2y^{''}-2y^'+4y=0

Lösung

2 y^{^{''}} - 2 y^{^{'}} + 4 y = 0

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

Schritte zur Lösung

2 y^{''} (t) - 2 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (0) = - 3, y^{^{'}} (0) = 21

x^{^{''}} + 6 x^{^{'}} + 10 x = 0

y^{^{''}} + 4 y^{^{'}} + 7 y = 0, y (0) = 3, y^{^{'}} (0) = 7

2 y^{^{''}} - 4 y^{^{'}} - 5 y = 0

y^{^{''''}} + 2 y^{^{'''}} - 5 y^{^{''}} - 5 y^{^{'}} + \frac{1}{2} y = 0