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Differential equations substitution problem

  1. Jun 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve using an appropriate substitution
    ydx+(1+ye^x)dy=0


    2. Relevant equations
    N/A


    3. The attempt at a solution
    u=y
    du/dx=dy/dx
    u+(1+ue^x)du/dx=0

    This is where it gets really sticky for me. I can't see it being a separable variable problem because it seems impossible to separate the u and the x. It also seems very unlikely that it is a linear d.e. or a bernoulli d.e. because the integrating factor would be e^∫(1/(1+ue^x)dx ) which seems impossible to do. The answer is e^(-x)=ylny+cy.
     
  2. jcsd
  3. Jun 3, 2012 #2

    tiny-tim

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    welcome to pf!

    hi lat77! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)
    since they give you the answer,

    a "cheat" way is to get the constant on its own

    e-x/y = lny + C …

    does that help? :smile:
     
  4. Jun 3, 2012 #3
    Not really. But on top of that, I need to know how to do it because I have a test Tuesday, so I need to be able to work the problem forwards since the answers obviously won't be given on the test.
     
  5. Jun 3, 2012 #4
    I tried many things, integrating factor did not seem practical (how to discover integrating factor is e^(-x)/y^2?)). Bernoulii equation does not apply. So, since they say subsititution, I tried what i thought might be natural:

    Your substitution u=y is not great, can you see why?

    How about u=1+ye^x or u=ye^x.

    Unfortunately, these got me close to a Bernoulli differential equation, but no cigar. Please let us know what the method ends up being!!!

    EDIT: Did this come from a textbook or another source, it may help if we have a context of similar problems, a chapter, an ambient discussion et cetera.
     
  6. Jun 3, 2012 #5

    SammyS

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    The substitution u = y does nothing !

    While you won't be given a solution on your test, following tim's advice to work the problem "backwards" may give you insight into the process of finding a substitution.
     
  7. Jun 3, 2012 #6
    Try v = ye^x.


    The way I came with the substitution was to try separating, but it failed. So I tried v = 1 + ye^x, but it failed. So I tried v = ye^x and it worked. A lot of differential equations don't have immediately obvious methods for solving them and you just have to try methods.

    Also, you should really consider why you tried u = y and why there is no hope that could of lead to something.
     
    Last edited: Jun 3, 2012
  8. Jun 4, 2012 #7

    tiny-tim

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    Nice, Joriss! :smile:

    lat77, that's the way you have to solve these things …

    you try something, it doesn't work, you try something else, it doesn't work, you … get the picture? :wink:

    just keep bashing your head against that brick wall until something gives! :biggrin:

    Just what makes that little old ant
    Think he'll move that rubber tree plant?
    Anyone knows an ant, can't
    Move a rubber tree plant

    But he's got high hopes, he's got high hopes
    He's got high apple pie, in the sky hopes

    So any time you're gettin' low
    'stead of lettin' go
    Just remember that ant

    Oops there goes another rubber tree
    Oops there goes another rubber tree
    Oops there goes another rubber tree plant!​

    Good luck on Tuesday! :smile:
     
  9. Jun 4, 2012 #8

    SammyS

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    Is there somewhere on PF to nominate the post of the day ... or week ... or month ... ?

    I think tim deserves a nomination !
     
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