{005}\mathit{h}\mathbf{,}\mathit{R}\mathbf{=}\mathbf{1}\mathit{\Omega }\mathbf{,}\mathit{C}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{02}\mathit{f} \\ \mathit{E}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{100}\mathbf{[}\mathit{t}\mathbf{-}\mathbf{(}\mathit{t}\mathbf{-}\mathbf{1}\mathbf{\left)}\mathit{U}\mathbf{\right(}\mathit{t}\mathbf{-}\mathbf{1}\mathbf{\left)}\mathbf{\right]} \end{gathered}$$

Short answer

The Solution is$i\left(t\right)=\left\{\begin{array}{l}2-2e^{-100t}-200t{e}^{-100t},0⩽t\\ -2e^{-100t}-200t{e}^{-100t}+2e^{-100(t-1)}+200(t-1)e^{-100(t-1)},t⩾1\end{array}\right.$

Step-by-step solution

Use the Formula to solve:

$\text{L}\frac{\text{d}i}{\text{d}t}+Ri\left(t\right)+\frac{1}{C}{\int }_0^{t}i\left(\tau \right)\text{d}\tau =E\left(t\right)$

Replace the given values in the above formula,

$$\begin{gathered} 0.005\frac{\text{d}i}{\text{d}t}+i+\frac{1}{0.02}{\int }_0^{t}i\left(\tau \right)\text{d}\tau =100[t-(t-1\left)U\right(t-1\left)\right] \\ \frac{\text{d}i}{\text{d}t}+200i+10000{\int }_0^{t}i\left(\tau \right)\text{d}\tau =20000[t-(t-1\left)U\right(t-1\left)\right] \end{gathered}$$

Take Laplace transform of both sides in the above equation,

$\mathcal{L}\left\{\frac{\text{d}i}{\text{d}t}\right\}+\mathcal{L}\left\{200i\right\}+\mathcal{L}\left\{10000{\int }_0^{t}i\left(\tau \right)\text{d}\tau \right\}=\mathcal{L}\left\{20000\right[t-(t-1)U(t-1)\left]\right\}$

e$\mathcal{L}\left\{10000{\int }_0^{t}i\left(\tau \right)\text{d}\tau \right\}=20000[\mathcal{L}\{t\}-\mathcal{L}\{(t-1)U(t-1)\left\}\right]-\mathcal{L}\left\{\frac{\text{d}i}{\text{d}t}\right\}-200\mathcal{L}\left\{i\right\}$

$10000\mathcal{L}\left\{{\int }_0^{t}i\left(\tau \right)\text{d}\tau \right\}\stackrel{**}{=}20000\left[\frac{1}{s^2}-\frac{1}{s^2}{e}^{-s}\right]-s\mathcal{L}\left\{i\right\}-200\mathcal{L}\left\{i\right\}$

$\left(**\right)\left\{\begin{array}{l}\mathcal{L}\left\{\frac{\text{d}i}{\text{d}t}\right\}=s\mathcal{L}\left\{i\right\}-i\left(0\right),i\left(0\right)=0\\ \mathcal{L}\left\{t\right\}=\frac{1}{s^2}\\ IfF\left(s\right)=L\{f(t\left)\right\}anda>0,then\mathcal{L}\left\{f\right(t-a\left)U\right(t-a\left)\right\}=e^{-as}F\left(s\right)\end{array}\right.$

Simplify the above Equation:

We get, $10000\mathcal{L}\left\{{\int }_0^{t}i\left(\tau \right)\text{d}\tau \right\}=20000\left[\frac{1}{s^2}-\frac{1}{s^2}{e}^{-s}\right]-s\mathcal{L}\left\{i\right\}-200\mathcal{L}\left\{i\right\}$ ------ (1)

Finding $\mathcal{L}\left\{{\int }_0^{t}i\left(\tau \right)\text{d}\tau \right\}$ in the following way,

$$\begin{gathered} \mathcal{L}\left\{{\int }_0^{t}i\left(\tau \right)\text{d}\tau \right\}=\mathcal{L}\left\{i\right(t)*1\}\left({\int }_0^{t}i\left(\tau \right)\text{d}\tau =i\left(t\right)*1\right) \\ =\mathcal{L}\left\{i\right(t\left)\right\}\mathcal{L}\left\{1\right\}(\mathcal{L}\{f*g\}=\mathcal{L}\{f\}\mathcal{L}\{g\} \\ =\mathcal{L}\{i\left(t\right)\}·\frac{1}{s} \end{gathered}$$

