What is the solution for (y+xy)+(x-xy)y^'=0 ?

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

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

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

y = - W (- \frac{1}{e ^{x - c_{1}} x})

Solution steps

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

Solve separable ODE:

y = - W (- \frac{1}{e ^{x - c_{1}} x})

y = - W (- \frac{1}{e ^{x - c_{1}} x})

\frac{d y}{d x} + y = y x e^{x + 8}

y^{^{'}} = 1 - y, y (0) = 0

x y \frac{d x}{d y} = y + 2

y^{^{'}} = \frac{cos ( x ) e ^{s i n (x)}}{2 y}

\frac{d P}{d t} = 5 Pt, P (1) = 5

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