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

解答

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

y = c_{1} + c_{2} x + c_{3} e^{2 x} + c_{4} x e^{2 x} - \frac{3}{25} sin (x) - \frac{4}{25} cos (x)

求解步骤

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

求解线性方程:

y = c_{1} + c_{2} x + c_{3} e^{2 x} + c_{4} x e^{2 x} - \frac{3}{25} sin (x) - \frac{4}{25} cos (x)

y = c_{1} + c_{2} x + c_{3} e^{2 x} + c_{4} x e^{2 x} - \frac{3}{25} sin (x) - \frac{4}{25} cos (x)

y^{^{'''}} - 25 y^{^{'}} = t e^{- 5 t} + 6 cos (5 t)

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{t}}{t ^{2} + 1}

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 13 t

y^{^{''}} + y^{^{'}} - 6 y = - 3 sin (x) + 4 e^{2 x}

\frac{d ^{2} y}{d t ^{2}} + 4 \frac{d y}{d t} + 4 y = t e^{2 t}