y^{''}-2y+2=0

Lösung

y^{^{''}} - 2 y + 2 = 0

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + 1

Schritte zur Lösung

y^{''} (t) - 2 y + 2 = 0

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + 1

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + 1

y^{^{''}} - 9 y^{^{'}} = 81 x^{2} + 7, y (0) = - 1, y^{^{'}} (0) = 2

y^{^{''}} - y^{^{'}} - 6 y = 14 sin (3)

y^{^{''}} + 2 y^{^{'}} + 5 y = 2 e^{(- t)} sin (2 t)

x^{^{''}} - 4 x^{^{'}} = 20 cos (5 t) - 4 sin (5 t) - 7

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 = 0