y^{''}-3y^'=(3t+8)sin(3t)

解

y^{^{''}} - 3 y^{^{'}} = (3 t + 8) sin (3 t)

y = c_{1} e^{3 t} + c_{2} + \frac{t cos ( 3 t )}{6} - \frac{t sin ( 3 t )}{6} + \frac{7}{18} cos (3 t) - \frac{5}{9} sin (3 t)

解答ステップ

y^{''} (t) - 3 y^{'} (t) = (3 t + 8) sin (3 t)

線形 ODE を解く:

y = c_{1} e^{3 t} + c_{2} + \frac{t cos ( 3 t )}{6} - \frac{t sin ( 3 t )}{6} + \frac{7}{18} cos (3 t) - \frac{5}{9} sin (3 t)

y = c_{1} e^{3 t} + c_{2} + \frac{t cos ( 3 t )}{6} - \frac{t sin ( 3 t )}{6} + \frac{7}{18} cos (3 t) - \frac{5}{9} sin (3 t)