y^{''}+16y=8tan(4x)

解

y^{^{''}} + 16 y = 8 tan (4 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{- 2 cos ( 4 x ) ( ln ( tan ( 4 x ) + sec ( 4 x ) ) - sin ( 4 x ) ) - sin ( 8 x )}{4}

解答ステップ

y^{''} (x) + 16 y = 8 tan (4 x)

線形 ODE を解く:

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{- 2 cos ( 4 x ) ( ln ( tan ( 4 x ) + sec ( 4 x ) ) - sin ( 4 x ) ) - sin ( 8 x )}{4}

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{- 2 cos ( 4 x ) ( ln ( tan ( 4 x ) + sec ( 4 x ) ) - sin ( 4 x ) ) - sin ( 8 x )}{4}