(D^2+D+1)y=xe^x

Lösung

(D^{2} + D + 1) y = x e^{x}

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

Schritte zur Lösung

(D^{2} + D + 1) y = x e^{x}

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} - 8 y = 8 x^{2} - 3, y (0) = - 1, y^{^{'}} (0) = - 3

y^{^{''}} + y = sec^{3} (x), y (0) = 2, y^{^{'}} (0) = \frac{5}{2}

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 3 y = 27 x

9 y^{^{''}} + 6 y^{^{'}} + 8 y = 37 sin (x)

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