Question

2.13 помогите пожалуйста

Answer 1

$\left \{ {{x'= 8x - 3y} \atop {y' = 2x + y} } \right. \\ \\ \left \{ {{y = \frac{8x - x'}{3} } \atop {y' = 2x + y} } \right. \\ y' = \frac{8x' - x''}{3} \\ \\ \frac{8x' - x''}{3} = 2x + \frac{8x - x'}{3} \\ 8x'- x''= 6x + 8x - x'\\ 8x'- x''- 14x + x' = 0 \\ - x'' + 9x' - 14x = 0 \\ x'' - 9x'+ 14x = 0 \\ \\ x = {e}^{kt} \\ \\ k {}^{2} - 9 k + 14 = 0 \\ D= 81 - 56 = 25 \\ k_1 = \frac{9 + 5}{2} = 7 \\ k_2 = 2 \\ \\ x = C_1 {e}^{2t} + C_2 {e}^{7t} \\ \\ \\ \left \{ {{x = C_1 {e}^{2t} + C_2 {e}^{7t} } \atop {y = \frac{8x - x'}{3} } } \right. \\ \\ x' = 2C_1 {e}^{2t} + 7 C_2 {e}^{7t} \\ \\ y = \frac{1}{3} (8(C_1 {e}^{27} + C_2 {e}^{7t} ) - (2C_1 {e}^{2t} + 7C_2 {e}^{7t} ) = \\ = \frac{1}{3} (8C_1 {e}^{2t} + 8C_2 {e}^{7t} - 2C_1 {e}^{2t} - 7C_2 {e}^{7t} ) = \\ = \frac{1}{3} (6C_1 {e}^{2t} + C_2 {e}^{7t} ) = 2C_1 {e}^{2t} + \frac{1}{3} C_2 {e}^{7t}$

Ответ:

$\left \{ {{x = C_1 {e}^{2t} + C_2 {e}^{7t} } \atop {y = 2C_1 {e}^{2t} + \frac{1}{3} C_2 {e}^{7t} } } \right.$