What is the solution for x(y-1)dx+(x+1)dy=0,x=2 ?

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

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x(y-1)dx+(x+1)dy=0,x=2

x (y - 1) d x + (x + 1) d y = 0, x = 2

y = e^{- x - 1 + c_{1}} x + e^{- x - 1 + c_{1}} + 1

Solution steps

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

Solve separable ODE:

y = e^{- x - 1 + c_{1}} x + e^{- x - 1 + c_{1}} + 1

y = e^{- x - 1 + c_{1}} x + e^{- x - 1 + c_{1}} + 1

y^{^{'}} = \frac{e ^{- x} - e ^{x}}{3 + 4 y}, y (0) = 1

y^{2} - 1 - (2 y + x y) y^{^{'}} = 0

\frac{d y}{d x} x^{3} 4 y = 5

y^{5} + 9 y^{^{'}} = 0

\frac{d y}{d x} = 4 e^{x - y}, y (0) = ln (6)

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