A differencial equation,

P(x)y''+Q(x)y'+R(x)y=0

Is exact if and only if,

P''(x)-Q'(x)+R(x)=0.

~~~

Now, you are given, a second order linear differencial equation

P(x)y''+Q(x)y'+R(x)y=0

You multiply by a function µ(x) called "integrating factor":

µ(x)P(x)y''+µ(x)Q(x)y'+µ(x)R(x)y =0

Now, for it to be exact we use the neccesarry and sufficient condition stated above,

[µ(x)P(x)]''-[µ(x)Q(x)]'+[µ(x)R(x)]=0

Now, if you use the product rules and expand this differencial equation and combine the µ's you should get (I am hoping):

P(x)µ''(x)+[2P'(x)-Q]µ'(x)+[P''(x)-Q'(x)+R(x)]µ(x)=0