# More differential equations

1. Feb 16, 2006

### 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…

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

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

2. Feb 16, 2006

### HallsofIvy

Staff Emeritus
You have my permission to kick yourself in the behind!

(2y- ex)y'= yex is the same as

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