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Homework Help: Simple seperable Diff EQ w/ I.C., did i make an algebra error?

  1. Jan 15, 2006 #1
    I'm lost on this one, i checked my problem over a few times and i don't see any algebraic errors, did i break some rule? Thanks! :smile:

    Find the solution of the differential equation
    (ln(y))^4*{dy}/{dx} = x^4 y
    which satisfies the initial condition y(1)=e^2.
    y= ?

    Here is my work:
    http://img80.imageshack.us/img80/2868/lastscan9oj.jpg [Broken]
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jan 15, 2006 #2

    arildno

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    Dearly Missed

    Line 4 is bad notation; line 5 is utterly wrong.
    Learn the difference between + and * and when they appear.
     
  4. Jan 15, 2006 #3

    HallsofIvy

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    Again! ea+b is NOT ea+ eb!!!

    By the way, it is easier to find C by putting y= e2 and x= 1 in
    (1/5)(ln y)5= (1/5)x5+ C, before trying to solve for y.
     
    Last edited by a moderator: Jan 15, 2006
  5. Jan 15, 2006 #4
    Thanks for the tips guys, I thought I redid it right this time but no! :bugeye:
    Here is what I did, i also took Ivey's advice and solved for C, and yes it was easier! thanks!
    http://img75.imageshack.us/img75/674/lastscan4ux.jpg [Broken]
     
    Last edited by a moderator: May 2, 2017
  6. Jan 16, 2006 #5

    TD

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    Your general solution looks ok, so we have that:

    [tex]y^5 = e^{x^5 + 5c} \Leftrightarrow y = \left( {e^{x^5 + 5c} } \right)^{1/5} [/tex]

    Now this is where you determine your c, I don't really understand why you used an expression of a few steps before, c = 0 isn't right.

    [tex]y\left( 1 \right) = e^2 \Leftrightarrow e^2 = \left( {e^{1 + 5c} } \right)^{1/5} \Leftrightarrow 2 = \frac{{1 + 5c}}
    {5} \Leftrightarrow c = \frac{9}{5}[/tex]
     
  7. Jan 16, 2006 #6
    In your post #4, (page(13)

    3 rd line is not right.

    You have used [tex] ln (y^5) [/tex]

    instead of [tex] (ln y)^5 [/tex]

    [tex] (ln y)^5 = x^5 + 5C [/tex]

    [tex] (ln y) = (x^5 + 5C)^{1/5} [/tex]

    [tex] y = exp( x^5 + 5C)^{1/5}[/tex]
     
  8. Jan 16, 2006 #7
    simplifying e^(x^5)*e^(5C) may help you easier.
     
  9. Jan 18, 2006 #8
    Ahhh, thank you everyone! it finally worked out. Sorry I didn't post sooner, i've been busy busy! :)
     
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