Question

Solve the system in Problem 7 subject to the initial condition

13$\mathit{X}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\left(\begin{array}{c}1\\ -4\\ 6\end{array}\right)$.

Short answer

The solution with the initial condition is$X=\left(\begin{array}{c}t+1\\ t\\ -2t\end{array}\right)-4\left(\begin{array}{c}t\\ t+1\\ -2t\end{array}\right)+6\left(\begin{array}{c}t\\ t\\ -2t+1\end{array}\right)$

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):
• $e^{At}=L^{-1}{(sI-A)}^{-1}$
  

Determine the value with the initial condition.

We previously obtained

$X=C_1\left(\begin{array}{c}t+1\\ t\\ -2t\end{array}\right)+C_2\left(\begin{array}{c}t\\ t+1\\ -2t\end{array}\right)+C_3\left(\begin{array}{c}t\\ t\\ -2t+1\end{array}\right)$

Applying the initial condition

$X\left(0\right)=\left(\begin{array}{c}1\\ -4\\ 6\end{array}\right)$

$$\begin{gathered} \left(\begin{array}{c}1\\ -4\\ 6\end{array}\right)=C_1\left(\begin{array}{c}0+1\\ 0\\ -2\left(0\right)\end{array}\right)+C_2\left(\begin{array}{c}0\\ 0+1\\ -2\left(0\right)\end{array}\right)+C_3\left(\begin{array}{c}0\\ 0\\ -2\left(0\right)+1\end{array}\right) \\ =C_1\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)+C_2\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)+C_3\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right) \end{gathered}$$

So here we have,

C₁=1, C₂=-4 and C₃=6

Therefore,

$X=\left(\begin{array}{c}t+1\\ t\\ -2t\end{array}\right)-4\left(\begin{array}{c}t\\ t+1\\ -2t\end{array}\right)+6\left(\begin{array}{c}t\\ t\\ -2t+1\end{array}\right)$

Hence, the solution with the initial condition is

$X=\left(\begin{array}{c}t+1\\ t\\ -2t\end{array}\right)-4\left(\begin{array}{c}t\\ t+1\\ -2t\end{array}\right)+6\left(\begin{array}{c}t\\ t\\ -2t+1\end{array}\right)$