After taking dy/dx on other side it can be written as
dy/dx + y/√x = e-2√x
This is an common liner differential equation of form
dy/dx + Py = Q
Here P is 1/√x and Q is e-2√x
We solve this equation by this formula
ye∫Pdx = ∫(Qe∫pdx )dx +c
Therefore
ye∫dx/√x = ∫(e-2√x )(e∫dx/√x)
ye2√x = x