y^{''}-4y^'+8y=0,y(0)=2,y^'(0)=2

Lösung

y^{^{''}} - 4 y^{^{'}} + 8 y = 0, y (0) = 2, y^{^{'}} (0) = 2

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

Schritte zur Lösung

y^{''} (t) - 4 y^{'} (t) + 8 y = 0

Solve linear ODE:

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

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

4 y^{^{''}} - 4 y^{^{'}} + 7 y = 0

(D^{2} + 3 D + 2) y = 0

y^{^{''}} - \frac{10}{3} y^{^{'}} + a y = 0

x^{^{''}} + a \cdot x^{^{'}} = 0

y^{^{'''}} + y^{^{''}} - 5 y^{^{'}} + 3 y = 0