y^{''}+4y=0,y(0)=6,y^'(0)=-6

Lösung

y^{^{''}} + 4 y = 0, y (0) = 6, y^{^{'}} (0) = - 6

y = 6 cos (2 t) - 3 sin (2 t)

Schritte zur Lösung

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

Solve linear ODE:

y = 6 cos (2 t) - 3 sin (2 t)

y = 6 cos (2 t) - 3 sin (2 t)

y^{^{''}} + 7 y^{^{'}} + 20 y = 0

2 y^{^{''}} + 14 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} - 8 y = 0

y^{^{''}} + 5 y^{^{'}} = 0, y (0) = 9, y^{^{'}} (0) = - 20

\frac{d ^{2} x}{d t ^{2}} + 6 \cdot \frac{d x}{d t} + 3 \cdot x = 0, x (0) = 0