y^{''}+3y^'-4y=-32t^2

解答

y^{^{''}} + 3 y^{^{'}} - 4 y = - 32 t^{2}

y = c_{1} e^{t} + c_{2} e^{- 4 t} + 8 t^{2} + 12 t + 13

求解步骤

y^{''} (t) + 3 y^{'} (t) - 4 y = - 32 t^{2}

求解线性方程:

y = c_{1} e^{t} + c_{2} e^{- 4 t} + 8 t^{2} + 12 t + 13

y = c_{1} e^{t} + c_{2} e^{- 4 t} + 8 t^{2} + 12 t + 13

y^{^{''}} + y^{^{'}} = (2 t)

y^{^{''}} + 3 y + 2 = 0, y (2) = - 1, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = e - t, y (0) = 0, y^{^{'}} (0) = 1

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

y^{^{''}} + 6 y^{^{'}} + 8 y = t H (t - 3)