Question

Multiple Choice: Let A = \begin{bmatrix} -1 & 0 & -1\\ 0 & 1 & 0\\ 4 & 0 & -1 \end{bmatrix} . Let x be the solution of the following initial value problem:

x' = Ax, x(0) = \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix} .

What is the value of ln(x(\pi))?

(a) \begin{bmatrix} -\pi \\ -\pi \\ \pi \end{bmatrix}    (b) \begin{bmatrix} 2\pi \\ 0 \\ -\pi \end{bmatrix}    (c) \begin{bmatrix} -\pi \\ 0 \\ 1 \end{bmatrix}    (d) \begin{bmatrix} \pi \\ -\pi \\ 3\pi \end{bmatrix}    (e) \begin{bmatrix} -\pi \\ \pi \\ -\pi \end{bmatrix}

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Answer #1

The answer is option(d). Look at the handwritten detailed solution.

Few Points:

1. If Roots of complimentary equation is of the form (a \pm i*b) (i.e complex), then the solution is of the form

   x(t) = e^{-at}(Fsin(bt)+Dcos(bt)) . This is used in finding the solution.

We have A-[-o-17 I 4 0 - 1 differential We have the equation. X Co x = AX . I withe initial value Let then xs dat 92 I dal 9Now Cup have two coupled differential equation aj: -4,-23 – ☺ az = 48,-43.- Now, Jorz = -a, -i, -o n Putting in we have (-2,Thus, x, (t) = pt (cosat - Sinat , as (t) - ae* (cosat +Sigat) I general Solution NOU *, (*)=en az (a) = en az (a) = 26 x 7 =

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