20y^{''}+3y^'-2y=0

Lösung

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

y = c_{1} e^{\frac{t}{4}} + c_{2} e^{- \frac{2 t}{5}}

Schritte zur Lösung

20 y^{''} (t) + 3 y^{'} (t) - 2 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{4}} + c_{2} e^{- \frac{2 t}{5}}

y = c_{1} e^{\frac{t}{4}} + c_{2} e^{- \frac{2 t}{5}}

3 \frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 8 y = 0

625 y^{^{''}} - 81 y = 0, y (0) = 9, y^{^{'}} (0) = 6

(D^{3} - D^{2} - 4 D + 4) y = 0

4 y^{^{''}} + 16 y^{^{'}} + \frac{1}{16} y = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = 0, y (0) = 10, y^{^{'}} (0) = 0