Solve 1st order linear ODE

  1. sharks

    sharks 837
    Gold Member

    The problem statement, all variables and given/known data
    dy/dx + y/x = e^(x^2)
    Express y in terms of x and arbitrary constant.

    The attempt at a solution
    It is in the standard 1st order linear ODE form.

    P(x) = 1/x
    Q(x) = e^(x^2)

    u(x) = x (after calculation)

    So, d(uy)/dx = uQ

    d(uy)/dx = x.e^(x^2)
    I have to integrate both sides w.r.t.x

    Finding the R.H.S is problematic though. As it seems infinite, from what i've understood from my calculations.

    Integral of x.e^(x^2) (done by partial integration)
    Let U = x, so dU/dx = 1
    Let dV = e^(x^2), so V = [e^(x^2)]/2x (is this correct?)

    xy = x.[e^(x^2)]/2x - integral of [e^(x^2)]/2x.(1).dx

    Then i have to integrate the R.H.S again and again and again, as i can't get rid of e^(x^2) with a multiple of x always in the denominator.

    Any advice?
     
  2. jcsd
  3. [tex]y'+a(x)y = b(x)[/tex]
    [tex]y=e^{-\int a(x)dx} \int e^{\int a(x)dx}b(x)dx[/tex]
     
  4. sharks

    sharks 837
    Gold Member

    Hi Quinzio

    I know about that formula. I used it in the first place, and the spot where i'm stuck is due to b(x) in your equation being e^(x^2). The power of e contains x which is giving me quite some trouble with this problem.
     
    Last edited: Nov 30, 2011
  5. Ok, but you got an x which is its derivative (of [itex]x^2[/itex]).
    It goes:
    [tex]a(x)=1/x[/tex]
    [tex]\int a(x)dx=ln|x|[/tex]
    [tex]e^{\int a(x)dx}=|x|[/tex]
    [tex]\int e^{\int a(x)dx}b(x)dx=|x|e^{x^2}=\frac{1}{2}e^{x^2}*sgn(x)[/tex]
     
  6. sharks

    sharks 837
    Gold Member

    I did some research and it appears that the integral of e^(x^2) is infinity and cannot be expressed in normal terms.

    However, it is curious that my calculus notes give this as the answer to this problem:

    y= [e^(x^2)]/2x + A/x
     
  7. Your notes show the right answer.
    I just wrote a part of it, what I thought was the harder.
    The answer in your notes is for x>=0 only, anyway.
     
  8. sharks

    sharks 837
    Gold Member

    This is the last line where i've been able to reach:

    xy = x.[e^(x^2)]/2x - integral of [e^(x^2)]/2x.(1).dx

    I'm stuck at the R.H.S. where i need to integrate [e^(x^2)]/2x

    I realize that i know nothing about the sign function. If that is a requirement to solve this problem, then i won't be able to do this.

    Is there any easy guide that you could possibly recommend to learn about the sign function?
     
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