Question

Solve the system in Problem 7 subject to the initial condition

14.$X\left(0\right)=\left(\begin{array}{c}4\\ 3\end{array}\right)$

Short answer

The solution with the initial condition is$X=7\left(\begin{array}{c}1\\ 0\end{array}\right)e^t+\frac{5}{2}\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t}+\left(\begin{array}{c}-3\\ \frac{1}{2}\end{array}\right)$

Step-by-step solution

Definition of Laplace transform.

• The integral transform of a given derivative function with real variable t into a complex function with variable s is known as the Laplace transform.
• Let f(t) be supplied for t 0, and assume that the function meets certain constraints that will be presented subsequently.
• The Laplace transform formula defines the Laplace transform of f(t), which is indicated by Lf(t) or F(s):
• $e^{At}=L^{-1}\left\{{(sI-A)}^{-1}\right\}$
  

Determine the value with the initial condition.

We previously obtained

$X=7\left(\begin{array}{c}1\\ 0\end{array}\right)e^t+\frac{5}{2}\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t}+\left(\begin{array}{c}-3\\ \frac{1}{2}\end{array}\right)$

Applying the initial condition,$X\left(0\right)=\left(\begin{array}{c}4\\ 3\end{array}\right)$

$$\begin{gathered} \left(\begin{array}{c}4\\ 3\end{array}\right)=C_3\left(\begin{array}{c}1\\ 0\end{array}\right)e^0+C_4\left(\begin{array}{c}0\\ 1\end{array}\right)e^0+\left(\begin{array}{c}-3\\ \frac{1}{2}\end{array}\right) \\ =C_3\left(\begin{array}{c}1\\ 0\end{array}\right)+C_4\left(\begin{array}{c}0\\ 1\end{array}\right)+\left(\begin{array}{c}-3\\ \frac{1}{2}\end{array}\right) \end{gathered}$$

Here we have,

$$\begin{gathered} 4=C_3-3\to C_3=7 \\ 3=C_4+\frac{1}{2}\to C_4=\frac{5}{2} \end{gathered}$$

Therefore,

$X=7\left(\begin{array}{c}1\\ 0\end{array}\right)e^t+\frac{5}{2}\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t}+\left(\begin{array}{c}-3\\ \frac{1}{2}\end{array}\right)$

Hence, the solution with the initial condition is

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