# Thread: Help with some Diff EQ

1. ## Help with some Diff EQ

Ok here's my problem, i'm really having trouble with:

"If the equation P(x)y" + Q(x)y' + R(x)y = 0 is not exact, it can be made exact by multiplying by a suitable integrating factor µ(x). Thus, µ(x) must satisfy the condition that the equation µ(x)P(x)y" +µ(x)Q(x)y' + µ(x)R(x)y = 0 is expressible in the form [µ(x)P(x)y']' + [S(x)y]' = 0 for some function S(x)."

So, what i need to do then is show that µ(x) must be a solution of the adjoint equation P(x)µ'' + [2P'(x) - Q(x)]µ' +[P"(x) - Q'(x) + R(x)]µ = 0

Thanks for any and all help!

2. 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 &#181;(x) called "integrating factor":
&#181;(x)P(x)y''+&#181;(x)Q(x)y'+&#181;(x)R(x)y =0
Now, for it to be exact we use the neccesarry and sufficient condition stated above,
[&#181;(x)P(x)]''-[&#181;(x)Q(x)]'+[&#181;(x)R(x)]=0
Now, if you use the product rules and expand this differencial equation and combine the &#181;'s you should get (I am hoping):
P(x)&#181;''(x)+[2P'(x)-Q]&#181;'(x)+[P''(x)-Q'(x)+R(x)]&#181;(x)=0