By substituting the above equation in (1) we get,

$\frac{10000}{s}\mathcal{L}\left\{i\right\}=20000\left[\frac{1}{s^2}-\frac{1}{s^2}{e}^{-s}\right]-s\mathcal{L}\left\{i\right\}-200\mathcal{L}\left\{i\right\}$

$$\begin{gathered} \left[s^2+200s+10000\right]\mathcal{L}\left\{i\right\}=\frac{20000}{s}\left[1-e^{-s}\right] \\ \mathcal{L}\left\{i\right\}=\frac{20000}{s{(s+100)}^2}\left(1-e^{-s}\right) \end{gathered}$$ (*)

Decompose the fraction$\frac{20000}{s{(s+100)}^2}$Using the partial fraction technique,

$*\frac{20000}{s{(s+100)}^2}=\frac{A}{s}+\frac{B}{s+100}+\frac{C}{{(s+100)}^2}$

$=\frac{A{(s+100)}^2+Bs(s+100)+Cs}{s{(s+100)}^2} \to \left(2\right)$

Since the fractions in above equation have the same denominators, it takes their numerators to be equal.

$$\begin{gathered} 20000=A{s}^2+200As+10000A+B{s}^2+100Bs+Cs \\ 20000=(A+B)s^2+(200A+100B+C)s+10000A \end{gathered}$$

Compare the fractions:

Comparing the coefficient on both sides, we get

$$\begin{gathered} A+B=0 \\ 200A+100B+C=0 \\ 10000A=20000 \end{gathered}$$

Based on the previous system we get it $A=2,B=-2andC=-200$.

Substituting A, B and C in (2),

$\frac{20000}{s{(s+100)}^2}=\frac{2}{s}-\frac{2}{s+100}-\frac{200}{{(s+100)}^2}$

The Equation (*) becomes,

$\mathcal{L}\left\{i\right\}=\left[\frac{2}{s}-\frac{2}{s+100}-\frac{200}{{(s+100)}^2}\right]\left(1-e^{-s}\right)$

$=\frac{2}{s}-\frac{2}{s+100}-\frac{200}{{(s+100)}^2}-\frac{2}{s}{e}^{-s}+\frac{2}{s+100}{e}^{-s}+\frac{200}{{(s+100)}^2}{e}^{-s}$

Take Inverse Laplace Transform on both sides:

$$\begin{gathered} i\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{2}{s}\right\}-{\mathcal{L}}^{-1}\left\{\frac{2}{s+100}\right\}-{\mathcal{L}}^{-1}\left\{\frac{200}{{(s+100)}^2}\right\} \\ -{\mathcal{L}}^{-1}\left\{\frac{2}{c}{e}^{-s}\right\}+{\mathcal{L}}^{-1}\left\{\frac{2}{s\perp 100}{e}^{-s}\right\}+{\mathcal{L}}^{-1}\left\{\frac{200}{{(e+100)}^2}{e}^{-s}\right\} \\ -{\mathcal{L}}^{-1}\left\{\frac{2}{c}{e}^{-s}\right\}+{\mathcal{L}}^{-1}\left\{\frac{2}{s\perp 100}{e}^{-s}\right\}+{\mathcal{L}}^{-1}\left\{\frac{200}{{(e+100)}^2}{e}^{-s}\right\} \end{gathered}$$

$$\begin{gathered} \stackrel{\left(**\right)}{=}2-2e^{-100t}-200t{e}^{-100t}-2U(t-1) \\ +2e^{-100(t-1)}U(t-1)+200(t-1)e^{-100(t-1)}U(t-1) \end{gathered}$$

$\left(**\right)=\left\{\begin{array}{l}If{\mathcal{L}}^{-1}\left\{F\right(s\left)\right\}=f\left(t\right)anda>0,thenf(t-a)U(t-a)={\mathcal{L}}^{-1}\left\{e^{-as}F\left(s\right)\right\}\\ U(t-a)={\mathcal{L}}^{-1}\left\{\frac{e^{-as}}{s}\right\}\\ 1={\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\},e^{at}={\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\},t^n{e}^{at}={\mathcal{L}}^{-1}\left\{\frac{n!}{{(s-a)}^{n+1}}\right\}\end{array}\right.$

So,

$i\left(t\right)=\left\{\begin{array}{l}2-2e^{-100t}-200t{e}^{-100t},0⩽t\\ -2e^{-100t}-200t{e}^{-100t}+2e^{-100(t-1)}+200(t-1)e^{-100(t-1)},t⩾1\end{array}\right.$

The Graph of the above solution: