4y^{''}+16y^'+25y=0

Lösung

4 y^{^{''}} + 16 y^{^{'}} + 25 y = 0

y = e^{- 2 t} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

Schritte zur Lösung

4 y^{''} (t) + 16 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

y = e^{- 2 t} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

y^{^{''}} - 14 y^{^{'}} + 65 y = 0

y^{^{''''}} - 81 y^{^{''}} = 0

3.5 y^{^{''}} - 5.6 y^{^{'}} + 2.2 y = 0

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

\frac{1}{4} y^{^{''}} + 2 y^{^{'}} + 2 y = 0