## Chapter 2

## Q. 2.7.1

## Q. 2.7.1

A Third-Order Equation

Obtain the solution of the following problem:

10 \frac{d^3x}{dt^3}+100 \frac{d^2x}{dt^2}+310 \frac{dx}{dt}+300x=750 u_s(t) \\ x(0)=2 \quad \overset{.}{x}(0)=4 \quad \overset{..}{x}=3

## Step-by-Step

## Verified Solution

Using the Laplace transform method, we have

10[s^3X(s)-\overset{..}{x}(0)-s\overset{.}{x}(0)-s^2x(0)]+100[s^2X(s)-\overset{.}{x}(0)-sx(0)] + 310 [sX(s)-x(0)]+300 X(s)=\frac{750}{s}

Solving for X(s) using the given initial values, we obtain

X(s)=\frac{2s^3+24s^2+105s+75}{s(s^3+10s^2+31s+30)}= \frac{2s^3+24s^2+105s+75}{s(s+2)(s+3)(s+5)}

Since the roots of the cubic are s= −2, −3, and −5, the partial-fraction expansion is

X(s)=\frac{C_1}{s}+\frac{C_2}{s+2}+\frac{C_3}{s+3}+\frac{C_4}{s+5}

For this problem the Least Common Denominator (LCD) method requires a lot of algebra, and the coefficients can be more easily obtained from the formula (2.7.4). They are

C_i=\underset{s \rightarrow -r_i}{\text{lim}}[X(s)(s+r_i)] \quad (2.7.4)

C_1=\underset{s \rightarrow 0}{\text{lim}} sX(s)=\frac{5}{2} \\ C_2=\underset{s \rightarrow -2}{\text{lim}} (s+2)X(s)=\frac{55}{6} \\ C_3=\underset{s \rightarrow -3}{\text{lim}} (s+3)X(s)= -13 \\ C_4=\underset{s \rightarrow -5}{\text{lim}} (s+5)X(s)=\frac{10}{3}

Thus, the answer is

x(t)=\frac{5}{2}+\frac{55}{6} e^{-2t}-13e^{-3t}+\frac{10}{3} e^{-5t}

The plot of the response is shown in Figure 2.7.1. The response contains three exponentials. The terms e^{−3t} and e^{−5t} die out faster than e^{−2t}, so for t > 4/3, the response is approximately given by x(t) = 5/2+(55/6) e^{−2t}. For t > 2, the response is approximately constant at x = 5/2. The “hump” in the response is produced by the positive values of ẋ(0) and ẍ(0).