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Homogenous Differential Equation Solution

  1. Sep 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve the differential equation:
    [itex] \frac{dy}{dx} = x + y [/itex]



    2. Relevant equations
    Integrating factor equation


    3. The attempt at a solution
    Ok, I am aware that in order to solve this equation, I need to make a substitution:
    [itex] v = x + y [/itex]
    However, at this point I am unsure about what to do. Unfortunately, my DiffEq textbook does not cover this technique thoroughly. I am hoping that perhaps somebody can explain how this technique works, thanks.
     
  2. jcsd
  3. Sep 25, 2014 #2

    Ray Vickson

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    The substitution ##v = x+y## leads to another inhomogeneous DE, but simpler than the original. Have you tried it?

    I cannot believe that your textbook does not cover the methods for solving
    [tex] \frac{dy}{dx} + f(x) y = g(x) [/tex]
    You say it does not cover this technique thoroughly, but surely it covers it at least a little bit. You need to show your work, and explain where/why you are stuck.
     
  4. Sep 25, 2014 #3

    LCKurtz

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    @N3WTON: Alternatively, doesn't your text discuss constant coefficient homogeneous and non-homogeneous equations and the method of undetermined coefficients for NH equations?
     
  5. Sep 25, 2014 #4

    ehild

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    With the substitution, y=v-x, y'=v'-1 and the differential equation becomes v'-1=v, that is, v'=v+1. Is it not a separable equation?

    ehild
     
  6. Sep 26, 2014 #5
    It does discuss constant coefficient homogenous and non-homogenous equations, however; no mention of the method of undetermined coefficients for NH equations. Perhaps we have just not reached that point in the text yet
     
  7. Sep 26, 2014 #6
    I have tried it but I don't understand what I need to do. The equation you listed could be solved using the integrating factor method, no? I'm confused about how that relates to the equation I listed...
     
  8. Sep 26, 2014 #7
    ok...so picking up from there I did:
    [itex] \int\frac{v'}{v+1} = \int 0 [/itex]
    [itex] ln(v+1) = C [/itex]
    [itex] v + 1 = e^c [/itex]
    [itex] v = e^{c} - 1 [/itex]
    [itex] x + y = e^{c} - 1 [/itex]
    Here I am stuck because the solution in the book is:
    [itex] y = 2e^x - x - 1 [/itex]
    Also the book notes that this problem can be "solved quite easily" using the integrating factor method, but I do not see it...
     
  9. Sep 26, 2014 #8

    pasmith

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    This should be [tex]\int \frac{v'}{v +1}\,dx = \int 1\,dx.[/tex]

    You appear to have omitted an initial condition (perhaps y(0) = 1) from your problem; otherwise '2' should instead be an arbitrary constant.
     
  10. Sep 26, 2014 #9

    ehild

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    What is v+1 divided by (v+1)???????
    ehild
     
  11. Sep 26, 2014 #10
    nvm
     
  12. Sep 26, 2014 #11
     
  13. Sep 26, 2014 #12

    Orodruin

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    You seem to have assumed that ##v'/(v+1) = 0##. This is not the case so I suggest you recheck that algebra.

    Another option is to find a particular solution (there is a fairly obvious one) and then add the solution to the homogeneous equation to that (which works since the ODE is linear).
     
  14. Sep 26, 2014 #13
    you're right, I do apologize for that....I'll make the edit now
     
  15. Sep 26, 2014 #14
    ok, I see the mistake now...thank you
     
  16. Sep 26, 2014 #15

    pasmith

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    You are integrating both sides with respect to [itex]x[/itex]. It is crucial to state what variable you are integrating with respect to, because shortly you will use substitution to change [itex]\int \frac{v'}{v +1}\,dx[/itex] into an integral with respect to [itex]v[/itex].
     
  17. Sep 26, 2014 #16
    thank you, so here is my second attempt:
    [itex] \int\frac{v'}{v+1} dx = \int dx [/itex]
    [itex] ln(v+1) = x + C [/itex]
    [itex] v+1 = e^x + C [/itex]
    [itex] x + y = e^x + C [/itex]
     
  18. Sep 26, 2014 #17

    ehild

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    the last line is wrong. It should be [itex] v+1 = e^{x + C} [/itex] instead.
     
  19. Sep 26, 2014 #18
    thank you....
     
  20. Sep 26, 2014 #19

    Ray Vickson

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    In your example, is it not the case that ##f(x) = -1## and ##g(x) = x##?
     
  21. Sep 26, 2014 #20

    ehild

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    And [itex] v+1 = e^{x + C} = Ae^x [/itex] . The constants C or A depend on the initial condition.

    ehild
     
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