Chapter 8, Problem 29P

Solve the following differential equations subject to the specified initial conditions

1. (a) d²v/dt² + 4v = 12, v(0) = 0, dv(0)/dt = 2
2. (b) d²i/dt² + 5 di/dt + 4i = 8, i(0) = −1, di(0)/dt = 0
3. (c) d²v/dt² + 2 dv/dt + v = 3, v(0) = 5, dv(0)/dt = 1
4. (d) d²i/dt² + 2 di/dt + 5i = 10, i(0) = 4, di(0)/dt = −2

To determine

Find the expression of $v\left(t\right)$ for the given differential equations.

Answer to Problem 29P

The expression of $v\left(t\right)$ for the given differential equation is $\left[3-3\cos 2t+\sin 2t\right]\text{V}$.

Explanation of Solution

Given data:

The differential equation is,

$\frac{d^{2}v}{d{t}^2}+4v=12$        (1)

The values of initial conditions are,

$\begin{array}{l}v\left(0\right)=0\\ \frac{dv\left(0\right)}{dt}=2\end{array}$

Formula used:

Write an expression to find the voltage response with the step input, if the roots of characteristic equation are real and imaginary.

$v\left(t\right)=V_s+\left(A_1\cos {\omega }_{d}t+A_2\sin {\omega }_{d}t\right)e^{-\alpha t}$        (2)

Here,

$V_s$ is the step response voltage,

$\alpha$ is the neper frequency,

${\omega }_d$ is the damped natural frequency, and

$A_1\text{and}{A}_2$ are constants.

Write a general expression for the roots of characteristic equation when the roots are real and imaginary.

$s_1=-\alpha +j{\omega }_d$        (3)

$s_2=-\alpha -j{\omega }_d$        (4)

Write an expression to solve quadratic equation.

$s_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2}$        (5)

Here,

$a$ is the coefficient of second order term,

$b$ is the coefficient of first order term, and

$c$ is the coefficient of constant term.

Calculation:

From equation (1), the characteristic equation is written as follows.

$s^2+4=0$        (6)

From the equation (6),

$\begin{array}{l}a=1\\ b=0\\ c=4\end{array}$

Substitute $1$ for $a$, $0$ for $b$, and $4$ for $c$ in equation (5) to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=\frac{-0\pm \sqrt{{\left(0\right)}^2-4\left(1\right)\left(4\right)}}{2}\\ =\frac{-0\pm \sqrt{-16}}{2}\\ =\frac{\pm j4}{2}\end{array}$

Simplify the equation.

$\begin{array}{c}{s}_{1,2}=\frac{j4}{2},\frac{-j4}{2}\\ =j2,-j2\end{array}$

Therefore, the roots of characteristic equations are real and imaginary.

$\begin{array}{l}{s}_1=+j2\\ s_2=-j2\end{array}$

Compare the values of roots with the equation (3) and equation (4).

$\begin{array}{l}\alpha =0\\ {\omega }_d=2\end{array}$

Substitute $0$ for $\alpha$, and $2$ for ${\omega }_d$ in equation (2) to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=V_s+\left(A_1\cos 2t+A_2\sin 2t\right)e^{-\left(0\right)t}\\ =V_s+\left(A_1\cos 2t+A_2\sin 2t\right)\left(1\right)\left\{\because e^0=1\right\}\end{array}$

$v\left(t\right)=V_s+\left(A_1\cos 2t+A_2\sin 2t\right)$        (7)

At $t=\infty$, the steady state condition is attained and rate of change of voltage with time equals to zero.

$\frac{dv\left(\infty \right)}{dt}=0$

Therefore,

$\frac{d{v}^2\left(\infty \right)}{d{t}^2}=0$

Substitute $v\left(\infty \right)$ for $v$ in equation (1).

$\frac{d^{2}v\left(\infty \right)}{d{t}^2}+4v\left(\infty \right)=12$        (8)

For the step response $RLC$ circuit,

$v\left(\infty \right)=V_s$

Substitute $0$ for $\frac{dv\left(\infty \right)}{dt}$, $0$ for $\frac{d{v}^2\left(\infty \right)}{d{t}^2}$ and $V_s$ for $v\left(\infty \right)$ in equation (8) to find $V_s$.

