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

The solution for (x-y)dx+(x-y+1)dy=0 is y=(-\W(-1/(e^{-c_{1)-4x}})+1)/2+x

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

(x - y) d x + (x - y + 1) d y = 0

y = \frac{- W ( - \frac{1}{e ^{- c_{1} - 4 x}} ) + 1}{2} + x

Solution steps

(x - y) d x + (x - y + 1) d y = 0

Solve using substitution:

y = \frac{- W ( - \frac{1}{e ^{- c_{1} - 4 x}} ) + 1}{2} + x

y = \frac{- W ( - \frac{1}{e ^{- c_{1} - 4 x}} ) + 1}{2} + x

x \to \infty + lim (x \cdot + \frac{1}{x})

\frac{1}{4} (x^{8} + 8)

\frac{\partial}{\partial y} (y + 3 x y^{2} - 18)

f (s) = \frac{s}{4}

\frac{\partial}{\partial x} (x^{2} - 2 x y + 3 y^{2})

What is the solution for (x-y)dx+(x-y+1)dy=0 ?
The solution for (x-y)dx+(x-y+1)dy=0 is y=(-\W(-1/(e^{-c_{1)-4x}})+1)/2+x