14y^{''}+y^'+y=x^2-5x

解答

14 y^{^{''}} + y^{^{'}} + y = x^{2} - 5 x

y = e^{- \frac{x}{28}} (c_{1} cos (\frac{55 x}{28}) + c_{2} sin (\frac{55 x}{28})) + x^{2} - 7 x - 21

求解步骤

14 y^{''} (x) + y^{'} (x) + y = x^{2} - 5 x

求解线性方程:

y = e^{- \frac{x}{28}} (c_{1} cos (\frac{55 x}{28}) + c_{2} sin (\frac{55 x}{28})) + x^{2} - 7 x - 21

y = e^{- \frac{x}{28}} (c_{1} cos (\frac{55 x}{28}) + c_{2} sin (\frac{55 x}{28})) + x^{2} - 7 x - 21

y^{^{''}} + 106 y^{^{'}} + 26 y = 2 t^{3} e^{- 5 t} sin (t)

y^{^{''}} - 7 y^{^{'}} = 5 x - 3

\frac{d ^{2} y}{d x ^{2}} - y = e^{x}

y^{^{''}} + 25 y^{^{'}} = - 24 t e^{- t}

y^{^{''}} + 4 y^{^{'}} + y = (\frac{e ^{- \frac{x}{2}}}{x ^{2}})