(1+x^2)y^'+xe^{-y}=0

Lösung

(1 + x^{2}) y^{^{'}} + x e^{- y} = 0

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

Schritte zur Lösung

(1 + x^{2}) y^{'} (x) + x e^{- y} = 0

Solve separable ODE:

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

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

x^{^{'}} = a x (1 - \frac{x}{c})

(x^{2} - 20 x + 75) y^{^{'}} = y

\frac{d y}{d x} = - \frac{7}{2} x^{4} y^{5}

2 x y (\frac{d y}{d x}) = 3

\frac{d y}{d x} = 5 x (y + 4)^{3}