$\begin{array}{l}4V_s=12\\ V_s=\frac{12}{4}\\ V_s=3\text{V}\end{array}$

Substitute $3\text{V}$ for $V_s$ in equation (7) to find $v\left(t\right)$.

$v\left(t\right)=3\text{V}+\left(A_1\cos 2t+A_2\sin 2t\right)$        (9)

Substitute $0$ for $t$ in equation (9) to find $v\left(0\right)$.

$\begin{array}{c}v\left(0\right)=3\text{V}+\left(A_1\cos 2\left(0\right)+A_2\sin 2\left(0\right)\right)\\ =3\text{V+}\left(A_1\left(1\right)+A_2\left(0\right)\right)\left(1\right)\left\{\because \cos 0°=1,\text{ and}\sin 0°=0\right\}\end{array}$

$v\left(0\right)=\left[3+A_1\right]\text{V}$        (10)

Substitute $0\text{V}$ for $v\left(0\right)$ in equation (10) to find $A_1$.

$\begin{array}{c}0\text{V}=\left[3+A_1\right]\text{V}\\ A_1=-3\end{array}$

Substitute $-3$ for $A_1$ in equation (9) to find $v\left(t\right)$.

$v\left(t\right)=\left[3+\left(-3\cos 2t+A_2\sin 2t\right)\right]\text{V}$

$v\left(t\right)=\left[3-3\cos 2t+A_2\sin 2t\right]\text{V}$        (11)

Differentiate equation (11) with respect to $t$ to find $\frac{dv\left(t\right)}{dt}$.

$\frac{dv\left(t\right)}{dt}=\left[0-3\left(-\sin 2t\right)\left(2\right)+A_2\cos 2t\left(2\right)\right]\frac{\text{V}}{\text{s}}$

$\frac{dv\left(t\right)}{dt}=\left[6\text{sin 2}t++2A_2\cos 2t\right]\frac{\text{V}}{\text{s}}$        (12)

Substitute $0$ for $t$ in equation (12) to find $\frac{dv\left(0\right)}{dt}$.

$\begin{array}{c}\frac{dv\left(0\right)}{dt}=\left[6\text{sin 2}\left(0\right)+2A_2\cos 2\left(0\right)\right]\frac{\text{V}}{\text{s}}\\ =\left[\left(6\text{V}\right)\left(0\right)+2A_2\left(1\right)\right]\frac{\text{V}}{\text{s}}\left\{\because \cos 0°=1,\text{ and}\sin 0°=0\right\}\end{array}$

$\frac{dv\left(0\right)}{dt}=\text{2}{A}_2\frac{\text{V}}{\text{s}}$        (13)

Substitute $2\frac{\text{V}}{\text{s}}$ for $\frac{dv\left(0\right)}{dt}$ in equation (13) to find $A_2$.

$\begin{array}{l}2\frac{\text{V}}{\text{s}}=\text{2}{A}_2\frac{\text{V}}{\text{s}}\\ 2A_2=2\\ A_2=\frac{2}{2}\\ A_2=1\end{array}$

Substitute $1$ for $A_2$ in equation (11) to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=\left[3+\left(-3\right)\cos 2+\left(1\right)\sin 2t\right]\text{V}\\ =\left[3-3\cos 2t+\sin 2t\right]\text{V}\end{array}$

Conclusion:

Thus, the expression of $v\left(t\right)$ for the given differential equation is $\left[3-3\cos 2t+\sin 2t\right]\text{V}$.

To determine

Find the expression of $i\left(t\right)$ for the given differential equations.

Answer to Problem 29P

The expression of $i\left(t\right)$ for the given differential equation is $\left(2-4e^{-t}+e^{-4t}\right)\text{A}$.

