9y^{''}+25y=0

Lösung

9 y^{^{''}} + 25 y = 0

y = c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3})

Schritte zur Lösung

9 y^{''} (t) + 25 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3})

y = c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3})

y^{^{''}} - b y = 0

49 y^{^{''}} - 168 y^{^{'}} + 144 y = 0

(D^{3} - 8) y = 0

4 \frac{d ^{2} y}{d x ^{2}} + 36 y = 0

2 y - 3 y^{^{''}} = 0