y^{'''}+2y^'-3y=0,y(0)=e^x

Lösung

y^{^{'''}} + 2 y^{^{'}} - 3 y = 0, y (0) = e^{x}

y = c_{1} e^{x} + e^{- \frac{x}{2}} (c_{2} cos (\frac{11 x}{2}) + c_{3} sin (\frac{11 x}{2}))

Schritte zur Lösung

y^{'''} (x) + 2 y^{'} (x) - 3 y = 0

Solve linear ODE:

y = c_{1} e^{x} + e^{- \frac{x}{2}} (c_{2} cos (\frac{11 x}{2}) + c_{3} sin (\frac{11 x}{2}))

y = c_{1} e^{x} + e^{- \frac{x}{2}} (c_{2} cos (\frac{11 x}{2}) + c_{3} sin (\frac{11 x}{2}))

y^{^{''}} + 4 y^{^{'}} = 2 y

(d^{4} - i^{2}) y = 0

(d^{3} - d^{2} + 2 d - d) y = 0

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

(d^{3} + d^{2} + d) \cdot y = 0