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

解

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

y = c_{1} e^{t} + c_{2} cos (2^{1 + \frac{1}{2}} t) + c_{3} sin (2^{1 + \frac{1}{2}} t)

解答ステップ

y^{'''} (t) - y^{''} (t) + 8 y^{'} (t) - 8 y = 0

線形 ODE を解く:

y = c_{1} e^{t} + c_{2} cos (2^{1 + \frac{1}{2}} t) + c_{3} sin (2^{1 + \frac{1}{2}} t)

y = c_{1} e^{t} + c_{2} cos (2^{1 + \frac{1}{2}} t) + c_{3} sin (2^{1 + \frac{1}{2}} t)