y^{'''}-2y^{''}+4y^'-8y=8sin(2x)

解答

y^{^{'''}} - 2 y^{^{''}} + 4 y^{^{'}} - 8 y = 8 sin (2 x)

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

求解步骤

y^{'''} (x) - 2 y^{''} (x) + 4 y^{'} (x) - 8 y = 8 sin (2 x)

求解线性方程:

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

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

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + y = x ln (x)

y^{^{''}} + 4 y = 16 t, y (0) = 3

y^{^{''}} + y = e^{x} + x^{3}

y^{^{''}} + y = 1 + 2 t sin (t)

y^{^{''}} + 16 y = sec (4)