y^{''}-10/3 y^'+ay=0

解答

y^{^{''}} - \frac{10}{3} y^{^{'}} + a y = 0

y = c_{1} e^{\frac{t ( 5 + - 9 a + 25 )}{3}} + c_{2} e^{\frac{t ( 5 - - 9 a + 25 )}{3}}

求解步骤

y^{''} (t) - \frac{10}{3} y^{'} (t) + a y = 0

求解线性方程:

y = c_{1} e^{\frac{t ( 5 + - 9 a + 25 )}{3}} + c_{2} e^{\frac{t ( 5 - - 9 a + 25 )}{3}}

y = c_{1} e^{\frac{t ( 5 + - 9 a + 25 )}{3}} + c_{2} e^{\frac{t ( 5 - - 9 a + 25 )}{3}}

x^{^{''}} + a \cdot x^{^{'}} = 0

y^{^{'''}} + y^{^{''}} - 5 y^{^{'}} + 3 y = 0

y^{^{'''}} - 5 y^{^{''}} + 9 y^{^{'}} - 5 y = 0

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = - 4, y^{^{'}} (0) = - \frac{71}{5}

25 y^{^{''}} + 10 y^{^{'}} + 1 y = 0