Explanation of Solution

Given data:

The differential equation is,

$\frac{d^{2}i}{d{t}^2}+5\frac{di}{dt}+4i=8$        (14)

The values of initial conditions are,

$\begin{array}{l}i\left(0\right)=-1\\ \frac{di\left(0\right)}{dt}=0\end{array}$

Formula used:

Write a general expression for the current response with the step input, if the roots of characteristic equation are real and negative.

$i\left(t\right)=\left(I_s+A_1{e}^{s_{1}t}+A_2{e}^{s_{2}t}\right)\text{A}$        (15)

Here,

$I_s$ is the step response current, and

$s_1$ and $s_2$ are the roots of characteristic equation.

Calculation:

From equation (14), the characteristic equation is written as follows.

$s^2+5s+4=0$        (16)

Compare the equation (16) with the quadratic equation $\left(a{s}^2+bs+c=0\right)$.

$\begin{array}{l}a=1\\ b=5\\ c=4\end{array}$

Substitute $1$ for $a$, $0$ for $b$, and $4$ for $c$ in equation (5) to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=\frac{-5\pm \sqrt{{\left(5\right)}^2-4\left(1\right)\left(4\right)}}{2}\\ =\frac{-5\pm \sqrt{9}}{2}\\ =\frac{-5\pm 3}{2}\end{array}$

Simplify the equation.

$\begin{array}{c}{s}_{1,2}=\frac{-5+3}{2},\frac{-5-3}{2}\\ =-1,-4\end{array}$

Therefore, the roots of characteristic equations are real and negative.

$\begin{array}{l}{s}_1=-1\\ s_2=-4\end{array}$

Substitute $-1$ for $s_1$, and $-4$ for $s_2$ in equation (15) to find $i\left(t\right)$.

$i\left(t\right)=\left(I_s+A_1{e}^{-t}+A_2{e}^{-4t}\right)\text{A}$        (17)

At $t=\infty$, the steady state condition is attained and rate of change of current with time equals to zero.

$\frac{di\left(\infty \right)}{dt}=0$

Therefore,

$\frac{d{i}^2\left(\infty \right)}{d{t}^2}=0$

Substitute $i\left(\infty \right)$ for $i$ in equation (14).

$\frac{d^{2}i\left(\infty \right)}{d{t}^2}+5\frac{di\left(\infty \right)}{dt}+4i\left(\infty \right)=8$        (18)

For the step response $RLC$ circuit,

$i\left(\infty \right)=I_s$

Substitute $0$ for $\frac{di\left(\infty \right)}{dt}$, $0$ for $\frac{d{i}^2\left(\infty \right)}{d{t}^2}$ and $I_s$ for $i\left(\infty \right)$ in equation (18) to find $I_s$.

$\begin{array}{l}0+5\left(0\right)+4I_s=8\\ I_s=\frac{8}{4}\\ I_s=2\text{A}\end{array}$

Substitute $2$ for $I_s$ in equation (17) to find $i\left(t\right)$.

$i\left(t\right)=\left(2+A_1{e}^{-t}+A_2{e}^{-4t}\right)\text{A}$        (19)

Substitute $0$ for $t$ in equation (19) to find $i\left(0\right)$.

$\begin{array}{c}i\left(0\right)=\left(2+A_1{e}^{-\left(0\right)}+A_2{e}^{-4\left(0\right)}\right)\text{A}\\ \text{=}\left(2+A_1\left(1\right)+A_2\left(1\right)\right)\text{A}\left\{\because e^0=1\right\}\\ =\left[2+A_1+A_2\right]\text{A}\end{array}$

Substitute $-1\text{A}$ for $i\left(0\right)$ in above equation to find $A_1$.

$\begin{array}{l}-1\text{A}=\left[2+A_1+A_2\right]\text{A}\\ -1=2+A_1+A_2\end{array}$

Simplify the above equation to find $A_1$.

$A_1=-1-2-A_2$

$A_1=-3-A_2$        (20)

Differentiate equation (19) with respect to $t$ to find $\frac{di\left(t\right)}{dt}$.

