y^{''}-y^'+14y=2+e^{x/2}

解

y^{^{''}} - y^{^{'}} + 14 y = 2 + e^{\frac{x}{2}}

y = e^{\frac{x}{2}} (c_{1} cos (\frac{5 5 ^{\frac{1}{2}} x}{2}) + c_{2} sin (\frac{5 5 ^{\frac{1}{2}} x}{2})) + \frac{1}{7} + \frac{4}{55} e^{\frac{x}{2}}

解答ステップ

y^{''} (x) - y^{'} (x) + 14 y = 2 + e^{\frac{x}{2}}

線形 ODE を解く:

y = e^{\frac{x}{2}} (c_{1} cos (\frac{5 5 ^{\frac{1}{2}} x}{2}) + c_{2} sin (\frac{5 5 ^{\frac{1}{2}} x}{2})) + \frac{1}{7} + \frac{4}{55} e^{\frac{x}{2}}

y = e^{\frac{x}{2}} (c_{1} cos (\frac{5 5 ^{\frac{1}{2}} x}{2}) + c_{2} sin (\frac{5 5 ^{\frac{1}{2}} x}{2})) + \frac{1}{7} + \frac{4}{55} e^{\frac{x}{2}}