0.05y^{''}-0.3y^'+0.4y=11

Lösung

0.05 y^{^{''}} - 0.3 y^{^{'}} + 0.4 y = 11

y = c_{1} e^{4 t} + c_{2} e^{2 t} + \frac{55}{2}

Schritte zur Lösung

0.05 y^{''} (t) - 0.3 y^{'} (t) + 0.4 y = 11

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{2 t} + \frac{55}{2}

y = c_{1} e^{4 t} + c_{2} e^{2 t} + \frac{55}{2}

y^{^{''}} + 4 y = - 6

y^{^{''}} + y = sec^{2} (t)

y^{^{''}} - 12 y^{^{'}} + 36 y = 6 e^{6 x}

y^{^{'''}} - y^{^{'}} = 3 cos (x)

y^{^{''}} - y^{^{'}} - 2 y = 3 x + 4