(D^3-3D^2+3D-1)y=e^x+1

Lösung

(D^{3} - 3 D^{2} + 3 D - 1) y = e^{x} + 1

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + \frac{e ^{x} x ^{3}}{6} - 1

Schritte zur Lösung

(D^{3} - 3 D^{2} + 3 D - 1) y = e^{x} + 1

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + \frac{e ^{x} x ^{3}}{6} - 1

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + \frac{e ^{x} x ^{3}}{6} - 1

y^{^{''}} - 4 y^{^{'}} = (x^{2} - 3) sin (2 x)

2 y^{^{''}} + 3 y^{^{'}} + y = sin (x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 3

x^{^{''}} + 4 x = cos (t)

y^{^{''}} + 9 y = cos (8 x)