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

Lösung

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

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{2} e^{x}

Schritte zur Lösung

y^{''} (x) - 2 y^{'} (x) + 3 y = e^{x}

Solve linear ODE:

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{2} e^{x}

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{2} e^{x}

y^{^{''}} - 3 y^{^{'}} = 8 e^{2 t} - 2 e^{- t}

y^{^{''}} - 2 y^{^{'}} + y = t

y^{^{''}} - 4 y^{^{'}} = 8 e^{6 x} (x - 1)

y^{^{''}} - y = e^{x} cos (x)

y^{^{'''}} - y^{^{'}} = 7 t