Solving $2(y+e^x)dy + (y^2+4y e^x)dx = 0 $ and understanding integrating-factors

Question

We want to solve $2(y+e^x)dy + (y^2+4y e^x)dx = 0 $ which, across the spectrum is the standard format of the integrating factor technique for ODE. My book, however, covers only the integrating factors of the forms \begin{align} \dfrac{1}{N(x,y)} \cdot \left( \dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} \right)\\ \dfrac{1}{M(x,y)} \cdot \left( \dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial y} \right) \end{align} where of course we want the first expression to be a function of $x$ only or the second to be a function of $y$ only.

In the above example however, doing these calculations yields nothing like that, and cannot be covered by my book. How can one proceed further?

Answer 1

$$2(y+e^x)dy + (y^2+4y e^x)dx = 0$$ $$(2ydy+y^2dx) + (2e^xdy+4y e^xdx) = 0$$ $$(dy^2+y^2dx )+ 2(e^xdy+2y e^xdx) = 0$$ Multiply by $e^x$: $$(e^xdy^2+y^2de^x) + 2(e^{2x}dy+y de^{2x}) = 0$$ $$d(y^2e^x) + 2d(e^{2x}y) = 0$$ Integrate: $$y^2e^x + 2e^{2x}y = C$$

Answer 2

Define: $u=e^x$

$$\frac{du}{dy}=\frac{2\frac{u}{y}\left( 1+\frac{u}{y}\right)}{1+4\frac{u}y}$$

Define: $w=u/y$

$$\frac{dw}{dy}=\frac{w(1-2w)}{(1+4w)y}$$

So it becomes a separable differential equation. now you can proceed.