y^{''}+6y^'+13y=e^{-3x}(cos(2x))

解答

y^{^{''}} + 6 y^{^{'}} + 13 y = e^{- 3 x} (cos (2 x))

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

求解步骤

y^{''} (x) + 6 y^{'} (x) + 13 y = e^{- 3 x} (cos (2 x))

求解线性方程:

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

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

y^{^{''}} + 8 y^{^{'}} + 12 = 1 - H (t - 3)

y^{^{''}} - 2 y^{^{'}} + y = t \cdot e^{t} + 4, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - y^{^{'}} - 6 y = 33 sin (3 x)

y^{^{'''}} + y^{^{''}} - 2 y = e^{3 x} (9 + 67 x + 17 x^{2})

y^{^{''}} - 8 y^{^{'}} + 80 y = - 2 sin (3 t)