y^{''}-2y=0,y(0)=1,y^'(0)=2

Lösung

y^{^{''}} - 2 y = 0, y (0) = 1, y^{^{'}} (0) = 2

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

Schritte zur Lösung

y^{''} (t) - 2 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 16 y^{^{'}} + 63 y = 0

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

y^{^{''}} - 10 y + 26 y = 0

6 x^{^{''}} + 8 x^{^{'}} + 35 x = 0

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