4y^{'''}-9y^{''}+14y^'-3y=xe^x

解

4 y^{^{'''}} - 9 y^{^{''}} + 14 y^{^{'}} - 3 y = x e^{x}

y = c_{1} e^{\frac{x}{4}} + e^{x} (c_{2} cos (2 x) + c_{3} sin (2 x)) + \frac{e ^{x} x}{6} - \frac{2}{9} e^{x}

解答ステップ

4 y^{'''} (x) - 9 y^{''} (x) + 14 y^{'} (x) - 3 y = x e^{x}

線形 ODE を解く:

y = c_{1} e^{\frac{x}{4}} + e^{x} (c_{2} cos (2 x) + c_{3} sin (2 x)) + \frac{e ^{x} x}{6} - \frac{2}{9} e^{x}

y = c_{1} e^{\frac{x}{4}} + e^{x} (c_{2} cos (2 x) + c_{3} sin (2 x)) + \frac{e ^{x} x}{6} - \frac{2}{9} e^{x}