y^{''''}-y^{'''}=x+e^x

解答

y^{^{''''}} - y^{^{'''}} = x + e^{x}

y = c_{1} e^{x} + c_{2} + c_{3} x + c_{4} x^{2} - x - 1 + e^{x} x + \frac{- x ^{4} - 4 x ^{3} - 12 x ^{2} - 72 e ^{x}}{24}

求解步骤

y^{(4)} (x) - y^{'''} (x) = x + e^{x}

求解线性方程:

y = c_{1} e^{x} + c_{2} + c_{3} x + c_{4} x^{2} - x - 1 + e^{x} x + \frac{- x ^{4} - 4 x ^{3} - 12 x ^{2} - 72 e ^{x}}{24}

y = c_{1} e^{x} + c_{2} + c_{3} x + c_{4} x^{2} - x - 1 + e^{x} x + \frac{- x ^{4} - 4 x ^{3} - 12 x ^{2} - 72 e ^{x}}{24}

y^{^{''}} - y^{^{'}} + 14 y = \frac{4 + e x}{2}

y^{^{''}} - 2 y^{^{'}} + 3 y = 2

y^{^{'''}} + 2 y^{^{''}} - 5 y^{^{'}} - 6 y = sin (6 t)

y^{^{''}} + 9 y = 7 sec (3 x)

y^{^{''}} - 36 y = - 42 e^{- 5 t}, y (0) = 7, y^{^{'}} (0) = 0