y^{''}+4y^'+13y=0,y(0)=2,y^'(0)=0

Lösung

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y = e^{- 2 t} (2 cos (3 t) + \frac{4}{3} sin (3 t))

Schritte zur Lösung

y^{''} (t) + 4 y^{'} (t) + 13 y = 0

Solve linear ODE:

y = e^{- 2 t} (2 cos (3 t) + \frac{4}{3} sin (3 t))

y = e^{- 2 t} (2 cos (3 t) + \frac{4}{3} sin (3 t))

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = - 2, y^{^{'}} (0) = 16

4 y^{^{''}} + 4 y^{^{'}} + y = 0, y (0) = 3, y^{^{'}} (0) = - 2

y^{^{''}} + 12 y^{^{'}} + 16 y = 0

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

\frac{d ^{2} y}{d x ^{2}} + 16 \frac{d y}{d x} + 64 y = 0