y^{''}-4y+5=0

Lösung

y^{^{''}} - 4 y + 5 = 0

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{5}{4}

Schritte zur Lösung

y^{''} (t) - 4 y + 5 = 0

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{5}{4}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{5}{4}

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

y^{^{''}} - y^{^{'}} - 20 = 0

\frac{d ^{2} y}{d x ^{2}} - 9 y = e^{3 x} cos (2 x) + e^{2 x}

y^{^{''}} + 9 y = cos (t)

y^{^{''}} + 4 y = t^{3} + 3 e^{t}, y (0) = 0, y^{^{'}} (0) = 2