y^{''}-3y^'+6y=xe^x

Lösung

y^{^{''}} - 3 y^{^{'}} + 6 y = x e^{x}

y = e^{\frac{3 x}{2}} (c_{1} cos (\frac{15 x}{2}) + c_{2} sin (\frac{15 x}{2})) + \frac{e ^{x} x}{4} + \frac{1}{16} e^{x}

Schritte zur Lösung

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

Solve linear ODE:

y = e^{\frac{3 x}{2}} (c_{1} cos (\frac{15 x}{2}) + c_{2} sin (\frac{15 x}{2})) + \frac{e ^{x} x}{4} + \frac{1}{16} e^{x}

y = e^{\frac{3 x}{2}} (c_{1} cos (\frac{15 x}{2}) + c_{2} sin (\frac{15 x}{2})) + \frac{e ^{x} x}{4} + \frac{1}{16} e^{x}

y^{^{''}} - 4 y = t H (t - 3), y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} + y^{^{'}} = 2 cos (t)

x^{^{''}} + 2 x^{^{'}} + 1 x = e^{- 2 t}, x (0) = 2, x^{^{'}} (0) = 1

y^{^{''}} + 4 y = cos (2 x) + 1

\frac{d ^{2} y}{d x ^{2}} + y = 4 x cos (x)