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Math Help - Help with some Diff EQ

  1. #1
    asdf123
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    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 (x). Thus, (x) must satisfy the condition that the equation (x)P(x)y" +(x)Q(x)y' + (x)R(x)y = 0 is expressible in the form [(x)P(x)y']' + [S(x)y]' = 0 for some function S(x)."

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

    Thanks for any and all help!
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  2. #2
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    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
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