2y^{''}+4y^'+3y=0

Lösung

2 y^{^{''}} + 4 y^{^{'}} + 3 y = 0

y = e^{- t} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

Schritte zur Lösung

2 y^{''} (t) + 4 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = e^{- t} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

y = e^{- t} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

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

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

so l v e f or u, \frac{d ^{2} u}{d x ^{2}} = u

16 y^{^{''}} + 8 y^{^{'}} + 5 y = 0

y^{^{''}} + 20 y = 0