y^{''}+4y=sin(x)*sin(2x)

解

y^{^{''}} + 4 y = sin (x) \cdot sin (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{2} cos (2 x) (- \frac{1}{5} cos (5 x) - \frac{1}{3} cos (3 x) - sin^{2} (2 x) cos (x)) + \frac{1}{8} sin (2 x) (- \frac{1}{5} sin (5 x) + \frac{1}{3} sin (3 x))

解答ステップ

y^{''} (x) + 4 y = sin (x) \cdot sin (2 x)

線形 ODE を解く:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{2} cos (2 x) (- \frac{1}{5} cos (5 x) - \frac{1}{3} cos (3 x) - sin^{2} (2 x) cos (x)) + \frac{1}{8} sin (2 x) (- \frac{1}{5} sin (5 x) + \frac{1}{3} sin (3 x))

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{2} cos (2 x) (- \frac{1}{5} cos (5 x) - \frac{1}{3} cos (3 x) - sin^{2} (2 x) cos (x)) + \frac{1}{8} sin (2 x) (- \frac{1}{5} sin (5 x) + \frac{1}{3} sin (3 x))