y^{''}+2y^'-y=0,y(0)=xe^{-x}

Lösung

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

y = c_{1} e^{(- 1 + 2) x} + c_{2} e^{(- 1 - 2) x}

Schritte zur Lösung

y^{''} (x) + 2 y^{'} (x) - y = 0

Solve linear ODE:

y = c_{1} e^{(- 1 + 2) x} + c_{2} e^{(- 1 - 2) x}

y = c_{1} e^{(- 1 + 2) x} + c_{2} e^{(- 1 - 2) x}

y^{^{'''}} + y^{^{'}} + 2 y = 0

y^{^{''}} = \frac{- y ^{^{'}} - 5 y}{4}

f^{^{''}} (x) = - f (x)

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 3 y = 0

(D^{2} + 4)^{2} y = 0