y^{''}+8y^'+41y=0,y^'(0)=2,y(0)=1

解答

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y^{^{'}} (0) = 2, y (0) = 1

y = e^{- 4 t} (cos (5 t) + \frac{6}{5} sin (5 t))

求解步骤

y^{''} (t) + 8 y^{'} (t) + 41 y = 0

求解线性方程:

y = e^{- 4 t} (cos (5 t) + \frac{6}{5} sin (5 t))

y = e^{- 4 t} (cos (5 t) + \frac{6}{5} sin (5 t))

2 y^{^{''}} - 17 y^{^{'}} - 9 y = 0, y (0) = 4, y^{^{'}} (0) = - 7

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

u^{^{''}} = H (t)

0.25 \cdot 0.04 x^{^{''}} + 2 \cdot 0.04 x^{^{'}} + x = 0

y^{^{''}} + 20 y^{^{'}} + 92 y = 0