24y^{''}+14y^'-3y=0

Lösung

24 y^{^{''}} + 14 y^{^{'}} - 3 y = 0

y = c_{1} e^{\frac{t}{6}} + c_{2} e^{- \frac{3 t}{4}}

Schritte zur Lösung

24 y^{''} (t) + 14 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{6}} + c_{2} e^{- \frac{3 t}{4}}

y = c_{1} e^{\frac{t}{6}} + c_{2} e^{- \frac{3 t}{4}}

y^{^{''}} + 5 \cdot y^{^{'}} + a \cdot y = 0, y (0) = 1, y^{^{'}} (0) = 7

5 y^{^{''}} + c y^{^{'}} + 45 y = 0, y (0) = - 8, y^{^{'}} (0) = 0

(d^{5} - d^{3}) \cdot y = 0

y^{^{''}} = k y (x)

4 \frac{d ^{2} x}{d t ^{2}} + c \frac{d x}{d t} + k (x) = 0