Why is this first-order ODE amenable to the method of integrating factor/separable?

Question

Background

We know that the first order differential equation in the form $$\frac{dy}{dx}+p(x)y=q(x)$$ is amenable to the method of integrating factor

Question

How is this first order equation amenable to the method of integrating factor? The equation seems not to be in the desired form.

$$y^{\prime}+y=\frac{2 x e^{-x}}{1+y e^{x}}$$

Answer 1

$$y^{\prime}+y=\frac{2 x e^{-x}}{1+y e^{x}}$$ This ODE is not on the form $$\frac{dy}{dx}+p(x)y=q(x)$$ because $q(x)\neq q(x,y)=\frac{2 x e^{-x}}{1+y e^{x}}$ . $$y^{\prime}+y=\frac{2 x e^{-2x}}{e^{-x}+y}$$ Change of function : $$u(x)=e^{-x}+y(x)\quad\implies\quad y=u-e^{-x}$$ $$y'=u'+e^{-x}$$ $$y^{\prime}+y=(u'+e^{-x})+(u-e^{-x})=\frac{2 x e^{-2x}}{u}$$ Now the equation is a separable. $$(u'+u)u=2 x e^{-2x}$$ Let $\quad u^2=v(x)$ $$\frac12\: \frac{dv}{dx}+v(x)=2 x e^{-2x}$$ This is a first order linear ODE.

Answer 2

The left side is related to the derivative of $u(x)=e^xy(x)$. More specifically, $$ u'(x)=e^x(y'+y)=\frac{2x}{1+u(x)} $$ which is separable, $$ (1+u(x))^2=2x^2+C $$