(d^2+4d+20)^2y=0

Lösung

(d^{2} + 4 d + 20)^{2} y = 0

y = e^{- 2 x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + x e^{- 2 x} (c_{3} cos (4 x) + c_{4} sin (4 x))

Schritte zur Lösung

(d^{2} + 4 d + 20)^{2} y = 0

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + x e^{- 2 x} (c_{3} cos (4 x) + c_{4} sin (4 x))

y = e^{- 2 x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + x e^{- 2 x} (c_{3} cos (4 x) + c_{4} sin (4 x))

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 6, y^{^{'}} (0) = 3

y^{^{''}} + 3 y^{^{'}} - 6 y = 0

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{'''}} + 5 y^{^{''}} - 13 y^{^{'}} + 7 y = 0

0.05 y^{^{''}} + 2 y^{^{'}} + 100 y = 0, y (0) = 5, y^{^{'}} (0) = 0