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1st order diff. eq.

  1. May 29, 2009 #1
    Okay I'm sorry. I'm sure this is very easy but I just can't figure it out. The exam is on Monday. If any of you remember me you will probably be shaking you heads in disgust. Again, I'm sorry, but please please I need help.

    [tex]y' + y/(x+1) = x^2[/tex]

    My first instinct with 1st order DEs is to seperate the variables.

    But I can't.

    Help.

    Edit: that's "y/(x+1)" not "(y/x)+1". My LaTeX is rusty
     
  2. jcsd
  3. May 29, 2009 #2

    Cyosis

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    Sorry for asking a question? Anyway what do you know about integrating factors? Multiply the differential equation by u(x) first then try to write the LHS as d(uy)/dx.
     
    Last edited: May 29, 2009
  4. May 29, 2009 #3
    I've heard of them but I can't get my head round them. I assume the u(x) is (x+1)^(-1).

    So the integrating factor is e^ln(x+1)=x+1

    Multiplying both sides gives me [tex](x+1)y' + y = (x+1)x^2[/tex]

    And that's where I get stuck.

    Is there a reasonably short guide to this anywhere?
     
  5. May 29, 2009 #4

    Cyosis

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    You are pretty close. I am sure you can find a guide somewhere, but I have not much to do currently so I'll try to help you out.

    Lets multiply the differential equation by u(x) this yields [itex]u(x)y'(x)+\frac{y(x)u(x)}{x+1}=u(x)x^2[/itex]. We want to write the left hand side as [itex]d(uy)/dx[/itex].

    [tex]
    \frac{duy}{dx}=u \frac{dy}{dx}+y\frac{du}{dx}
    [/tex]

    This must be equal to the LHS so we get.

    [tex]
    u \frac{dy}{dx}+y\frac{du}{dx}=u\frac{dy}{dx}+\frac{yu}{x+1}
    [/tex]

    For this to be true [tex]\frac{du}{dx}=\frac{u}{x+1} \Rightarrow u(x)=c(x+1)[/tex].

    We can now write the differential equation as:

    [tex]
    \frac{duy}{dx}=x^2 u
    [/tex]

    You know u can you solve for y now?
     
    Last edited: May 29, 2009
  6. May 29, 2009 #5

    CompuChip

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    While Cyosis is editing, you could try that last part?
     
  7. May 29, 2009 #6
    With you so far.

    And now you've lost me.

    *thinks hard*

    Okay given that expression for [tex]\frac{duy}{dx}[/tex] I can substitute in the u(x), split the variables and integrate but I can't see how you arrived there in the first place.

    In an ideal world I'd probably like to also know exactly what [tex]duy/dx[/tex] actually means, but just learning how to do these should be enough for now.
     
  8. May 29, 2009 #7
    Ohh... the penny has dropped.

    Thankyou so much.

    Edit: and I think I figured out what duy/dx means too!
     
    Last edited: May 29, 2009
  9. May 29, 2009 #8

    Cyosis

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    Hehe you're welcome.
     
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