y^{''}-2y^'+5y=0,y(0)=1,y^'(0)=3

Lösung

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 3

y = e^{t} (cos (2 t) + sin (2 t))

Schritte zur Lösung

y^{''} (t) - 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{t} (cos (2 t) + sin (2 t))

y = e^{t} (cos (2 t) + sin (2 t))

(d^{3} + 1) y = 0

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

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

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

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