y^{''}+40y^'+2000y=0,y(0)=5,y^'(0)=0

Lösung

y^{^{''}} + 40 y^{^{'}} + 2000 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y = e^{- 20 t} (5 cos (40 t) + \frac{5}{2} sin (40 t))

Schritte zur Lösung

y^{''} (t) + 40 y^{'} (t) + 2000 y = 0

Solve linear ODE:

y = e^{- 20 t} (5 cos (40 t) + \frac{5}{2} sin (40 t))

y = e^{- 20 t} (5 cos (40 t) + \frac{5}{2} sin (40 t))

y^{^{'''}} - 2 y^{^{''}} - 25 y^{^{'}} + 50 y = 0

(d^{3} + 3 d^{2} + 3 d + 1) y = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''''}} - 38 y^{^{'''}} + 361 y^{^{''}} = 0

\frac{d ^{4} y}{d x ^{4}} + 5 \frac{d ^{2} y}{d x ^{2}} + 4 y = 0