x^{''}+2ax^'+bx=0

Lösung

x^{^{''}} + 2 a x^{^{'}} + b x = 0

x = c_{1} e^{(- a + a^{2} - b) t} + c_{2} e^{(- a - a^{2} - b) t}

Schritte zur Lösung

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

Solve linear ODE:

x = c_{1} e^{(- a + a^{2} - b) t} + c_{2} e^{(- a - a^{2} - b) t}

x = c_{1} e^{(- a + a^{2} - b) t} + c_{2} e^{(- a - a^{2} - b) t}

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

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

y^{^{'''}} + 9 y^{^{'}} = 0, y (0) = 5, y^{^{'}} (0) = 4, y^{^{''}} (0) = 3

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

y^{^{''''}} + 2 y^{^{'''}} - 4 y^{^{''}} - 2 y^{^{'}} + \frac{1}{2} y = 0