y^{''}+2y^'+5y=0,y(0)=2,y^'(0)=4

Lösung

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = 4

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d t ^{2}} - 14 \frac{d y}{d t} + 149 y = 0

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

\frac{1}{8} y^{^{''}} + 16 y = 0

y^{^{''''}} - 4 y = 0

y^{^{''}} = 6 y^{^{'}} - 8 y