y^{''}+6y^'+9y=0,y(0)=-4,y^'(0)=14

解答

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = - 4, y^{^{'}} (0) = 14

y = - 4 e^{- 3 t} + 2 e^{- 3 t} t

求解步骤

y^{''} (t) + 6 y^{'} (t) + 9 y = 0

求解线性方程:

y = - 4 e^{- 3 t} + 2 e^{- 3 t} t

y = - 4 e^{- 3 t} + 2 e^{- 3 t} t

3 y^{^{''}} - 24 y^{^{'}} + 48 y = 0

4 \frac{d ^{2} y}{d t ^{2}} - y = 0

y^{^{'''}} - 20 y^{^{''}} + 111 y^{^{'}} - 180 y = 0

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 2, y^{^{'}} (0) = 0

\frac{d ^{4} y}{d x ^{4}} + 2 y^{^{'''}} - 2 y^{^{'}} - y = 0