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

The solution for x(x+y)y^'=y(x-y) is y= x/(\W(\frac{x^2){c_{1}})}

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

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

y = \frac{x}{W ( \frac{x ^{2}}{c _{1}} )}

Solution steps

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

Solve using substitution:

y = \frac{x}{W ( \frac{x ^{2}}{c _{1}} )}

y = \frac{x}{W ( \frac{x ^{2}}{c _{1}} )}

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What is the solution for x(x+y)y^'=y(x-y) ?
The solution for x(x+y)y^'=y(x-y) is y= x/(\W(\frac{x^2){c_{1}})}