y^{''}-3y^'+8y=xe^x

Lösung

y^{^{''}} - 3 y^{^{'}} + 8 y = x e^{x}

y = e^{\frac{3 x}{2}} (c_{1} cos (\frac{23 x}{2}) + c_{2} sin (\frac{23 x}{2})) + \frac{e ^{x} x}{6} + \frac{1}{36} e^{x}

Schritte zur Lösung

y^{''} (x) - 3 y^{'} (x) + 8 y = x e^{x}

Solve linear ODE:

y = e^{\frac{3 x}{2}} (c_{1} cos (\frac{23 x}{2}) + c_{2} sin (\frac{23 x}{2})) + \frac{e ^{x} x}{6} + \frac{1}{36} e^{x}

y = e^{\frac{3 x}{2}} (c_{1} cos (\frac{23 x}{2}) + c_{2} sin (\frac{23 x}{2})) + \frac{e ^{x} x}{6} + \frac{1}{36} e^{x}

y^{^{'}} + 9 (tan (9 x)) y = 5 cos (9 x)

\int x^{5} e^{x^{6 + 1}} d x

\frac{d ^{2}}{d x ^{2}} ((e^{x} + x^{π}) (x^{2} + x + 1))

a re a y = x, y = - x + 2, [0, 2]

a re a x + y = 13, x + 7 = y^{2}