y^{(4)}-8y^'=0

Lösung

y^{(4)} - 8 y^{^{'}} = 0

y = c_{1} + c_{2} e^{2 t} + e^{- t} (c_{3} cos (3 t) + c_{4} sin (3 t))

Schritte zur Lösung

y^{(4)} - 8 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{2 t} + e^{- t} (c_{3} cos (3 t) + c_{4} sin (3 t))

y = c_{1} + c_{2} e^{2 t} + e^{- t} (c_{3} cos (3 t) + c_{4} sin (3 t))

y^{^{'}} = \frac{1}{sin ( y )}

\frac{d x}{d t} = 8 x t^{4}

d y = \frac{( y + 7 )}{( x + 4 )} d x

y^{^{'}} = (x + 1) e^{y}

\frac{d y}{d t} = \frac{- c ^{2} 2 g y}{a ^{2}}