## Chapter 2

## Q. 2.7.8

## Q. 2.7.8

Free Response of a Second-Order Model with Complex Roots

Use the Laplace transform to solve the following problem:

4 \overset{..}{x}+24\overset{.}{x}+136x=0 \quad x(0)=\frac{7}{4} \quad \overset{.}{x}(0)=-\frac{11}{4}

## Step-by-Step

## Verified Solution

Taking the transform of both sides of the equation, we obtain

4[s^2X(s)-x(0)s-\overset{.}{x}(0)]+24[sX(s)-x(0)]+136X(s)=0

Solve for X(s) using the given values of x(0) and ẋ(0).

X(s)=\frac{4[x(0)s+\overset{.}{x}(0)]+24x(0)}{4s^2+24s+136}= \frac{3s+7}{4(s^2+6s+34)}

The expansion was obtained in part (a) of Example 2.7.7. It is

X(s)=-\frac{1}{10}[\frac{5}{(s+3)^2+5^2}] + \frac{3}{4}[\frac{s+3}{(s+3)^2+5^2}]

and the inverse transform is

x(t)=-\frac{1}{10}e^{-3t} \sin 5t+\frac{3}{4}e^{-3t} \cos 5t