y^{''}+2y^'+1.25y=0

Lösung

y^{^{''}} + 2 y^{^{'}} + 1.25 y = 0

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

Schritte zur Lösung

y^{''} (t) + 2 y^{'} (t) + 1.25 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}))

y^{^{''}} + 20 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = 0

3 \frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 3 y = 0

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

y^{^{'''}} - 7 y^{^{''}} + y^{^{'}} + 12 y = 0

y^{^{''}} - 2 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = - 2