$\begin{array}{c}\frac{di\left(t\right)}{dt}=\left[0+A_1{e}^{-t}\left(-1\right)+A_2{e}^{-4t}\left(-4\right)\right]\frac{\text{A}}{\text{s}}\\ =\left[-A_1{e}^{-t}-4A_2{e}^{-4t}\right]\frac{\text{A}}{\text{s}}\end{array}$

Substitute $0$ for $t$ in above equation to find $\frac{di\left(0\right)}{dt}$.

$\begin{array}{c}\frac{di\left(0\right)}{dt}=\left[-A_1{e}^{-\left(0\right)}-4A_2{e}^{-4\left(0\right)}\right]\frac{\text{A}}{\text{s}}\\ =\left[-A_1\left(1\right)-4A_2\left(1\right)\right]\frac{\text{A}}{\text{s}}\left\{\because e^0=1\right\}\\ =\left[-A_1-4A_2\right]\frac{\text{A}}{\text{s}}\end{array}$

Substitute $0\frac{\text{A}}{\text{s}}$ for $\frac{di\left(0\right)}{dt}$ in above equation to find $A_2$.

$\begin{array}{l}0\frac{\text{A}}{\text{s}}=\left[-A_1-4A_2\right]\frac{\text{A}}{\text{s}}\\ -A_1-4A_2=0\end{array}$

Substitute equation (20) in above equation to find $A_2$.

$\begin{array}{l}-\left(-3-A_2\right)-4A_2=0\\ 3+A_2-4A_2=0\\ 3A_2=3\end{array}$

Simplify the above equation to find $A_2$.

$\begin{array}{c}{A}_2=\frac{3}{3}\\ =1\end{array}$

Substitute $1$ for $A_2$ in equation (20) to find $A_1$.

$\begin{array}{c}{A}_1=-3-1\\ =-4\end{array}$

Substitute $-4$ for $A_1$, and $1$ for $A_2$ in equation (19) to find $i\left(t\right)$.

$i\left(t\right)=\left(2-4e^{-t}+e^{-4t}\right)\text{A}$

Conclusion:

Thus, the expression of $i\left(t\right)$ for the given differential equation is $\left(2-4e^{-t}+e^{-4t}\right)\text{A}$.

To determine

Find the expression of $v\left(t\right)$ for the given differential equations.

Answer to Problem 29P

The expression of $v\left(t\right)$ for the given differential equation is $\left[3+\left(2+3t\right)e^{-t}\right]\text{V}$.

Explanation of Solution

Given data:

The differential equation is,

$\frac{d^{2}v}{d{t}^2}+2\frac{dv}{dt}+v=3$        (21)

The values of initial conditions are,

$\begin{array}{l}v\left(0\right)=5\\ \frac{dv\left(0\right)}{dt}=1\end{array}$

Formula used:

Write an expression to find the voltage response with the step input, if the roots of characteristic equation are real and equal.

$v\left(t\right)=\left[V_s+\left(A_1+A_{2}t\right)e^{-\alpha t}\right]\text{V}$        (22)

Here,

$A_1$ and $A_2$ are constants, and

Write a general expression for the roots of characteristic equation when the roots are real and equal.

$s_1=s_2=-\alpha$        (23)

Calculation:

From equation (21), the characteristic equation is written as follows.

$s^2+2s+1=0$        (24)

Compare the equation (24) with the quadratic equation $\left(a{s}^2+bs+c=0\right)$.

$\begin{array}{l}a=1\\ b=2\\ c=1\end{array}$

Substitute $1$ for $a$, $2$ for $b$, and $1$ for $c$ in equation (5) to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=\frac{-2\pm \sqrt{{\left(2\right)}^2-4\left(1\right)\left(1\right)}}{2}\\ =\frac{-2\pm \sqrt{4-4}}{2}\\ =\frac{-2\pm 0}{2}\\ =\frac{-2}{2}\end{array}$

Simplify the equation.

$s_{1,2}=-1,-1$

Therefore, the roots of characteristic equations are real and imaginary.

$\begin{array}{l}{s}_1=-1\\ s_2=-1\end{array}$

Substitute $-1$ for $s_1$ in equation (23) to find $\alpha$.

