y^{''}-4y^'+20y=0,y(0)=0

Lösung

y^{^{''}} - 4 y^{^{'}} + 20 y = 0, y (0) = 0

y = c_{2} e^{2 t} sin (4 t)

Schritte zur Lösung

y^{''} (t) - 4 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = c_{2} e^{2 t} sin (4 t)

y = c_{2} e^{2 t} sin (4 t)

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = - 2, y^{^{'}} (0) = 0

y^{^{''}} - 83 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 73

x^{^{''}} - 9 x = 0

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

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