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

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

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(x+y)y^'=y

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

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

Solution steps

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

Solve using substitution:

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

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

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