y^{'''}-4y^{''}+5y^'=40t-7-6e^{2t}

解答

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} = 40 t - 7 - 6 e^{2 t}

y = c_{1} + e^{2 t} (c_{2} cos (t) + c_{3} sin (t)) + 4 t^{2} + 5 t - 3 e^{2 t}

求解步骤

y^{'''} (t) - 4 y^{''} (t) + 5 y^{'} (t) = 40 t - 7 - 6 e^{2 t}

求解线性方程:

y = c_{1} + e^{2 t} (c_{2} cos (t) + c_{3} sin (t)) + 4 t^{2} + 5 t - 3 e^{2 t}

y = c_{1} + e^{2 t} (c_{2} cos (t) + c_{3} sin (t)) + 4 t^{2} + 5 t - 3 e^{2 t}

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

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{- 2 x} cos (x)

y^{^{''}} - 5 y + 6 y = e^{3 x} - x^{2}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 3 y = e

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