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Homework Help: Working Out Answer Checks - DE

  1. Sep 12, 2010 #1
    1. The problem statement, all variables and given/known data

    The symbolic solution of y'=-2xy, y(0)=2 is [tex]y(x)=2e^{-x^2}[/tex]. Display the details of the linear integrating factor method derivation of this symbolic solution, plus a full answer check.

    2. Relevant equations

    Linear integration factor method uses the standard for y'+p(x)y=q(x)

    3. The attempt at a solution

    I can solve this no problem. I am having trouble doing the answer checks on these. With previous sections you could just take the derivative of both sides and end up with the original statement. However with this, I cannot seem to make it anywhere close to the original problem.

    This for example taking the solution from above:
    [tex]y(x)=2e^{-x^2}[/tex]

    If I take the derivative of both sides I end up with:

    [tex]y'=-4e^{-x^2}[/tex]

    Which isn't even remotely close.

    What concept am I missing here?
     
  2. jcsd
  3. Sep 12, 2010 #2

    LCKurtz

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    You are forgetting the chain rule when you differentiate.
     
  4. Sep 12, 2010 #3
    Hmm...I thought I did.

    Starting with:

    [tex]y=2e^{-x^{2}}[/tex]

    Without simplifying I get [tex]y'=2e^{-x^{2}}(-2x)[/tex]

    Am I really doing the derivative of something that simple wrong?:cry:

    I threw it into wolframalpha.com and maple and got the same answers.
     
  5. Sep 12, 2010 #4

    LCKurtz

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    No, that is right but it isn't what you had in your earlier post. Anyway, all you have left to do is to verify that that value of y' equals -2xy and that y satisfies the BC's. Pretty easy now.
     
  6. Sep 12, 2010 #5
    Oh yeah, I left off the extra x when I typed that out.

    I guess the part I am confused with is how I go from something with an exponential and no y to no exponential and a y.

    This is the first class I've had to do answer checks so I've never learned exactly how to do them.
     
  7. Sep 13, 2010 #6

    LCKurtz

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    You have the DE y' = -2xy and you are trying to show [itex]y=2e^{-x^2}[/itex] is a solution. You have calculated [itex]y'=-4xe^{-x^2}[/itex]. All you have to do is plug your expressions for y and y' into the DE and see it reduces it to an identity, which means the equation is satisfied.
     
  8. Sep 13, 2010 #7
    So I tried this...

    Here is what I have.

    DE = [tex]y'=-2xy[/tex]
    [tex]y=2e^{-x^{2}}[/tex]
    [tex]y'=-4xe^{-x^{2}}[/tex]

    Taking the original DE and subbing in what I think is correct.

    [tex]-4xe^{-x^{2}}=-2x(2e^{-x^{2}})[/tex]

    Then.
    [tex]-4xe^{-x^{2}}=-4xe^{-x^{2}}[/tex]

    So both values are equal but I don't see an identity.
     
  9. Sep 13, 2010 #8

    LCKurtz

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    Here's a definition for you:

    Identity Equation: An equation which is true for every value of the variable is called an identity equation.

    Now do you see one?
     
  10. Sep 13, 2010 #9
    Ooooh, yes I get it now!

    I was thinking something like trig identity.

    Perfect. Thanks for the help. I am glad this class requires these on a lot of the problems. One tends to learn A LOT more when they do the answer one way then back again.
     
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