y^{''}-5y^'+9y=xe^x

解答

y^{^{''}} - 5 y^{^{'}} + 9 y = x e^{x}

y = e^{\frac{5 x}{2}} (c_{1} cos (\frac{11 x}{2}) + c_{2} sin (\frac{11 x}{2})) + \frac{e ^{x} x}{5} + \frac{3}{25} e^{x}

求解步骤

y^{''} (x) - 5 y^{'} (x) + 9 y = x e^{x}

求解线性方程:

y = e^{\frac{5 x}{2}} (c_{1} cos (\frac{11 x}{2}) + c_{2} sin (\frac{11 x}{2})) + \frac{e ^{x} x}{5} + \frac{3}{25} e^{x}

y = e^{\frac{5 x}{2}} (c_{1} cos (\frac{11 x}{2}) + c_{2} sin (\frac{11 x}{2})) + \frac{e ^{x} x}{5} + \frac{3}{25} e^{x}

\frac{d}{d x} ((x^{2} - 2 x - 1)^{\frac{2}{3}})

\int tan^{5} (x) d x

(x + y) e^{\frac{x}{y}} d x + (x - \frac{x ^{2}}{y}) e^{\frac{x}{y}} d y = 0

\int (5 x^{14} - 2 x^{5}) d x

\int (y^{2} - 1)^{2} d y