y^{''}-2y^'+y=t^{-4}e^t

解答

y^{^{''}} - 2 y^{^{'}} + y = t^{- 4} e^{t}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{6 t ^{2}}

求解步骤

y^{''} (t) - 2 y^{'} (t) + y = t^{- 4} e^{t}

求解线性方程:

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{6 t ^{2}}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t}}{6 t ^{2}}

y^{^{''}} - 2 y^{^{'}} - 3 y = x - e^{x}

\frac{d ^{2} y}{d x ^{2}} - 6 y = e^{3 x} (x^{2} + 2)

2 y^{^{''}} + y^{^{'}} + 2 y = 1, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y = (sec (x))^{2}

2 q^{^{''}} + 10 q^{^{'}} + 12 q = e^{- 3 t}