$\begin{array}{l}-1=-\alpha \\ \alpha =1\end{array}$

Substitute $1$ for $\alpha$ in equation (22) to find $v\left(t\right)$.

$v\left(t\right)=\left[V_s+\left(A_1+A_{2}t\right)e^{-t}\right]\text{V}$        (25)

Substitute $v\left(\infty \right)$ for $v$ in equation (21).

$\frac{d^{2}v\left(\infty \right)}{d{t}^2}+2\frac{dv\left(\infty \right)}{dt}+v\left(\infty \right)=3$        (26)

For the step response,

$v\left(\infty \right)=V_s$

Substitute $0$ for $\frac{dv\left(\infty \right)}{dt}$, $0$ for $\frac{d{v}^2\left(\infty \right)}{d{t}^2}$ and $V_s$ for $v\left(\infty \right)$ in equation (26) to find $V_s$.

$\begin{array}{c}0+2\left(0\right)+V_s=3\\ V_s=3\text{V}\end{array}$

Substitute $3$ for $V_s$ in equation (25) to find $v\left(t\right)$.

$v\left(t\right)=\left[3+\left(A_1+A_{2}t\right)e^{-t}\right]\text{V}$        (27)

Substitute $0$ for $t$ in equation (27) to find $v\left(0\right)$.

$\begin{array}{c}v\left(0\right)=\left[3+\left(A_1+A_2\left(0\right)\right)e^{-\left(0\right)}\right]\text{V}\\ \text{=}\left[3+\left(A_1+0\right)\left(1\right)\right]\text{V}\left\{\because e^0=1\right\}\\ =\left[3+A_1\right]\text{V}\end{array}$

Substitute $5\text{V}$ for $v\left(0\right)$ in above equation to find $A_1$.

$\begin{array}{l}5\text{V}=\left[3+A_1\right]\text{V}\\ 5=3+A_1\end{array}$

Simplify the above equation to find $A_1$.

$\begin{array}{c}{A}_1=5-3\\ =2\end{array}$

Expand the equation (27) as follows:

$v\left(t\right)=\left[3+A_1{e}^{-t}+A_{2}t{e}^{-t}\right]\text{V}$

Differentiate the above with respect to $t$ to find $\frac{dv\left(t\right)}{dt}$.

$\begin{array}{c}\frac{dv\left(t\right)}{dt}=\left[0+A_1{e}^{-t}\left(-1\right)+A_2\left(1\right)e^{-t}+A_{2}t{e}^{-t}\left(-1\right)\right]\frac{\text{V}}{\text{s}}\\ =\left[-A_1{e}^{-t}+A_2{e}^{-t}-A_{2}t{e}^{-t}\right]\frac{\text{V}}{\text{s}}\end{array}$

Substitute $0$ for $t$ in above equation to find $\frac{dv\left(0\right)}{dt}$.

$\begin{array}{c}\frac{dv\left(0\right)}{dt}=\left[-A_1{e}^{-\left(0\right)}+A_2{e}^{-\left(0\right)}-A_2\left(0\right)e^{-\left(0\right)}\right]\frac{\text{V}}{\text{s}}\\ =\left[-A_1\left(1\right)+A_2\left(1\right)-A_2\left(0\right)\left(1\right)\right]\frac{\text{V}}{\text{s}}\left\{\because e^0=1\right\}\\ =\left[-A_1+A_2\right]\frac{\text{V}}{\text{s}}\end{array}$

Substitute $1\frac{\text{V}}{\text{s}}$ for $\frac{dv\left(0\right)}{dt}$ in above equation to find $A_2$.

$\begin{array}{l}1\frac{\text{V}}{\text{s}}=\left[-A_1+A_2\right]\frac{\text{V}}{\text{s}}\\ 1=-A_1+A_2\end{array}$

Simplify the above equation to find $A_2$.

$A_2=1+A_1$

Substitute $2$ for $A_1$ in above equation to find $A_2$.

$\begin{array}{c}{A}_2=1+2\\ =3\end{array}$

Substitute $2$ for $A_1$, and $3$ for $A_2$ in equation (27) to find $v\left(t\right)$.

