y^{''}+(a-1)y^'+by=0

解答

y^{^{''}} + (a - 1) y^{^{'}} + b y = 0

y = c_{1} e^{\frac{t ( - a + 1 + ( a - 1 ) ^{2} - 4 b )}{2}} + c_{2} e^{\frac{t ( - a + 1 - ( a - 1 ) ^{2} - 4 b )}{2}}

求解步骤

y^{''} (t) + (a - 1) y^{'} (t) + b y = 0

求解线性方程:

y = c_{1} e^{\frac{t ( - a + 1 + ( a - 1 ) ^{2} - 4 b )}{2}} + c_{2} e^{\frac{t ( - a + 1 - ( a - 1 ) ^{2} - 4 b )}{2}}

y = c_{1} e^{\frac{t ( - a + 1 + ( a - 1 ) ^{2} - 4 b )}{2}} + c_{2} e^{\frac{t ( - a + 1 - ( a - 1 ) ^{2} - 4 b )}{2}}

8 y^{^{''}} - 9 y^{^{'}} + y = 0, y (0) = 8, y^{^{'}} (0) = \frac{1}{8}

y^{^{''''}} - 2 y^{^{'''}} - 10 y^{^{''}} + 6 y^{^{'}} + 45 y = 0

16 \frac{d ^{4} y}{d x ^{4}} + 16 \frac{d ^{2} y}{d x ^{2}} + 4 y = 0

y^{^{''}} + y^{^{'}} = 6 y

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