y^{''}+3y^'+2y=3e^{-4t},y(0)=1

Lösung

y^{^{''}} + 3 y^{^{'}} + 2 y = 3 e^{- 4 t}, y (0) = 1

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} = e^{- 3 t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 8 y^{^{'}} + 16 = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} = 6 x + 6 - 2 e^{4 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 15 sin (x) - 35 cos (x)

y^{^{''}} + 3 y^{^{'}} + 2 y = 2 x + x^{2}