y^{''}-2y=0,y(0)=1,y^'(0)=6

解答

y^{^{''}} - 2 y = 0, y (0) = 1, y^{^{'}} (0) = 6

y = (1 + \frac{6 - 2}{2 2}) e^{2 t} - \frac{6 - 2}{2 2} e^{- 2 t}

求解步骤

y^{''} (t) - 2 y = 0

求解线性方程:

y = (1 + \frac{6 - 2}{2 2}) e^{2 t} - \frac{6 - 2}{2 2} e^{- 2 t}

y = (1 + \frac{6 - 2}{2 2}) e^{2 t} - \frac{6 - 2}{2 2} e^{- 2 t}

y^{^{''}} + 25 y = 0, y^{^{'}} (π) = 0, y^{^{'}} (0) = 0

y^{^{''''}} - k^{2} y^{^{''}} = 0

y^{^{''''}} - y^{^{''}} = 2 (4 y^{^{'''}})

y^{^{''}} - 4 y = 0, y^{^{'}} (- 1) = 1, y (2) = 2

y^{^{'''}} + y^{^{''}} + y^{^{'}} - 39 y = 0