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Solving ODE y' = ye^(x+y)

  1. Nov 4, 2009 #1
    1. The problem statement, all variables and given/known data
    dy/dx=y*e^(x+y)
    solve for y

    3. The attempt at a solution
    move the y's over to the other side:
    dy/(y*e^y)=e^x.dx
    then you integrate and this is where i get screwed up.
    i use inverse product rule where u=y and v'=e^y, and i also tried u=e^y and v'=y. but this didnt get me anywhere
    so i did it again for the integral on the right hand side for each and it still didnt get me anywhere.
    HELP!
    also, i doubt if id be able to put it back into y= forms so yea could you help with that too? if not it doesnt really matter. primarily its just the integrating i want help with.
     
  2. jcsd
  3. Nov 4, 2009 #2

    Gib Z

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    Re: dy/dx=y*e^(x+y)

    It's not solvable in terms of elementary functions, but can be done with a special Function. You may have seen the the more common problem [tex]\int \frac{e^x}{x} dx[/tex]. It is defined to be equal to the Exponential Integral function. It's most common definition is [tex]Ei (x) = \int^x_{-\infty} \frac{e^t}{t} dt[/tex].

    Can you see how the integral you have is a certain case of that function?
     
  4. Nov 4, 2009 #3
    Re: dy/dx=y*e^(x+y)

    well yea its e^-t
    but im a dummy, i have to solve something this in an exam.
    i still dont know how to do it.

    whats with the Ei(x) part? and how does this give you the integral
     
  5. Nov 4, 2009 #4

    Gib Z

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    Re: dy/dx=y*e^(x+y)

    Basically the function you want to integrate, as well as the one I showed you with the e^x on the numerator, both can't actually be done by any normal methods! It has no simple answer, but since it comes up so often we just denote it with a shorter symbol because its useful! Think about it, using the Fundamental theorem of calculus, differentiate Ei (x), what do you get? =]

    So now, express the Integral you have in terms of the Ei function.
     
  6. Nov 5, 2009 #5
    Re: dy/dx=y*e^(x+y)

    blarg! i'm sorry, i still dont get it. i think i need a bigger hint as to how to relate it and then what to do from there. im sorry im so slow, i really am.
     
  7. Nov 5, 2009 #6

    Gib Z

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    Re: dy/dx=y*e^(x+y)

    Its ok, many people seem perturbed by the idea of just "naming" a solution. The short story is, the integral you have to work out Can't be done in any normal way. So if you got that one in a test, you should just leave it because its not doable. They will probably make sure in the actual exam they give you possible ones.
     
  8. Nov 5, 2009 #7
    Re: dy/dx=y*e^(x+y)

    oh wait, so if you differentiate Ei(x) then you get ex/x
    but what about when its e-x/x
    i'm not sure i understand how to relate it.

    but you'd just write that the integral of ex/x=Ei(x)??? or = that thing you wrote. and would you keep it with respect to t or x.
     
  9. Nov 5, 2009 #8
    Re: dy/dx=y*e^(x+y)

    and we will be getting that exact question in the test, i am 110% sure of this. so :S
     
  10. Nov 5, 2009 #9

    Gib Z

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    Re: dy/dx=y*e^(x+y)

    Well, the integral you have isn't perfectly Ei(x), but Ei(-x). Can you see why?
     
  11. Nov 5, 2009 #10
    Re: dy/dx=y*e^(x+y)

    yea i can see why.
    so would it be
    [tex]\int e^(-y) /y * dx [/tex]
    = Ei (-y) = [tex]\int [/tex] [tex]^{-y}_{-\infty}[/tex] et /t *dt]

    sorry, i dont really know how to use the latex stuff, i have never really used them before.
     
  12. Nov 5, 2009 #11

    Gib Z

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    Re: dy/dx=y*e^(x+y)

    Well not exactly =[ We're getting variables a bit mixed up here. Look at the definition of Ei i gave, x is a variable in the bounds, not the integrand.

    So [tex]
    Ei (-x) = \int^{-x}_{-\infty} \frac{e^t}{t} dt
    [/tex]

    To get that to look like yours (remember indefinite integrals vary by additive constants) let u=-t and simplify.
     
  13. Nov 5, 2009 #12
    Re: dy/dx=y*e^(x+y)

    huh? i didnt put it in the integrand. how can you simplfy that?
    and would it become e^-u /-u becaues its e^t/t or because i have a negative x its e^-t /t and therefore it becomes e^u / -u

    i'm so sorry, i apologise for my inept brain uut you are really helping me so thankyou. verily much
     
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