Question

Решить систему дифференциальных уравнений

Answer 1

$\left \{ {{x' = 2x + 3y} \atop {y' = x + 4y} } \right. \\ \\ \left \{ {{x' = 2x + 3y} \atop {x= y' - 4y} } \right. \\ x'= y'' - 4y'\\ \\ y''- 4y'= 2(y' - 4y) + 3y \\ y'' - 4y' = 2y' - 8y + 3y \\ y''- 6y' + 5y = 0 \\ \\ y = {e}^{kt} \\ \\ k {}^{2} - 6k + 5 = 0\\ D = 36 - 20 = 16\\ k_1 = \frac{6 + 4}{2} = 5 \\ k_2 = 1 \\ \\ y = C_1 {e}^{5t} + C_2 {e}^{t}$

Получаем:

$\left \{ {{y = C_1 {e}^{5t} + C_2e {}^{t} } \atop { x = y'- 4y} } \right. \\ \\ y' = 5C_1 {e}^{5t} + C_2 {e}^{t} \\ \\ x = 5C_1 {e}^{5t} + C_2e {}^{t} - 4 C_1 {e}^{5t} - 4C_2e {}^{t} = \\ = C_1e {}^{5t} - 3C_2e {}^{t}$

$\left \{ {{y = C_1e {}^{5t} + C_2e {}^{ {t}^{} } } \atop {x = C_1e {}^{5t} - 3C_2e {}^{t} } } \right. \\$

- общее решение