What is the solution for (y+1)dx+(y-1)(1+x^2)dy=0 ?

The solution for (y+1)dx+(y-1)(1+x^2)dy=0 is y=-2\W(-1/(2e^{(1-c_{1)-arctan(x))/2}})-1

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(y+1)dx+(y-1)(1+x^2)dy=0

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

y = - 2 W (- \frac{1}{2 e ^{\frac{1 - c _{1} - a r c t a n ( x )}{2}}}) - 1

Solution steps

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

Solve separable ODE:

y = - 2 W (- \frac{1}{2 e ^{\frac{1 - c _{1} - a r c t a n ( x )}{2}}}) - 1

y = - 2 W (- \frac{1}{2 e ^{\frac{1 - c _{1} - a r c t a n ( x )}{2}}}) - 1

\frac{\partial}{\partial y} (- 3 y^{4} sin (3 x))

\frac{d}{d x} (4 x^{2} + 5 x - 10)

x^{^{''}} + 9 x^{^{'}} = 27, x (0) = 4, x^{^{'}} (0) = 6

x \to 0 lim (\frac{cos ( 1 )}{x})

\frac{d}{d x} (a ln (x^{2}))

What is the solution for (y+1)dx+(y-1)(1+x^2)dy=0 ?
The solution for (y+1)dx+(y-1)(1+x^2)dy=0 is y=-2\W(-1/(2e^{(1-c_{1)-arctan(x))/2}})-1