y^{''}+9y=6cos(3x)+9sin(3x)

解答

y^{^{''}} + 9 y = 6 cos (3 x) + 9 sin (3 x)

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

求解步骤

y^{''} (x) + 9 y = 6 cos (3 x) + 9 sin (3 x)

求解线性方程:

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

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

y^{^{'''}} + 7 y^{^{''}} + 8 y^{^{'}} - 16 y = - 2 e^{x} - 2

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

y^{^{''}} + 3 y^{^{'}} + 15.25 y = e^{t}

y^{^{''}} + 5 y = cos (7 t)

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{- x} + sin (x)