y^{''}-4y^'+5y=0,y^1(0)=e^xcos(2x)

Lösung

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y^{1} (0) = e^{x} cos (2 x)

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

Schritte zur Lösung

y^{''} (x) - 4 y^{'} (x) + 5 y = 0

Solve linear ODE:

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

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

4 y^{^{''}} + 24 y^{^{'}} + 37 y = 0, y (π) = 1

l in e a r y^{^{'''}} - 3 y^{^{''}} - 5 y^{^{'}} + 39 y = 0

y^{^{''}} + 29.4 y = 0

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

y^{^{''}} + 2 y^{^{'}} - 24 y = 0, y (0) = 1, y^{^{'}} (0) = - 2