y^{''}-2y^'+5y=0,y(0)=1,y^'(0)=-3

Lösung

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = - 3

y = e^{t} (cos (2 t) - 2 sin (2 t))

Schritte zur Lösung

y^{''} (t) - 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{t} (cos (2 t) - 2 sin (2 t))

y = e^{t} (cos (2 t) - 2 sin (2 t))

\frac{1}{8} y^{^{''}} + 4 y^{^{'}} + 12 y = 0, y (0) = 0, y^{^{'}} (0) = 4

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

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 3, y^{^{'}} (0) = - 11

y^{^{''}} - 6 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = - 1, y^{^{'}} (0) = 0