Is the Differential Equation (2y - e^x)y' = ye^x Exact or Solvable?

[Pengwuino]

I hate this stuff!

The next problem I must face upon my path of enlightenment is the following:

Solve the initial value problem:

$$(2y - e^x )y' = ye^x ,y(0) = 1$$

I thought that it looked exact but I ended up with…

$$\begin{array}{l}<br /> M = (2y - e^x )dy,N = (ye^x )dx \\ <br /> \frac{{dM}}{{dx}} = - e^x ,\frac{{dN}}{{dy}} = e^x \\ <br /> \end{array}$$

So it's not exact… how should I go about solving this? I really suck at these differentials by the way!

 

[HallsofIvy]

You have my permission to kick yourself in the behind!

(2y- e^x)y'= ye^x is the same as

(2y- e^x)dy= ye^xdx but in order to check if it is an exact differential, you must write it as Mdy+ Ndx= 0.
That is: (2y-e^x)dy- ye^xdx= 0