y^{''}+y=sec(x^3)

解

y^{^{''}} + y = sec (x^{3})

y = c_{1} cos (x) + c_{2} sin (x) + cos (x) \cdot \int - sin (x) sec (x^{3}) d x + sin (x) \cdot \int cos (x) sec (x^{3}) d x

解答ステップ

y^{''} (x) + y = sec (x^{3})

線形 ODE を解く:

y = c_{1} cos (x) + c_{2} sin (x) + cos (x) \cdot \int - sin (x) sec (x^{3}) d x + sin (x) \cdot \int cos (x) sec (x^{3}) d x

y = c_{1} cos (x) + c_{2} sin (x) + cos (x) \cdot \int - sin (x) sec (x^{3}) d x + sin (x) \cdot \int cos (x) sec (x^{3}) d x