Question

y''-3y'-10y=sinx+3cos

Answer 1

Ответ:

$y = C_1e^{5x} + C_2e^{-2x} -\cfrac{3}{13}\cos{x} - \cfrac{2}{13}\sin{x}$

Пошаговое объяснение:

$y'' -3y' -10y = \sin{x}+3\cos{x} \\\\ y'' -3y' -10y = 0 \\\\ k^2 -3k - 10 = 0 \\\\ (k+2)(k-5) = 0\\\\k_1 = 5, \ k_2 = -2\\\\ Y = C_1e^{5x} + C_2e^{-2x}\\\\ \overline{y} = A\cos{x} + B\sin{x} \\\\ \overline{y}' = -A\sin{x} + B\cos{x} \\\\ \overline{y}'' = -A\cos{x} - B\sin{x} \\\\ \overline{y}'' - 3\overline{y}' - 10\overline{y} = \sin{x} +3\cos{x} \\\\ -A\cos{x} - B\sin{x} +3A\sin{x} - 3B\cos{x} -10A\cos{x} - \\\\ -10B\sin{x} = \sin{x} + 3\cos{x} \\\\ (-11A-3B)\cos{x} + (3A-11B)\sin{x} = \sin{x} + 3\cos{x}$

$\displaystyle \left \{ {{-11A-3B = 3} \atop {3A-11B = 1}} \right. \\\\\\ \left \{ {{-11A-3B = 3} \atop {9A-33B = 3}} \right. \\\\\\ \left \{ {{-11A-3B = 3} \atop {-20A+30B = 0}} \right. \\\\\\ \left \{ {{-11A-2A = 3} \atop {B=\cfrac{2}{3}A}} \right. \\\\\\ \left \{ {{A = -\cfrac{3}{13}} \atop {B=-\cfrac{2}{13}}} \right.$

$\overline{y} = -\cfrac{3}{13}\cos{x} - \cfrac{2}{13}\sin{x} \\\\ y = Y + \overline{y} = C_1e^{5x} + C_2e^{-2x} -\cfrac{3}{13}\cos{x} - \cfrac{2}{13}\sin{x}$