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