x^{''}+2x^'+0.75x=0

解答

x^{^{''}} + 2 x^{^{'}} + 0.75 x = 0

x = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{3 t}{2}}

求解步骤

x^{''} (t) + 2 x^{'} (t) + 0.75 x = 0

求解线性方程:

x = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{3 t}{2}}

x = c_{1} e^{- \frac{t}{2}} + c_{2} e^{- \frac{3 t}{2}}

y^{^{''}} = 7 y^{^{'}} - 12 y, y (0) = 0, y^{^{'}} (0) = - 1

y^{^{''}} - 10 y^{^{'}} + 2 y = 0, y (0) = 0

\frac{d ^{4} y}{d x ^{4}} - 2 y^{^{''}} + 1 y = 0

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

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 0, y^{^{'}} (0) = 5