y^{''}-36y=8,y(0)=1,y^'(0)=0

解答

y^{^{''}} - 36 y = 8, y (0) = 1, y^{^{'}} (0) = 0

y = (- \frac{11}{18} + \frac{11}{9}) e^{6 t} + \frac{11}{18} e^{- 6 t} - \frac{2}{9}

求解步骤

y^{''} (t) - 36 y = 8

求解线性方程:

y = (- \frac{11}{18} + \frac{11}{9}) e^{6 t} + \frac{11}{18} e^{- 6 t} - \frac{2}{9}

y = (- \frac{11}{18} + \frac{11}{9}) e^{6 t} + \frac{11}{18} e^{- 6 t} - \frac{2}{9}

y^{^{'''}} - 36 y^{^{'}} = t e^{- 6 t} + 7 cos (6 t)

2 y^{^{''}} + 3 y^{^{'}} + y = g (t)

y^{^{''}} + y^{^{'}} = - 4 x + 2

(D^{2} - 1) y = (1 + e^{- x})^{- 2}

so l v e f or d, d (t) = 12 t + 20