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Differential equation help.

  1. Feb 7, 2009 #1
    x(dy/dx)=y+e^(8x/y)
    I tried dividing both sides by x
    and multiply each term with e^lnx
    I kinda solve the problem with x=
    but i need to have y=...
    please help
    and i didnt use to hint of the question of letting u=y/x
     
  2. jcsd
  3. Feb 7, 2009 #2

    rock.freak667

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    Why did you not use the hint to substitute u=y/x ? Use it and see if it makes things easier.
     
  4. Feb 7, 2009 #3
    I tired and got a wrong answer.. like I take the derivative of my answer and it wouldnt go back to the question.
     
  5. Feb 7, 2009 #4

    Hootenanny

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    Perhaps if you showed your working we could help you out.
     
  6. Feb 7, 2009 #5
    ok
    so I did this
    x(dy/dx)=y+e^(8x/y)
    dy/dx=y/x+(e^(8x/y))/x
    dy/dx-y/x=(e^(8x/y))/x sub u=y/x
    du/dx-u=(e^8u)/x
    multiply both side by e^lnu
    d/dx(u-e^lnu)=(e^8u)/x * e^lnu

    this is basically what i did. i got an answer late last night and i am in school right now...
    couldnt find my work but i got a messy answer.
     
  7. Feb 7, 2009 #6

    Hootenanny

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    Are you sure that

    [tex]\frac{dy}{dx}=\frac{du}{dx}[/tex]
     
  8. Feb 7, 2009 #7
    oh i typed it out wrong... sorry... i have x du/dx
    cause i used the subsitution dy/dx=xdu/dx..
    since u=y/x dy/du=x?
    I dont know if it is right... my prof showed us something weird and i am just trying to learn this with my book.
     
  9. Feb 7, 2009 #8
    If u = y/x, what is du/dx ? You can use the answer to find dy/dx in terms of du/dx, u and x.
     
  10. Feb 7, 2009 #9
    ylnx.
    but i dont know how to use it..
    my idea was expressing the question in terms of u and x with the subsitution.
     
  11. Feb 7, 2009 #10

    Hootenanny

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    That's not quite right:

    [tex] u = \frac{y}{x} \Rightarrow \frac{du}{dx} = \frac{d}{dx}\frac{y}{x}[/tex]

    Note that here y=y(x), that is y is a function of x. So you need to differentiate a product (or quotient) of two functions of x. Do you know how to evaluate the derivative of this form?

    HINT:
    Use the product/quotient rule and the chain rule
     
  12. Feb 7, 2009 #11
    I am confused now. If i do du/dx = xdy/dx-y/x^2
    I will get an equation with 3 different variables
    let me show you what i did.
    x(dy/dx)=y+e^(8x/y)
    dy/dx=y/x+(e^(8x/y))/x
    dy/dx-y/x=(e^(8x/y))/x sub u=y/x

    since dy/dx= (dy/du)(du/dx) dy/du=x and du/dx=(xdy/dx-y)/x^2
    multiply to replace dy/dx. i get (xdy/dx-y)/x-u=(e^8u)/x
     
  13. Feb 7, 2009 #12
    Note that since u = y/x, we have y = ux, so your expression becomes u' = (y' - u)/x, so we can now get y' as a function of u', u, and x.
     
  14. Feb 7, 2009 #13
    I got an answer but the want it interms of y= no way i can separate it..
    i got y=u^2+1/8e^8u
    put u=y/x
     
  15. Feb 7, 2009 #14
    I may be being dense here but as someone else mentioned can you use the quotient rule on that?

    [tex]f(x) = \frac{g(x)}{h(x)}[/tex]

    [tex]\frac{d}{dx}f(x) = f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}[/tex]

    And as said the chain rule, have you studied this yet?
     
  16. Feb 7, 2009 #15
    well i know the quotient rule for derivatives...
    i or chain rule for derivatives...
    other than that.. I dont know any other chain rule...
     
  17. Feb 7, 2009 #16
    Ok have you tried plugging it in as g(x) and h(x) what do you get? Can you then use the chain rule also?

    Sorry I'm not really paying attention to this thread, doing 8 things at once. Hope that helps...
     
  18. Feb 7, 2009 #17
    well i used the chain rule to find u' and i am stuck from there
    i am ask to solve for the dfferential equation. i try to integral both sides and got an answer.
    but the answer is not solve for y=
     
  19. Feb 8, 2009 #18

    Hootenanny

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    That is not correct. Perhaps if you posted your working we could point out where you have made an error.
     
  20. Feb 8, 2009 #19
    k I just found out i type my question wrong here originally. but this is what i did and got another answer late last night.
    u'=(y'-u)/x
    y'=u+e^8u
    u'=(e^8u)/x
    ln(x)=-1/8e^8u

    ln(ln(x))=-u
    y=-ln(ln(x))/x
    the 1/8 is gone which makes my answer weird....
    since you will get the same answer for different e^ whatever x.....
     
  21. Feb 8, 2009 #20

    Hootenanny

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    Can I just clear one thing up first:
    Did you mean e^{8y/x}?
     
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