What is the differential equation for dy/dx = xe^(y-2x)?

dy/dx = xe^(y-2x) , i am asked to form differential equation using this equation . the ans given is (e^-y) = 0.5(e^-2x)(x+0.5) + a , how to get the answer?

Solution:

Given: dy/dx = xe^(y-2x)

dy/dx = x(e^y/e^2x)

We can arrange it as:

dy/e^y= (x /e^2x)dx
On integrating both the sides, we get;

∫(x /e^2x)dx = ∫(dy/e^y) 

∫(x e^-2x)dx = ∫(dy/e^y)

On solving we will get;

e^-y = 1/2  e^-2x (x+1/2) +A