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

Lösung

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

y = 2 cos (2 t) - \frac{1}{2} sin (2 t)

Schritte zur Lösung

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

Solve linear ODE:

y = 2 cos (2 t) - \frac{1}{2} sin (2 t)

y = 2 cos (2 t) - \frac{1}{2} sin (2 t)

y^{^{''''}} + y^{^{''}} - 12 y = 0

y^{^{''}} - 225 y = 0

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

2 x^{^{''}} + 10 x = 0

2 y^{^{'''}} - 6 y^{^{''}} = 0