y^{''}-3y=0,y(0)=1,y^'(0)=5

Lösung

y^{^{''}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 5

y = (1 + \frac{5 - 3}{2 3}) e^{3 t} - \frac{5 - 3}{2 3} e^{- 3 t}

Schritte zur Lösung

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

Solve linear ODE:

y = (1 + \frac{5 - 3}{2 3}) e^{3 t} - \frac{5 - 3}{2 3} e^{- 3 t}

y = (1 + \frac{5 - 3}{2 3}) e^{3 t} - \frac{5 - 3}{2 3} e^{- 3 t}

u^{^{''}} \frac{( x )}{x} + b^{2} H (x) \frac{( x )}{x} = 0

(d^{2} - 6 d + 9) y = 0

x^{^{'''}} - 2 x^{^{''}} = 0

(x + 1) y^{^{'}} = y^{2}

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