$v\left(t\right)=\left[3+\left(2+3t\right)e^{-t}\right]\text{V}$

Conclusion:

Thus, the expression of $v\left(t\right)$ for the given differential equation is $\left[3+\left(2+3t\right)e^{-t}\right]\text{V}$.

To determine

Find the expression of $i\left(t\right)$ for the given differential equations.

Answer to Problem 29P

The expression of $i\left(t\right)$ for the given differential equation is $\left[2+2\cos \left(2t\right)e^{-t}\right]\text{A}$.

Explanation of Solution

Given data:

The differential equation is,

$\frac{d^{2}i}{d{t}^2}+2\frac{di}{dt}+5i=10$        (28)

The values of initial conditions are,

$\begin{array}{l}i\left(0\right)=4\\ \frac{di\left(0\right)}{dt}=-2\end{array}$

Formula used:

Write an expression to find the current response with the step input, if the roots of characteristic equation are real and imaginary.

$i\left(t\right)=\left[I_s+\left(A_1\cos {\omega }_{d}t+A_2\sin {\omega }_{d}t\right)e^{-\alpha t}\right]\text{A}$        (29)

Calculation:

From equation (28), the characteristic equation is written as follows.

$s^2+2s+5=0$        (30)

From the equation (6),

$\begin{array}{l}a=1\\ b=2\\ c=5\end{array}$

Substitute $1$ for $a$, $2$ for $b$, and $5$ for $c$ in equation (5) to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=\frac{-2\pm \sqrt{{\left(2\right)}^2-4\left(1\right)\left(5\right)}}{2}\\ =\frac{-2\pm \sqrt{-16}}{2}\\ =\frac{-2\pm j4}{2}\end{array}$

Simplify the equation.

$\begin{array}{c}{s}_{1,2}=\frac{-2\pm j4}{2},\frac{-2\pm j4}{2}\\ =-1+j2,-1-j2\end{array}$

Therefore, the roots of characteristic equations are real and imaginary.

$\begin{array}{l}{s}_1=-1+j2\\ s_2=-1-j2\end{array}$

Compare the values of roots with the equation (3) and equation (4).

$\begin{array}{l}\alpha =1\\ {\omega }_d=2\end{array}$

Substitute $1$ for $\alpha$, and $2$ for ${\omega }_d$ in equation (29) to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=I_s+\left(A_1\cos 2t+A_2\sin 2t\right)e^{-t}\\ =I_s+\left(A_1\cos 2t+A_2\sin 2t\right)e^{-t}\end{array}$

$i\left(t\right)=\left[I_s+\left(A_1\cos \left(2t\right)e^{-t}+A_2\sin \left(2t\right)e^{-t}\right)\right]\text{A}$        (31)

Substitute $i\left(\infty \right)$ for $i$ in equation (28).

$\frac{d^{2}i\left(\infty \right)}{d{t}^2}+2\frac{di\left(\infty \right)}{dt}+5i\left(\infty \right)=10$        (32)

For the step response,

$i\left(\infty \right)=I_s$

Substitute $0$ for $\frac{di\left(\infty \right)}{dt}$, $0$ for $\frac{d{i}^2\left(\infty \right)}{d{t}^2}$ and $I_s$ for $i\left(\infty \right)$ in equation (32) to find $I_s$.

$\begin{array}{c}0+2\left(0\right)+5I_s=10\\ 5I_s=10\\ I_s=\frac{10}{5}\\ =2\text{A}\end{array}$

Substitute $2\text{A}$ for $I_s$ in equation (31) to find $i\left(t\right)$.

$i\left(t\right)=\left[2+\left(A_1\cos \left(2t\right)e^{-t}+A_2\sin \left(2t\right)e^{-t}\right)\right]\text{A}$        (33)

Substitute $0$ for $t$ in equation (33) to find $i\left(0\right)$.

