y^{''}+4y=5sin(2x)-7cos(2x)

解答

y^{^{''}} + 4 y = 5 sin (2 x) - 7 cos (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{7 x sin ( 2 x )}{4} - \frac{5 x cos ( 2 x )}{4}

求解步骤

y^{''} (x) + 4 y = 5 sin (2 x) - 7 cos (2 x)

求解线性方程:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{7 x sin ( 2 x )}{4} - \frac{5 x cos ( 2 x )}{4}

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{7 x sin ( 2 x )}{4} - \frac{5 x cos ( 2 x )}{4}

y^{^{''}} - 4 y^{^{'}} = 6 e^{3 x} - 3 e^{- x}

l in e a r 2 y^{^{''}} + 8 y^{^{'}} - y = 16

y^{^{'''}} + 4 y^{^{'}} = tan (2 x)

y^{^{''}} + y^{^{'}} = e^{5 x}

y^{^{''}} - 16 y^{^{'}} + 64 y = 6 e^{8 x}