A calculus problem by Rocco Dalto

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

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

$$

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

$$

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

$$

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

$\implies (\lambda + 1)^2 * (8 - \lambda) = 0 \implies \lambda = -1, \: \lambda = 8$

$$

For $\lambda = 8$ we obtain the system:

$$

$\left[ \begin{array}{ccc|c} -5 & 2 & 4 & 0\\ 2 & -8 & 2 & 0\\ 4 & 2 & -5 & 0 \\ \ \end{array} \right]$

Using row operations we obtain:

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

$\implies$

$A{1} - 4 B_{1} + C_{1} = 0$ $-2 B_{1} + C_{1} = 0$

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

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

For $\lambda = -1$ we obtain:

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

Using row operations we obtain:

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

$\implies$

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

We have two degrees of freedom so let $A_{2}$ and $C_{2}$ be arbitrary variables $\implies B_{2} = -2 A{2} - 2 C_{2}$ .

For $A_{2} = 1, \: C_{2} = 0 \implies B_{2} = -2$ and $\vec{v_{2}} = \begin{vmatrix}{1} \\{-2} \\ {0} \\ \end{vmatrix}$

For $A_{2} = 0, \: C_{2} = 1 \implies B_{2} = -2$ and $\vec{v_{3}} = \begin{vmatrix}{0} \\{-2} \\ {1} \\ \end{vmatrix}$

Every solution has the form:

$\begin{vmatrix}{A_{2}} \\{B_{2}} \\ {C_{2}} \\ \end{vmatrix} = A_{2} \begin{vmatrix}{1} \\{-2} \\ {0} \\ \end{vmatrix} + C_{2} \begin{vmatrix}{0} \\{-2} \\ {1} \\ \end{vmatrix}$

$\therefore$ The general solution is:

$x(t) = 2 c_{1} e^{8t} + c_{2} e^{-t}$ $y(t) = c_{1} e^{8t} - 2 c_{2} e^{-t} - 2 c_{3} e^{-t}$ $z(t) = 2 c_{1} e^{8t} + c_{3} e^{-t}$

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

$(1): 2 c_{1} + c_{2} = 0$ $(2): c_{1} - 2 c_{2} - 2 c_{3} = 1$ $(3): 2 c_{1} + c_{3} = 2$

Multiplying equation $(1)$ by $2$ and adding to equation $(2)$ we obtain:

$5 c_{1} - 2 c_{3} = 1$ $2 c_{1} + c_{3} = 2$

$\implies c_{1} = \dfrac{5}{9} \implies c_{3} = \dfrac{8}{9} \implies c_{2} = \dfrac{-10}{9} \implies c_{1} + c_{2} + c_{2} = \dfrac{3}{9} = \dfrac{1}{3} = \dfrac{a}{b}$

$\implies a + b = \boxed{4}$ .