# - ( 1 + x - y ) / (y - x) = dy/dx determine the solution of differential equation first order

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You need to solve the differential equation `- ( 1 + x- y ) / (y - x) = (dy)/(dx) =gt -1/(y-x) - (y-x)/(y-x) = (dy)/(dx)`

`-1/(y-x) - 1= (dy)/(dx)`

You should come up with the substitution `y - x = v` , differentiating both sides with respect to x yields, `(dy)/(dx) - 1 = (dv)/(dx) =gt (dy)/(dx) = (dv)/(dx) + 1.`

You need to substitute `(dv)/(dx) + 1` for `(dy)/(dx)` and `v` for `y- x` such that:

`-1/v - 1 = (dv)/(dx) + 1 =gt -1/v -2 = (dv)/(dx)`

You need to bring the terms from the left to a common denominator such that:

`(-1 - 2v)/v = (dv)/(dx) =gt v(dv)/(-1-2v) = dx`

Integrating both sides yields:

`int v(dv)/(-1-2v) = int dx =gt (-1/2)int (-2vdv)/(-1-2v) = x =gt`

`(-1/2)int ((-2v - 1)dv)/(-1-2v) - (1/2)int (dv)/(-1-2v)= x`

`x = (-1/2)int dv - (1/2)int (dv)/(-1-2v)`

You should come up with the substitution`-1-2v = u =gt -2dv = du =gt -dv = (du)/2`

`` `x = (-v/2) + (1/2)int (du)/(2u)`

`x = (-v/2) + (1/4)*ln|u| + c`

`x = (-v/2) + (1/4)*ln|-1-2v| + c`

You need to substitute y - x for v such that:

`x = (x-y)/2 + (1/4)*ln|-1+ 2x - 2y| + c`

`4x = 2x - 2y + ln|-1+ 2x - 2y| + c =gt 2x = ln|-1+ 2x - 2y| - 2y + c`

`x = ln sqrt(2x-2y-1) - y + c =gt y + x - c= ln sqrt(2x-2y-1)`

`sqrt(2x-2y-1) = e^(y + x - c) =gt 2x-2y-1 = e^(2y + 2x - c)`

`2y = 2x - 1 - e^(2y + 2x - c) =gty = x - 1/2 - (e^(2y + 2x - c))/2`

**Hence, the general solution to this ODE is `y = x - 1/2 - (e^(2y + 2x - c))/2.` **