Differential equation of form: $y^{\prime} + a(x)y = f(x,y)$ I have been working on the solution to the following differential equation:
$$y' + y = \frac{2x}{e^x(1+ye^x)}$$
I am not sure how to solve this type of problem because of the y on the right side... I tried to multiply both sides by the bottom. From there I tried to find the solution as though it were:
$$(1+ye^x)y' + y(1+ye^x) = 0$$
However, I thought that this method of solution required the $y'$ to be isolated. If that is not the case, then I am solving it fine. If that is the case, then what am I meant to be doing?
 A: Let $$y=ue^{-x}$$
$$y'=u'e^{-x}-ue^{-x}$$
Your differential equation transforms into 
$$u'=\frac {2x}{1+u}$$
 Which is separable.
$$\int (1+u)du = \int 2x dx$$
Solve for $u$ and substitute in $y=ue^{-x}$ to find $y$
A: $$y' + y = \frac{2x}{e^x(1+ye^x)}$$
$$\implies \{ye^x(1+ye^x)-2x\}dx+e^x(1+ye^x)dy=0$$
which is of the form $$M(x,y)dx+N(x,y)dy=0$$
where $$M(x,y)=ye^x(1+ye^x)-2x\qquad \text{and}\qquad N(x,y)=e^x(1+ye^x)$$
Now $$\frac{\partial M}{\partial y}=e^x+2ye^{2x}\qquad \text{and}\qquad \frac{\partial N}{\partial x}=e^x+2ye^{2x}$$
Clearly, $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
So the given differential equation is exact differential equation and its solution is 
$$\int_{\text{treating $~y~$} as constant}M~dx~+~\int(\text{terms in $~N~$not containing $~x~$})~dy~=~c$$
$$\implies \int_{\text{treating $~y~$} as constant}\{ye^x(1+ye^x)-2x\}~dx~+~\int0~dy~=~c$$
$$\implies ye^x~+~\frac{1}{2}y^2~e^{2x}-x^2~=~c$$
$$\implies y=\frac{-e^x\pm\sqrt{e^{2x}- 2~e^{2x}\cdot (-x^2-c)}}{e^{2x}}$$
$$\implies y=\frac{-1\pm\sqrt{1- 2(-x^2-c)}}{e^{x}}$$ 
where $~c~$is constant.
