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

Lösung

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

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

Schritte zur Lösung

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

Solve linear ODE:

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

y^{^{''}} - (8) y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 73

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 0, y^{^{'}} (0) = 1

8 y^{^{''}} + y^{^{'}} + 128 y = 0, y (0) = 5, y^{^{'}} (0) = 8

\frac{1}{8} x^{^{''}} + 36 x = 0

\frac{d ^{2} y}{d x ^{2}} = - 4 y