A calculus problem by Rocco Dalto

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

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

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

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$\begin{bmatrix}{1 - \lambda} && {2} && {4} && {3} \\ {0} && {- \lambda} && {2} && {1} \\ {0} && {0} && {1 - \lambda} && {1} \\ {0} && {0} && {0} && {2 - \lambda}\end{bmatrix} * \begin{vmatrix}{A} \\{B} \\ {C} \\ {D} \\ \end{vmatrix} = \begin{vmatrix}{0} \\{0} \\ {0} \\ {0} \end{vmatrix}$

We want a nontrivial solution so let

$\det\begin{bmatrix}{1 - \lambda} && {2} && {4} && {3} \\ {0} && {- \lambda} && {2} && {1} \\ {0} && {0} && {1 - \lambda} && {1} \\ {0} && {0} && {0} && {2 - \lambda}\end{bmatrix} = 0$

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

For $\lambda = 0$ we obtain the system:

$\left[ \begin{array}{cccc|c} 1 & 2 & 4 & 3 & 0\\ 0 & 0 & 2 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 & 0 \\ \ \end{array} \right]$

Applying row operations we obtain:

$\left[ \begin{array}{cccc|c} 1 & 2 & 4 & 3 & 0\\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ \ \end{array} \right]$

$D_{1} = 0 \implies C_{1} = 0 \implies$ $A_{1} + 2 B_{1} = 0$

Let $B_{1} = 1 \implies A_{1} = -2 \implies$

$\vec{v_{1}} = \begin{vmatrix}{-2} \\{1} \\ {0} \\ {0} \end{vmatrix}$

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

$\left[ \begin{array}{cccc|c} -1 & 2 & 4 & 3 & 0\\ 0 & -2 & 2 & 1 & 0 \\ 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \ \end{array} \right]$

Applying row operations we obtain:

$\left[ \begin{array}{cccc|c} -1 & 0 & 0 & 10 & 0\\ 0 & 1 & 2 & -\dfrac{3}{2} & 0 \\ 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \ \end{array} \right]$

$-A_{2} + 10 D_{2} = 0$ $B_{2} - \dfrac{3}{2} D_{2} = 0$ $C_{2} - D_{2} = 0$

Let $D_{2} = 1 \implies C_{2} = 1, B_{2} = \dfrac{3}{2}, A_{2} = 10$

$\implies \vec{v_{2}} = \begin{vmatrix}{10} \\{\dfrac{3}{2}} \\ {1} \\ {1} \end{vmatrix}$

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

$\left[ \begin{array}{cccc|c} 0 & 2 & 4 & 3 & 0\\ 0 & -1 & 2 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ \ \end{array} \right]$

$D_{3} = 0 \implies$ $2 B_{3} + 4 C_{3} = 0$ $-B_{3} + 2 C_{3} = 0$

$\implies B_{3} = C_{3} = 0$

Let $A_{3} = 1$

$\implies \vec{v_{3}} = \begin{vmatrix}{1} \\{0} \\ {0} \\ {0} \end{vmatrix}$

Let $A_{4 \: X \: 4} = \begin{bmatrix}{1} && {2} && {4} && {3} \\ {0} && {0} && {2} && {1} \\ {0} && {0} && {1} && {1} \\ {0} && {0} && {0} && {2}\end{bmatrix}$

For $(A_{4 \: X \: 4} - I) \vec{v_{4}} = \vec{v_{3}}$

$D_{4} = 0 \implies$ $2 B_{3} + 4 C_{3} = 1$ $-B_{3} + 2 C_{3} = 0$

$\implies C_{4} = \dfrac{1}{8} \implies \dfrac{1}{4}$

Let $A_{4} = 1$

$\implies \vec{v_{4}} = \begin{vmatrix}{1} \\{\dfrac{1}{4}} \\ {\dfrac{1}{8}} \\ {0} \end{vmatrix}$

$\therefore$ The general solution is:

$x(t) = -2 c_{1} + 10 c_{2} e^{2t} + c_{3} e^{t} + (t + 1) c_{4} e^{t}$ $y(t) = c_{1} + \dfrac{3}{2} c_{2} e^{2t} + \dfrac{1}{4} c_{4} e^{t}$ $z(t) = c_{2} e^{2t} + \dfrac{1}{8} c_{4} e^{t}$ $w(t) = c_{2} e^{2t}$

$x(0) = 0$ , $y(0) = 1$ , $z(0) = 2$ , and $w(0) = 0 \implies$

$-2 c_{1} + 10 c_{2} + + c_{3} + c_{4} = 0$ $c_{1} + \dfrac{3}{2} c_{2} + \dfrac{1}{4} c_{4} = 1$ $c_{2} + \dfrac{1}{8} c_{4} = 2$ $c_{2} = 0$

$c_{2} = 0 \implies c_{4} = 16 \implies c_{1} = -3 \implies c_{3} = -22$

$\implies c_{1} + c_{2} + c_{3} + c_{4} = \boxed{-9}.$