y^'=e^y

Lösung

y^{^{'}} = e^{y}

y = - ln (- t - c_{1})

Schritte zur Lösung

y^{'} (t) = e^{y}

Solve separable ODE:

y = - ln (- t - c_{1})

y = - ln (- t - c_{1})

\frac{d}{d x} (4 sin (2 x) - 3 cos (3 x))

\int_{- 2}^{0} - x^{2} - 2 x d x

d er i v a t i v e f (x) = 3 x - 5

\int_{- 2}^{2} (6 x) - (x^{2} - 7) d x

a re a y = x^{2} - 4 x, y = 4 x