$\begin{array}{c}i\left(0\right)=\left[2+\left(A_1\cos \left(2\left(0\right)\right)e^{-\left(0\right)}+A_2\sin \left(2\left(0\right)\right)e^{-\left(0\right)}\right)\right]\text{A}\\ =\left[2+\left(A_1\cos 2\left(0\right)\left(1\right)+A_2\sin 2\left(0\right)\left(1\right)\right)\right]\text{A}\left\{\because e^0=1\right\}\\ =\left[2+\left(A_1\left(1\right)\left(1\right)+A_2\left(0\right)\left(1\right)\right)\right]\text{A}\left\{\because \cos 0°=1,\text{ and}\sin 0°=0\right\}\\ =\left[2+A_1\right]\text{A}\end{array}$

Substitute $4\text{A}$ for $i\left(0\right)$ in above equation to find $A_1$.

$\begin{array}{c}4\text{A}=\left[2+A_1\right]\text{A}\\ 4=2+A_1\end{array}$

Simplify the above equation to find $A_1$.

$\begin{array}{c}{A}_1=4-2\\ =2\end{array}$

Substitute $2$ for $A_1$ in equation (33) to find $i\left(t\right)$.

$i\left(t\right)=\left[2+\left(2\cos \left(2t\right)e^{-t}+A_2\sin \left(2t\right)e^{-t}\right)\right]\text{A}$        (34)

Differentiate equation (34) with respect to $t$ to find $\frac{di\left(t\right)}{dt}$.

$\frac{di\left(t\right)}{dt}=\left[\begin{array}{l}0+2\left(-\sin 2t\right)\left(2\right)e^{-t}+2\left(\cos 2t\right)\left(-1\right)e^{-t}+A_2\cos 2t\left(2\right)e^{-t}\\ +A_2\sin 2t{e}^{-t}\left(-1\right)\end{array}\right]\frac{\text{A}}{\text{s}}$

$\frac{di\left(t\right)}{dt}=\left[-4\text{sin 2}t{e}^{-t}-2\cos 2t{e}^{-t}+2A_2\cos 2t{e}^{-t}-A_2\sin 2t{e}^{-t}\right]\frac{\text{A}}{\text{s}}$ 

Substitute $0$ for $t$ in above equation to find $\frac{di\left(0\right)}{dt}$.

$\begin{array}{c}\frac{di\left(0\right)}{dt}=\left[\begin{array}{l}-4\text{sin }\left(\text{2}\left(0\right)\right)e^{-\left(0\right)}-2\cos \left(2\left(0\right)\right)e^{-\left(0\right)}+2A_2\cos \left(2\left(0\right)\right)e^{-0}\\ -A_2\sin \left(2\left(0\right)\right)e^{-0}\end{array}\right]\frac{\text{A}}{\text{s}}\\ =\left[\begin{array}{l}-4\left(0\right)\left(1\right)-2\left(1\right)\left(1\right)+2A_2\left(1\right)\left(1\right)\\ -A_2\left(0\right)\left(1\right)\end{array}\right]\frac{\text{A}}{\text{s}}\left\{\because e^0=1,\cos 0°=1,\text{ and}\sin 0°=0\right\}\\ =\left[-2+2A_2\right]\frac{\text{A}}{\text{s}}\end{array}$

Substitute $-2\frac{\text{A}}{\text{s}}$ for $\frac{di\left(0\right)}{dt}$ in above equation to find $A_2$.

$\begin{array}{l}-2\frac{\text{A}}{\text{s}}=\left[-2+2A_2\right]\frac{\text{A}}{\text{s}}\\ -2=-2+2A_2\\ 2A_2=-2+2\\ 2A_2=0\end{array}$

$A_2=0$

Substitute $0$ for $A_2$ in equation (34) to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=\left[2+\left(2\cos \left(2t\right)e^{-t}+\left(0\right)\sin \left(2t\right)e^{-t}\right)\right]\text{A}\\ \text{=}\left[2+2\cos \left(2t\right)e^{-t}\right]\text{A}\end{array}$

Conclusion:

Thus, the expression of $i\left(t\right)$ for the given differential equation is $\left[2+2\cos \left(2t\right)e^{-t}\right]\text{A}$.