What is the solution for (1+xy)y^'+y^2=0 ?

The solution for (1+xy)y^'+y^2=0 is y=(\W(xc_{1}))/x

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(1+xy)y^'+y^2=0

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

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

Solution steps

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

Solve using substitution:

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

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

y = (arctan (5 x))^{2}

k = 0 \sum \infty (\frac{x - 4}{5})^{k}

\frac{\partial}{\partial x} (2 x sin (y - z))

\frac{3 e ^{(- 2 s)}}{( s ^{2} + 2 s + 5 )}

\int \frac{3 x ( 2 )}{( x )} d x

What is the solution for (1+xy)y^'+y^2=0 ?
The solution for (1+xy)y^'+y^2=0 is y=(\W(xc_{1}))/x