xy = 2x-y

y = 2x/(1+x)

Let integrate both sides with respect to x

∫ydx =∫2x/(1+x) dx

we can write 2x/(1+x) = A+B/(1+x)

Then 2x = A(x+1)+B = Ax+(A+B)

So A=2 and A+B = 0

Therefore finally A=2 and B=-2

∫ydx =∫2x/(1+x) dx

=∫ 2-2/(1+x) dx

=...

## See

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xy = 2x-y

y = 2x/(1+x)

Let integrate both sides with respect to x

∫ydx =∫2x/(1+x) dx

we can write 2x/(1+x) = A+B/(1+x)

Then 2x = A(x+1)+B = Ax+(A+B)

So A=2 and A+B = 0

Therefore finally A=2 and B=-2

∫ydx =∫2x/(1+x) dx

=∫ 2-2/(1+x) dx

= ∫2dx-∫2/(1+x) dx

**∫ydx = 2x-2log(1+x)+c**

You need to isolate all the terms that contain y to the left side such that:

`xy + y = 2x`

You need to factor out y to the left side such that:

`y(x + 1) = 2x =gt y = 2x/(x+1)`

Notice that you may write the equation such that:

`y = (x + 1 - 1 + x + 1 - 1)/(x+1)`

`y = (x+1)/(x+1) + (x+1)/(x+1) - 2/(x+1)`

`y = 1 + 1 - 2/(x+1)`

`y = 2 - 2/(x+1)`

Notice that the value of y will be integer if the fraction `2/(x+1)` is an integer number, hence, the denominator x + 1 needs to be oneof the divisors of the numerator 2 such that:

`x + 1 = {+-1 ; +-2}`

`x + 1 = 1 =gt x = 0 =gt y = 0`

`x + 1 = -1 =gt x = -2 =gt y = -4/(-1) =gt y = 4`

`x + 1 = 2 =gt x = 1 =gt y = 2/2 =gt y = 1`

`x + 1 = -2 =gt x = -3 =gt y = -6/(-2) =gty = 3`

**Hence, evaluating all integral solutions to the equation yields `(0;0) ; (-2;4) ; (1;1) ; (-3;3).` **