y^{'''}+y^'=sec(t),y^'(0)=3

Lösung

y^{^{'''}} + y^{^{'}} = sec (t), y^{^{'}} (0) = 3

y = c_{1} + c_{2} cos (t) + 3 sin (t) + ln (tan (t) + sec (t)) - t cos (t) + ln (cos (t)) sin (t)

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} + c_{2} cos (t) + 3 sin (t) + ln (tan (t) + sec (t)) - t cos (t) + ln (cos (t)) sin (t)

y = c_{1} + c_{2} cos (t) + 3 sin (t) + ln (tan (t) + sec (t)) - t cos (t) + ln (cos (t)) sin (t)

(d^{2} - 2 d) y = e^{x} sin (x)

\frac{1}{2} y^{^{''}} + 2 y = 3 tan (2 t) - \frac{1}{3} e^{4 t}

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2 x} (cos (x) + sin (x))

y^{^{''}} - 3 y^{^{'}} + 3 y^{^{'}} - y = x - 8 e^{x}

y^{^{''}} + 100 y^{^{'}} + \frac{1}{0.0004} y = 30, y (0) = 0