y^{'''}-y^{''}=2cos(x)

解答

y^{^{'''}} - y^{^{''}} = 2 cos (x)

y = c_{1} e^{x} + c_{2} + c_{3} x - sin (x) + cos (x)

求解步骤

y^{'''} (x) - y^{''} (x) = 2 cos (x)

求解线性方程:

y = c_{1} e^{x} + c_{2} + c_{3} x - sin (x) + cos (x)

y = c_{1} e^{x} + c_{2} + c_{3} x - sin (x) + cos (x)

y^{^{''}} - y^{^{'}} - 12 y = 98 (x + 2) e^{- 3 x}

y^{^{''}} + 2 y^{^{'}} + y = (\frac{1}{6}) e^{- x}

y^{^{'''}} + 2 y^{^{''}} - y^{^{'}} - 2 y = 5 H (t)

y^{^{''''}} + 3 y^{^{'''}} - 4 y^{^{''}} - 4 y^{^{'}} + \frac{1}{2} = 0

y^{^{''}} - 5 y^{^{'}} - 6 y = - 6 e^{t}