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

Lösung

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

y = - ln (- \frac{x ^{2}}{2} - x - c_{1})

Schritte zur Lösung

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

Solve separable ODE:

y = - ln (- \frac{x ^{2}}{2} - x - c_{1})

y = - ln (- \frac{x ^{2}}{2} - x - c_{1})

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

(v + 1) (v d x + x d v) = 0

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

\frac{d V}{d t} = - V

x^{3} y^{^{'}} = y (x^{3} + x^{2})