x^{''}-0.1x^'+1.0025x=0

解答

x^{^{''}} - 0.1 x^{^{'}} + 1.0025 x = 0

x = e^{0.05 t} (c_{1} cos (t) + c_{2} sin (t))

求解步骤

x^{''} (t) - 0.1 x^{'} (t) + 1.0025 x = 0

求解线性方程:

x = e^{0.05 t} (c_{1} cos (t) + c_{2} sin (t))

x = e^{0.05 t} (c_{1} cos (t) + c_{2} sin (t))

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

y^{^{''}} + 120 y^{^{'}} + 10000 y = 0

4 y^{^{''}} = - 25 y

35 y^{^{''''}} + 52 y^{^{''}} + 9 y = 0

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