(y^{''}-ay^')=by

解答

(y^{^{''}} - a y^{^{'}}) = b y

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

求解步骤

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

求解线性方程:

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

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

y^{^{''}} + 12 y^{^{'}} + 4 y = 0, y (0) = - 3, y^{^{'}} (0) = 0

y^{^{''}} + 16.2 y^{^{'}} = 0

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 4 y = 0

y^{^{''}} + 10 y^{^{'}} + 24 y = 0, y (0) = 0, y^{^{'}} (0) = - 5

y^{^{'''}} + 9 y^{^{''}} + 8 y^{^{'}} = 0