(x+xy)y^'+y=0

Lösung

(x + x y) y^{^{'}} + y = 0

y = W (\frac{c _{1}}{x})

Schritte zur Lösung

(x + x y) y^{'} (x) + y = 0

Solve separable ODE:

y = W (\frac{c _{1}}{x})

y = W (\frac{c _{1}}{x})

(7 + x) y^{^{'}} = (x + 8) (y - 7)

\frac{d y}{d x} = e^{x + y}, e^{x} + e^{y} = c

2 d x - e^{y - x} d y = 0

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

\frac{d S}{d t} = 0.06 - \frac{3 S}{1420}, t = 10