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

解答

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

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

求解步骤

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

求解线性方程:

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

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

y^{^{''}} + 14 y^{^{'}} + 49 y = x^{- 3} e^{- 7 x}

y^{^{''}} + 2 y^{^{'}} + 2 y = 3 t^{2}

y^{^{''}} - y = a x e^{- x}, y (0) = \frac{1}{4}, y^{^{'}} (0) = 0

y^{^{''}} - 5 y^{^{'}} + 4 y = 2 e^{4 t}

y^{^{''}} - 14 y^{^{'}} + 65 y = e^{7 x} sin (b x)