y^{''}-2y^'+10y=0,y(0)=0,y^'(0)=2

Lösung

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 0, y^{^{'}} (0) = 2

y = \frac{2}{3} e^{t} sin (3 t)

Schritte zur Lösung

y^{''} (t) - 2 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = \frac{2}{3} e^{t} sin (3 t)

y = \frac{2}{3} e^{t} sin (3 t)

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

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

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

(D^{3} + D^{2} + 4 D + 4) y = 0

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