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

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

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

(x + y) d x - (x + y - 2) d y = 0

y = - W (\frac{1}{2 e ^{2 x + c_{1}}}) + 1 - x

Solution steps

(x + y) d x - (x + y - 2) d y = 0

Solve using substitution:

y = - W (\frac{1}{2 e ^{2 x + c_{1}}}) + 1 - x

y = - W (\frac{1}{2 e ^{2 x + c_{1}}}) + 1 - x

\frac{d}{d x} ((x^{2} - 1) (x^{2} + 1))

sin^{2} (2 x)

\int \frac{1 - x}{e ^{x}} d x

\int t sec^{2} (2 t) d t

\frac{\partial}{\partial y} (\frac{y}{( x + y ) ^{2}})

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