2y^{''}+3y^'+5y=0

Lösung

2 y^{^{''}} + 3 y^{^{'}} + 5 y = 0

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{31 t}{4}) + c_{2} sin (\frac{31 t}{4}))

Schritte zur Lösung

2 y^{''} (t) + 3 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{31 t}{4}) + c_{2} sin (\frac{31 t}{4}))

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{31 t}{4}) + c_{2} sin (\frac{31 t}{4}))

y^{^{''}} + y^{^{'}} + \frac{5}{4} y = 0, y (0) = 0, y^{^{'}} (0) = 0

\frac{y ^{^{''}} + 2 y ^{^{'}}}{x + y} = 0

y^{^{''}} + a y = 0, y (c) = 0, y (0) = 0

x y^{^{''}} + x y^{^{'}} = 0

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