y^{'''}+4y^'=sec(2t)

Lösung

y^{^{'''}} + 4 y^{^{'}} = sec (2 t)

y = c_{1} + c_{2} cos (2 t) + c_{3} sin (2 t) + \frac{1}{8} ln (tan (2 t) + sec (2 t)) - \frac{1}{4} t cos (2 t) + \frac{1}{8} ln (cos (2 t)) sin (2 t)

Schritte zur Lösung

y^{'''} (t) + 4 y^{'} (t) = sec (2 t)

Solve linear ODE:

y = c_{1} + c_{2} cos (2 t) + c_{3} sin (2 t) + \frac{1}{8} ln (tan (2 t) + sec (2 t)) - \frac{1}{4} t cos (2 t) + \frac{1}{8} ln (cos (2 t)) sin (2 t)

y = c_{1} + c_{2} cos (2 t) + c_{3} sin (2 t) + \frac{1}{8} ln (tan (2 t) + sec (2 t)) - \frac{1}{4} t cos (2 t) + \frac{1}{8} ln (cos (2 t)) sin (2 t)

D (D - 1)^{3} (y) = (x^{2} + 2 x) e^{x}

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

y^{^{''}} + y^{^{'}} + 64 y = 5 cos (2 t)

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

y^{^{''}} - y^{^{'}} + 14 y = \frac{4 + e x}{2}