y^{''}-3y^'-5=e^{2x}-3cos(3x)

解答

y^{^{''}} - 3 y^{^{'}} - 5 = e^{2 x} - 3 cos (3 x)

y = c_{1} e^{3 x} + c_{2} - \frac{1}{2} e^{2 x} + \frac{1}{6} sin (3 x) + \frac{1}{6} cos (3 x) - \frac{5 x}{3}

求解步骤

y^{''} (x) - 3 y^{'} (x) - 5 = e^{2 x} - 3 cos (3 x)

求解线性方程:

y = c_{1} e^{3 x} + c_{2} - \frac{1}{2} e^{2 x} + \frac{1}{6} sin (3 x) + \frac{1}{6} cos (3 x) - \frac{5 x}{3}

y = c_{1} e^{3 x} + c_{2} - \frac{1}{2} e^{2 x} + \frac{1}{6} sin (3 x) + \frac{1}{6} cos (3 x) - \frac{5 x}{3}

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

y^{^{''}} + 10 y^{^{'}} + 16 y = 949 sin (3 t)

y^{^{''}} + 2 y^{^{'}} + 5 y = 10 cos (x)

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

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} - 2 y = - 6 e^{x} + 10 cos (x)