8y^{''}+2y^'=0

Lösung

8 y^{^{''}} + 2 y^{^{'}} = 0

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

Schritte zur Lösung

8 y^{''} (t) + 2 y^{'} (t) = 0

Solve linear ODE:

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

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

15 x^{^{''}} + 44 x = 0

x^{^{''}} - 7 x^{^{'}} + 10 x = 0

\frac{3}{4} x^{^{''}} + 9 x = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 6, y^{^{'}} (0) = 1

y^{^{''}} + 6 y^{^{'}} + 35 y = 0