y^{''}-3y^'+2y=3cos(3t),y(0)=2,y^'(0)=3

解

y^{^{''}} - 3 y^{^{'}} + 2 y = 3 cos (3 t), y (0) = 2, y^{^{'}} (0) = 3

y = (- \frac{7}{10} + \frac{281}{130}) e^{2 t} + \frac{7}{10} e^{t} - \frac{27}{130} sin (3 t) - \frac{21}{130} cos (3 t)

解答ステップ

y^{''} (t) - 3 y^{'} (t) + 2 y = 3 cos (3 t)

線形 ODE を解く:

y = (- \frac{7}{10} + \frac{281}{130}) e^{2 t} + \frac{7}{10} e^{t} - \frac{27}{130} sin (3 t) - \frac{21}{130} cos (3 t)

y = (- \frac{7}{10} + \frac{281}{130}) e^{2 t} + \frac{7}{10} e^{t} - \frac{27}{130} sin (3 t) - \frac{21}{130} cos (3 t)