A calculus problem by Rocco Dalto

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

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

$$

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

$$

$\begin{bmatrix}{5 - \lambda} && {-3} && {-2}\\ {8} && {-5 - \lambda} && {-4}\\ {-4} && {3} && {3 - \lambda} \end{bmatrix} * \begin{vmatrix}{A} \\{B} \\ {C} \\ \end{vmatrix} = \begin{vmatrix}{0} \\{0} \\ {0} \\ \end{vmatrix}$

We want a nontrivial solution so let $\det(\begin{bmatrix}{5 - \lambda} && {-3} && {-2}\\ {8} && {-5 - \lambda} && {-4}\\ {-4} && {3} && {3 - \lambda} \end{bmatrix}) = 0$

$\implies -(\lambda^3 - 3 \lambda^2 + 3 \lambda - 1) = 0 \implies$

$(\lambda - 1)^3 = 0 \implies \lambda = 1$

$$

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

$$

$\left[ \begin{array}{ccc|c} 4 & -3 & -2 & 0\\ 8 & -6 & -4 & 0\\ -4 & 3 & 2 & 0 \\ \ \end{array} \right]$

Applying the row operations:

$\dfrac{1}{2} * Row_{2}$

$-1 * Row_{1} + Row_{3}$

$Row_{1} + Row_{2}$

we obtain:

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

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

We have two degrees of freedom.

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

Let $B_{1} = 0$ and $C_{1} = 6 \implies A_{1} = 3$

$\implies$

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

$\vec{v_{2}} = \begin{vmatrix}{3} \\ 0 \\ 6 \\ \end{vmatrix}$

For the third solution we want:

$t e^{\lambda t} \vec{\alpha} + e^{\lambda t} \vec{\beta}$ that satisfies

$(A_{3 x 3} - \lambda I) \vec{\alpha} = \vec{0}$ and $(A_{3 x 3} - \lambda I) \vec{\beta} = \vec{0}$

where,

$\vec{\alpha} = a \vec{v_{1}} + b \vec{v_{2}} = \begin{vmatrix}{3 a + 3 b} \\ 2 a \\ 3 a + 6 b\\ \end{vmatrix}$ .

$\vec{\alpha}$ satisfies $(A_{3 x 3} - \lambda I) \vec{\alpha} = \vec{0}$

For $(A_{3 x 3} - \lambda I) \vec{\beta} = \vec{\alpha}$ we obtain:

$4 A^{*} - 3 B^{*} - 2 C^{*} = 3 a + 3 b$ $8 A^{*} - 6 B^{*} - 4 C^{*} = 2 a$ $-4 A^{*} + 3 B^{*} + 2 C^{*} = 3 a + 6 b$

$\implies 2 a + 3 b = 0$ .

Let $b = 2 \implies a = -3 \implies$

$\vec{\alpha} = \begin{vmatrix}{-3} \\ -6 \\ 3 \\ \end{vmatrix}$

$\implies$

$\begin{bmatrix}{4} && {-3} && {-2}\\ {8} && {-6} && {-4}\\ {-4} && {3} && {2} \end{bmatrix} * \vec{\beta} = \begin{vmatrix}{-3} \\{-6} \\ {3} \\ \end{vmatrix}$

Applying row operations:

$\dfrac{1}{2} * Row_{2}$

$-1 * Row_{1} + Row_{3}$

$Row_{1} + Row_{2}$

we obtain:

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

$\implies 4 A^{*} - 3 B^{*} - 2 C^{*} = -3$

Let $C^{*} = 0$ and $B^{*} = 3 \implies A^{*} = \dfrac{3}{2}$

$\implies$

$\vec{\beta} = \begin{vmatrix}{\dfrac{3}{2}} \\ 3 \\ 0 \\ \end{vmatrix}$ .

$\therefore$ The general solution is:

$\begin{cases} x(t) = (3 c_{1} + 3 c_{2} + (-3 t + \dfrac{3}{2}) c_{3}) e^{t} \\ y(t) = (2 c_{1} + c_{2} + (-6 t + 3) c_{3}) e^{t} \\ z(t) = (3 c_{1} + 6 c_{2} + 3 t c_{3}) e^{t} \end{cases}$

$$

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

$6 c_{1} + 6 c_{2} + 3 c_{3} = 0$ $2 c_{1} + 3 c_{3} = 1$ $3 c_{1} + 6 c_{2} = 2$

$\implies$

$-6 c_{2} + 3 c_{3} = -4$ $6 c_{2} - 6 c_{3} = -3$

$\implies$

$c_{3} = \dfrac{7}{3} \implies c_{2} = \dfrac{11}{6} \implies c_{1} = -3 \implies$

$c_{1} + c_{2} + c_{3} = \dfrac{7}{6} = \dfrac{m}{n} \implies m + n = \boxed{13}$