y^{''}+4y=8tan^2(2x)

解

y^{^{''}} + 4 y = 8 tan^{2} (2 x)

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

解答ステップ

y^{''} (x) + 4 y = 8 tan^{2} (2 x)

線形 ODE を解く:

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

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