Question

Помогите решить диф. уравнение

Answer 1

Ответ:

$x'= x - 4y \\ y'= x + y \\ \\ x' = x - 4y \\ x = y' - y \\ \\ x' = y'' - y'$

$y'' - y'= y' - y - 4y \\ y'' - 2y' + 5y = 0 \\ y = {e}^{kt} \\ {e}^{kt} ( {k}^{2} - 2k + 5) = 0 \\ d = 4 - 20 = - 16 \\ k1 = \frac{2 + 4i}{2} = 1 + 2i \\ k2 = 1 - 2i \\ y = {e}^{x} (C1 \sin(2t) + C2 \cos(2t) )$

получаем:

$x = y'- y \\ y = {e}^{x} (C1 \sin(2t) + C2 \cos(2t) ) \\ \\ y' = {e}^{x} (C1 \sin(2t) + C2 \cos(2t)) + {e}^{x} (2C1 \cos(2t) - 2C2 \sin(2t) ) = \\ = {e}^{x} (C1 \sin(2t) + C2 \cos(2t) + 2 C1 \cos(2t) - 2C2 \sin(2t) )$

$x = {e}^{x} (C1 \sin(2t) + C2 \cos(2t) - C1 \sin(2t) - C \cos(2t) - 2C1 \cos(2t) + 2C2 \sin(2t) ) \\ x = {e}^{x} (2C2 \sin(2t) - 2C1 \cos(2t))$

Ответ:

$y = {e}^{x} (C1 \sin(2t) + C2 \cos(2t) ) \\ x = {e}^{x} (2C2 \sin(2t) - 2C1 \cos(2t))$