A calculus problem by Rocco Dalto

Let $x(t) = A * e^{\lambda t}$ $y(t) = B * e^{\lambda t}$ $z(t) = C * e^{\lambda t}$

$\implies \dfrac{dx}{dt} = \lambda * A * e^{\lambda t}, \: \dfrac{dy}{dt} = \lambda * B * e^{\lambda t}, \: \dfrac{dx}{dt} = \lambda * C * e^{\lambda t}$

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Replacing the above in the differential system we obtain the system:

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$\begin{bmatrix}{3 - \lambda} && {4} && {2}\\ {-6} && {-7 - \lambda} && {-6}\\ {8} && {8} && {9 - \lambda} \end{bmatrix} * \begin{vmatrix}{A} \\{B} \\ {C} \\ \end{vmatrix} = \begin{vmatrix}{0} \\{0} \\ {0} \\ \end{vmatrix}$

$$

We want a nontrivial solution so let $\det(\begin{bmatrix}{3 - \lambda} && {4} && {2}\\ {-6} && {-7 - \lambda} && {-6}\\ {8} && {8} && {9 - \lambda} \end{bmatrix}) = 0$

$\implies$

$(\lambda - 5) * (\lambda - 1) * (\lambda + 1) = 0 \implies \lambda = 5, \: \lambda = 1, \: \lambda = -1,$

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For $\lambda = 5$ we obtain the system:

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$\left[ \begin{array}{ccc|c} -2 & 4 & 2 & 0\\ -6 & -12 & -6 & 0\\ 8 & 8 & 4 & 0 \\ \ \end{array} \right]$

Using row operations we obtain:

$A_{1} = 0$ , $B_{1} + \dfrac{1}{2} * C_{1} = 0$

Letting $C_{1} = -2 \implies B_{1} = 1 \implies$

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$x_{1}(t) = 0$ $y_{1}(t) = e^{5t}$ $z_{1}(t) = -2 * e^{5t}$

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For $\lambda = 1$ we obtain the system:

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$\left[ \begin{array}{ccc|c} 2 & 4 & 2 & 0\\ -6 & -8 & -6 & 0\\ 8 & 8 & 8 & 0 \\ \ \end{array} \right]$

Using row operations we obtain:

$A_{2} + C_{2} = 0$ $B_{2} = 0$

Letting $C _{2}= 1 \implies A_{2} = -1$ and $B_{2} = 0 \implies$

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$x_{2}(t) = -e^{t}$ $y_{2}(t) = 0$ $z_{2}(t) = e^{t}$

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For $\lambda = -1$ we obtain the system:

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$\left[ \begin{array}{ccc|c} 4 & 4 & 2 & 0\\ -6 & -6 & -6 & 0\\ 8 & -8 & 10 & 0 \\ \ \end{array} \right]$

Using row operations we obtain:

$A_{3} + B_{3} = 0$ $C_{3} = 0$

Letting $B_{3} = 1 \implies A_{3} = -1$ and $C_{3} = 0 \implies$

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$x_{3}(t) = -e^{-t}$ $y_{3}(t) = e^{-t}$ $z_{3}(t) = 0$

$\implies$

$x(t) = -c_{2} * e^{t} - c_{3} * e^{-t}$ $y(t) = c_{1} * e^{5t} + c_{3} * e^{-t}$ $z(t) = -2 * c_{1} * e^{5t} + c_{2} e^{t}$

$$

$x(0) = 0, y(0) = 2, z(0) = 3 \implies$

$-c_{2} - c_{3} = 0$ $c_{1} + c_{3} = 2$ $-2 * c_{1} + c_{2} = 3$

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$c_{3} = -c_{2} \implies$

$c_{1} - c_{2} = 2$ $-2 * c_{1} + c_{2} = 3$ .

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$\implies c_{1} = -5, \: c_{2} = -7 \implies \: c_{3} = 7 \implies$

$$

$c_{1} + c_{2} + c_{3} = \boxed{-5}$ .