y^{''}+ay^'+y=e^{-4t}

解

y^{^{''}} + a y^{^{'}} + y = e^{- 4 t}

y = c_{1} e^{\frac{t ( - a + a ^{2} - 4 )}{2}} + c_{2} e^{\frac{t ( - a - a ^{2} - 4 )}{2}} + \frac{e ^{- 4 t}}{17 - 4 a}

解答ステップ

y^{''} (t) + a y^{'} (t) + y = e^{- 4 t}

線形 ODE を解く:

y = c_{1} e^{\frac{t ( - a + a ^{2} - 4 )}{2}} + c_{2} e^{\frac{t ( - a - a ^{2} - 4 )}{2}} + \frac{e ^{- 4 t}}{17 - 4 a}

y = c_{1} e^{\frac{t ( - a + a ^{2} - 4 )}{2}} + c_{2} e^{\frac{t ( - a - a ^{2} - 4 )}{2}} + \frac{e ^{- 4 t}}{17 - 4 a}