# 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(ey/e2x)

We can arrange it as:

dy/ey= (x /e2x)dx
On integrating both the sides, we get;

∫(x /e2x)dx = ∫(dy/ey

∫(x e-2x)dx = ∫(dy/ey)

On solving we will get;

e-y = 1/2  e-2x (x+1/2) +A