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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$\begin{bmatrix}{3 - \lambda} && {1} && {-2}\\ {1} && {1 - \lambda} && {0}\\ {1} && {0} && {1 - \lambda} \end{bmatrix} * \begin{vmatrix}{A} \\{B} \\ {C} \\ \end{vmatrix} = \begin{vmatrix}{0} \\{0} \\ {0} \\ \end{vmatrix}$

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We want a nontrivial solution so let $\det(\begin{bmatrix}{3 - \lambda} && {1} && {-2}\\ {1} && {1 - \lambda} && {0}\\ {1} && {0} && {1 - \lambda} \end{bmatrix}) = 0$

$\implies \lambda^3 - 5 \lambda^2 + 8 \lambda - 4 = 0 \implies$

$(\lambda - 1) * (\lambda - 2)^2 = 0 \implies \lambda = 1, \: \lambda = 2$

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

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$\left[ \begin{array}{ccc|c} 2 & 1 & -2 & 0\\ 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 \\ \ \end{array} \right]$

$\implies$

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$A_{1} = 0$

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$B_{1} - 2 C_{1} = 0$

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

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

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For $\lambda = 2$ we obtain the system:

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$\left[ \begin{array}{ccc|c} 1 & 1 & -2 & 0\\ 1 & -1 & 0 & 0\\ 1 & 0 & -1 & 0 \\ \ \end{array} \right]$

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Let $A_{2} = B_{2} \implies A_{2} - C_{2} = 0$

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

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

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Replacing $\begin{vmatrix}{x_{3}(t)} \\{y_{3}(t)} \\ {z_{3}(t)} \\ \end{vmatrix} = \begin{vmatrix}{A_{3} + 2 * t} \\{B_{3} + 2 * t}\\ {C_{3} + 2 * t} \\ \end{vmatrix} * e^{\lambda_{2}t }$ into the differential system we obtain:

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$\begin{bmatrix}{1} && {1} && {-2}\\ {1} && {-1} && {0}\\ {1} && {0} && {-1} \end{bmatrix} * \begin{vmatrix}{A_{3}} \\{B_{3}} \\ {C_{3}} \\ \end{vmatrix} = \begin{vmatrix}{1} \\{1} \\ {1} \\ \end{vmatrix}$

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$\implies$

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

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Letting $A_{3} = 2 \implies C_{3} = 1 \implies B_{3} = 1$

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$x_{3}(t) = 2 e^{2t}$ $y_{3}(t) = e^{2t}$ $z_{3}(t) = e^{2t}$

$\therefore$ The general solution is:

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$\begin{cases} x(t) = c_{2} e^{2t} + c_{3} (2 + t) e^{2t} \\ y(t) = 2 c_{1} e^{t} + c_{2} e^{2t} + c_{3} (1 + t) e^{2t} \\ z(t) = c_{1} e^{t} + c_{2} e^{2t} + c_{3} (1 + t) e^{2t} \end{cases}$

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

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

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Let $c_{2} = -2 c_{3} \implies$ $2 c_{1} - c_{3} = 2$ $c_{1} - c_{3} = 3$

$\implies c_{1} = -1 \implies c_{3} = -4 \implies c_{2} = 8 \implies c_{1} + c_{2} + c_{3} = \boxed{3}$