4x^{''}+2x=0

Lösung

4 x^{^{''}} + 2 x = 0

x = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2})

Schritte zur Lösung

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

Solve linear ODE:

x = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2})

x = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2})

y^{^{''}} - 10 y^{^{'}} - 98 y = 0

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

y^{^{''}} - 9 y^{^{'}} + 7 y = 0

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

5 y^{^{''}} + 5 y^{^{'}} + 10 y = 0