Question # Find if the following first order differential equations seperable,linear,exact,almost exact,homogeneous.(x)(e^y)(dy/dx) = e^(-2x) + e^(y-2x)

Differential equations
ANSWERED Find if the following first order differential equations seperable, linear, exact, almost exact, homogeneous, or Bernoulli. Rewrite the equation into standard form for the classification it fits.
$$\displaystyle{\left({x}\right)}{\left({e}^{{y}}\right)}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}={e}^{{-{2}{x}}}+{e}^{{{y}-{2}{x}}}$$ 2021-09-16
Separation of variables: If in an equation, it is possible to get all the functions of x and dx to one side and all the functions of y and dy to the other, the variables are said to be separable.
Linear Differential Equation: A differential equation is called linear if every dependent variable and every derivative involved occurs in the first degree only, and no products of dependent variables and/or derivatives occur.
Exact Differential Equation: The necessary and efficient for the differential equation $$\displaystyle{\left({M}\right)}{\left.{d}{x}\right.}+{\left({N}\right)}{\left.{d}{y}\right.}={0}$$ to be exact is
$$\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}={\frac{{\partial{n}}}{{\partial{x}}}}$$
Homogeneous equation: A differential equation of first order and first N degree is said to be homogeneous if it can be put in the form
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={f{{\left({\frac{{{y}}}{{{x}}}}\right)}}}$$
Or, equations of the type $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{A}{\left({x},{y}\right)}}}{{{B}{\left({x},{y}\right)}}}}$$ where
$$\displaystyle{A}{\left(\lambda{x},\lambda{y}\right)}=\lambda^{{d}}{A}{\left({x},{y}\right)}$$
$$\displaystyle{B}{\left(\lambda{x},\lambda{y}\right)}=\lambda^{{d}}{B}{\left({x},{y}\right)}$$
are homogeneous equations of degree d.
Bernoulli’s Equation: An equation of the form
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{P}{y}={Q}{y}^{{n}}$$
where P and Q are constants or functions of x alone (amd not of y) and n is constant except 0 and 1, is called a Bernoulli’s differential equation.
The given differential equation is $$\displaystyle{\left({x}\right)}{\left({e}^{{y}}\right)}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}={e}^{{-{2}{x}}}+{e}^{{{y}-{2}{x}}}$$
It can be written as $$\displaystyle{\left({x}\right)}{\left({e}^{{y}}\right)}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}={e}^{{-{2}{x}}}+{e}^{{{y}-{2}{x}}}$$
implies $$\displaystyle{\left({x}\right)}{\left({e}^{{y}}\right)}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}={e}^{{-{2}{x}}}-{e}^{{y}}{e}^{{-{2}{x}}}$$
implies $$\displaystyle{\left({x}\right)}{\left({e}^{{y}}\right)}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}={e}^{{-{2}{x}}}{\left({1}-{e}^{{y}}\right)}$$
implies $$\displaystyle{\frac{{{e}^{{y}}}}{{{1}-{e}^{{y}}}}}{\left.{d}{y}\right.}=\frac{{1}}{{x}}{e}^{{-{2}{x}}}{\left.{d}{x}\right.}$$
The function of x, dx and y,dy can be written on different sides, so the equation is separable
It has products of dependent variables, so this is not a linear differential equation.
Here $$\displaystyle{M}=\frac{{1}}{{x}}{e}^{{-{2}{x}}}{\quad\text{and}\quad}{N}={\frac{{{e}^{{y}}}}{{{1}-{e}^{{y}}}}}$$
Now, $$\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}={0}{\quad\text{and}\quad}{\frac{{\partial{n}}}{{\partial{x}}}}={0}$$
So, $$\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}={\frac{{\partial{n}}}{{\partial{x}}}}$$. This is an exact equation
The given equation can not be written in the form $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={f{{\left({\frac{{{y}}}{{{x}}}}\right)}}}.$$
So, it is not a homogeneous equation
$$\displaystyle{\left({x}\right)}{\left({e}^{{y}}\right)}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}={e}^{{-{2}{x}}}+{e}^{{{y}-{2}{x}}}$$
implies $$\displaystyle{\left({x}\right)}{\left({e}^{{y}}\right)}{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}={e}^{{-{2}{x}}}+{e}^{{y}}{e}^{{{2}{x}}}$$
implies $$\displaystyle{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}-{\frac{{{1}}}{{{x}}}}{e}^{{-{y}}}{e}^{{-{2}{x}}}=-\frac{{1}}{{x}}{e}^{{-{2}{x}}}$$
which is not the required form. So this is not a Bernoulli's equation.