25y^{''}+20y^'+4y=0

Lösung

25 y^{^{''}} + 20 y^{^{'}} + 4 y = 0

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

Schritte zur Lösung

25 y^{''} (t) + 20 y^{'} (t) + 4 y = 0

Solve linear ODE:

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

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

4 \frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} - 10 y = 0

(4 D^{2} - 4 D + 5) y = 0

y^{^{''}} + 6 y^{^{'}} + 10 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 4 y^{^{'}} + 16 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = - 2, y^{^{'}} (0) = 1