How do I solve this differential equation: dy/dx=1/(y-x)

It's supposed to be linear in x. I tried to solve it like this:
dy/dx=1/(y-x)
(y-x)dy/dx=1
(y-x)=dx/dy
y=(dy/dx)+x
but that's where I get stuck. Am I not seeing something?

dy/dx = 1/(y - x)

dx/dy = y - x

dx/dy + x = y

Integrating factor = e^(∫dy) = e^(y)

Multiplying both sides by integrating factor:

e^(y)(dx/dy) + xe^(y) = ye^(y)

d/dy[e^(y)(x)] = ye^(y)

e^(y)(x) = ∫ye^(y) dy

e^(y)(x) = ye^(y) - e^(y) + C

x = y - 1 + C*e^(-y)

Note that in this case, you could've treated dy/dx as a ratio and then transform the first order linear differential equation of the form y' + P(x)y = Q(x) to x' + P(y)x = Q(y).

I'm glad this helped!

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dy/dx isn't a ratio. It can't be flipped like that.

The way to approach this is to make a substitution. Let u be a function of x. Then suppose
y - x = u
y = u + x
y' = u' + 1

Then
u' + 1 = 1/u

Now it's separable:
u' = (1/u) - 1 = (1 - u)/u
u du / (1 - u) = dx

Now you can integrate and back-substitute. Don't forget the constant of integration.

Let y = u + x so that dy/dx = du/dx + 1 and y - x = u:

dy/dx = 1/(y - x)

du/dx + 1 = 1/u

du/dx = 1/u - 1

du/(1/u - 1) = dx

u/(1 - u) du = dx

∫ u/(1 - u) du = ∫ dx

∫ [1/(1 - u) - 1] du = x + C

-ln(1 - u) - u = x + C

By the substitution u = y - x, the solution can finally be expressed in terms of x and y:

-ln(1 + x - y) + x - y = x + C

y = C - ln(1 + x - y).