0.5y^{''}+0.5y^'+y=0

Lösung

0.5 y^{^{''}} + 0.5 y^{^{'}} + y = 0

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

Schritte zur Lösung

0.5 y^{''} (t) + 0.5 y^{'} (t) + y = 0

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y^{^{''}} - 10 y^{^{'}} - 144 y = 0, y (0) = 11, y^{^{'}} (0) = 94

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (\frac{π}{4}) = 1, y (0) = 9

y^{^{''''}} + 3 y^{^{'''}} + 2 y^{^{''}} + 6 y^{^{'}} = 0

1 y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = - 7

(d^{3} - d^{2} - 12 d) y = 0