y^{''}-2y^'+2y=e^{xsec(x)}

解

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x s e c (x)}

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + e^{x} cos (x) \cdot \int - e^{- x + x s e c (x)} sin (x) d x + e^{x} sin (x) \cdot \int e^{- x + x s e c (x)} cos (x) d x

解答ステップ

y^{''} (x) - 2 y^{'} (x) + 2 y = e^{x s e c (x)}

線形 ODE を解く:

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + e^{x} cos (x) \cdot \int - e^{- x + x s e c (x)} sin (x) d x + e^{x} sin (x) \cdot \int e^{- x + x s e c (x)} cos (x) d x

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + e^{x} cos (x) \cdot \int - e^{- x + x s e c (x)} sin (x) d x + e^{x} sin (x) \cdot \int e^{- x + x s e c (x)} cos (x) d x