y^'=-2xe^y

Lösung

y^{^{'}} = - 2 x e^{y}

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

\frac{d x}{d t} = e^{2 x - t}

2 y y^{^{'}} = 9 t^{2}, y (0) = 8

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

(y^{2} + y) - x \frac{d y}{d x} = 0

y^{^{'}} - 3 y^{^{'}} + 2 y = 0