y^{'''}+6y^{''}=3-cos(x)

解答

y^{^{'''}} + 6 y^{^{''}} = 3 - cos (x)

y = c_{1} e^{- 6 x} + c_{2} + c_{3} x + \frac{x ^{2}}{4} + \frac{1}{37} sin (x) + \frac{6}{37} cos (x)

求解步骤

y^{'''} (x) + 6 y^{''} (x) = 3 - cos (x)

求解线性方程:

y = c_{1} e^{- 6 x} + c_{2} + c_{3} x + \frac{x ^{2}}{4} + \frac{1}{37} sin (x) + \frac{6}{37} cos (x)

y = c_{1} e^{- 6 x} + c_{2} + c_{3} x + \frac{x ^{2}}{4} + \frac{1}{37} sin (x) + \frac{6}{37} cos (x)

x^{^{''}} + 3 x^{^{'}} + \frac{61}{4} x = \frac{1}{2} e^{t}

y^{^{''}} + 4 y^{^{'}} = 3 sin (2 t)

y^{^{'''}} + y^{^{'}} = sec (t), y^{^{'}} (0) = 3

(d^{2} - 2 d) y = e^{x} sin (x)

\frac{1}{2} y^{^{''}} + 2 y = 3 tan (2 t) - \frac{1}{3} e^{4 t}