y^{''}+4y^'-5y=0,y(0)=1,y^'(0)=7,y(0)=y

解答

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (0) = 1, y^{^{'}} (0) = 7, y (0) = y

y = c_{1} e^{t} + c_{2} e^{- 5 t}

求解步骤

y^{''} (t) + 4 y^{'} (t) - 5 y = 0

求解线性方程:

y = c_{1} e^{t} + c_{2} e^{- 5 t}

y = c_{1} e^{t} + c_{2} e^{- 5 t}

y^{^{''''}} - 7 y^{^{''}} - 18 y = 0

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

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (0) = 2, y (0) = 26

64 y^{^{''}} - y = 0, y (- 8) = 1, y^{^{'}} (- 8) = - 1

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = 0, y^{^{'}} (0) = - 7