y^{''}-20y^'+96y=0

Lösung

y^{^{''}} - 20 y^{^{'}} + 96 y = 0

y = c_{1} e^{12 t} + c_{2} e^{8 t}

Schritte zur Lösung

y^{''} (t) - 20 y^{'} (t) + 96 y = 0

Solve linear ODE:

y = c_{1} e^{12 t} + c_{2} e^{8 t}

y = c_{1} e^{12 t} + c_{2} e^{8 t}

y^{^{''}} - y = 0, y (- 1) = 9, y^{^{'}} (- 1) = - 9

y^{^{'''}} - y^{^{''}} + 9 y^{^{'}} - 145 y = 0, y (0) = e^{5 x}

f^{^{''}} (x) - f^{^{'}} (x) - 2 f (x) = 0, f (0) = 2

100 (1 0^{- 6}) y^{^{''}} + \frac{1}{500} y^{^{'}} + y = 0

y^{^{''}} + 3 y = 0, y^{^{'}} (1) = 1