y^{''}-12y^'+20y=0

解答

y^{^{''}} - 12 y^{^{'}} + 20 y = 0

y = c_{1} e^{10 t} + c_{2} e^{2 t}

求解步骤

y^{''} (t) - 12 y^{'} (t) + 20 y = 0

求解线性方程:

y = c_{1} e^{10 t} + c_{2} e^{2 t}

y = c_{1} e^{10 t} + c_{2} e^{2 t}

y^{^{'''}} - 12 y^{^{''}} + 21 y^{^{'}} + 98 y = 0

y^{^{''}} + 2 y^{^{'}} + y = 0, y (1) = 6

y^{^{''''}} + 7 y^{^{'''}} + 14 y^{^{''}} - 22 y^{^{'}} - 84 y = 0

(6 d^{3} + 5 d^{2}) y = 0

\frac{d ^{4} y}{d t ^{4}} + 2 y^{^{''}} + y = 0