y^{'''}=y^'

Lösung

y^{^{'''}} = y^{^{'}}

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t}

Schritte zur Lösung

y^{'''} (t) = y^{'} (t)

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t}

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t}

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{4}{3}

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 0, y (0) = 3, y^{^{'}} (0) = - 1

y^{^{'''}} - 5 y^{^{''}} + 2 y^{^{'}} + 8 y = 0

y^{^{''}} - 4 y^{^{'}} - 3 y = 0