9y^{''}+y=0,y(0)=-2,y^'(0)=1

Lösung

9 y^{^{''}} + y = 0, y (0) = - 2, y^{^{'}} (0) = 1

y = - 2 cos (\frac{t}{3}) + 3 sin (\frac{t}{3})

Schritte zur Lösung

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

Solve linear ODE:

y = - 2 cos (\frac{t}{3}) + 3 sin (\frac{t}{3})

y = - 2 cos (\frac{t}{3}) + 3 sin (\frac{t}{3})

so l v e f or u, \frac{d ^{2} u}{d x ^{2}} = a u (x)

y^{^{''}} - 7 y^{^{'}} + 8 y = 0, y (0) = 2, y^{^{'}} (0) = 7

y^{^{''}} + 3 y^{^{'}} - 18 y = 0, y (0) = 3, y^{^{'}} (0) = 18

y^{^{''}} + y = 0, y (0) = - 1, y^{^{'}} (0) = 6

y^{^{''}} + 6 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = - 44