6y^{''}+11y^'-6y=0

解答

6 y^{^{''}} + 11 y^{^{'}} - 6 y = 0

y = c_{1} e^{\frac{( - 11 + 265 ) t}{12}} + c_{2} e^{\frac{( - 11 - 265 ) t}{12}}

求解步骤

6 y^{''} (t) + 11 y^{'} (t) - 6 y = 0

求解线性方程:

y = c_{1} e^{\frac{( - 11 + 265 ) t}{12}} + c_{2} e^{\frac{( - 11 - 265 ) t}{12}}

y = c_{1} e^{\frac{( - 11 + 265 ) t}{12}} + c_{2} e^{\frac{( - 11 - 265 ) t}{12}}

(D^{2} - D - 6) y = 0

y^{^{''}} - 16 y^{^{'}} = 0

x^{^{''}} + 5 x^{^{'}} + 4 x = 0

y^{^{''}} + 4 y = 0, y (\frac{π}{6}) = 1, y^{^{'}} (\frac{π}{6}) = - 2

y^{^{''}} + y^{^{'}} + \frac{5}{4} y = 0, y (0) = 3, y^{^{'}} (0) = 1