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$$ \begin{aligned} x'' &= 2 x' + 6y + 3 \\ y' &= - x' - 2y \end{aligned}$$

subject the the initial condition $x(0) = 0$, $x'(0) = 0$ and $y(0) = 1$.

The first part of the question is about finding the matrix exponential $e^{At}$ of the following matrix

$$A = \begin{bmatrix} 0 & 1& 0 \\[0.3em] 0 & 2 & 5 \\[0.3em] 0 & -1 & -2 \end{bmatrix}$$

and I already found $e^{At}$. I observed that entries of the matrix is similar to the question but not sure how it works. How to proceed after finding $e^{At}$?

Rodrigo de Azevedo
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Linh Phan
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  • Set $u = x'$ and rewrite the system as $u' = 2u + 6y + 3, y' = -u - 2y$. The matrix for this system is $2 \times 2$, its trace is zero and its determinant is 6. Therefore its eigenvalues are $\pm \sqrt{6}i$. What is the general form of the exponential of such a matrix? Once you have found $y(t)$ and $u(t)$, you can then find $x(t) = \int_0^t u(s) ds$. – Hans Engler Apr 28 '25 at 20:22
  • Possibly related – Rodrigo de Azevedo Apr 28 '25 at 23:01

1 Answers1

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we can look for a particular solution of the form $$x = a, x' = b, y = c \to 0 = 2b + 6c + 3, \, 0 = -b - 2c \to a = 0,b = 3, c = -\frac 32 $$

now, make a change of variable $u = x, v = x'-3, w = y + \frac 32.$ so that $$\pmatrix{u'&v'&w'}^T= A\pmatrix{u&v&w}^T, u = 0, v = -3, w = \frac 52 \text{ at } t = 0 \\\to (u,v,w)^T = e^{At}(0,-3,\frac 52)^T, (x,x',y) = (0,3, -\frac 52)^T + e^{At}(0,-3,\frac 52)^T.$$

abel
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