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

Lösung

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

y = c_{1} e^{2 t} + c_{2} e^{- 2 t}

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t}

f^{^{''}} (t) + 2 f^{^{'}} (t) + 5 f (t) = 0, f (0) = 3, f^{^{'}} (0) = - 7

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

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

\frac{d ^{2} x}{d t ^{2}} - 6 \frac{d x}{d t} + 9 x = 0

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = 0