y^{''}+2y=0,y(0)=sin(kt)

Lösung

y^{^{''}} + 2 y = 0, y (0) = sin (k t)

y = c_{1} cos (2 t) + c_{2} sin (2 t)

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t)

\frac{d ^{2} y}{d t ^{2}} = 5 \frac{d y}{d t} - 4 y

f^{^{''}} (x) - f^{^{'}} (x) - 2 f (x) = 0, f (0) = - 2, f^{^{'}} (0) = 2

9 y^{^{''}} - 12 y^{^{'}} + 11 y = 0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (- 1) = 5, y^{^{'}} (- 1) = 3

4 y^{^{''}} - 19 y^{^{'}} + 12 y = 0, y (0) = 5, y^{^{'}} (0) = 13.5