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Math Help - [SOLVED] Challenging differential equation

  1. #1
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    [SOLVED] Challenging differential equation

    I would really appreciate some assistance with the following problem. I barely know where to start!

    Let x+y=z

    Solve the equations a) and b):

    a) d/dx [ (1/3)x^3+x+2 ] = 0, where x is an element of the set C
    Solve for x.

    b) dy/du = ln(e), y(1)=1
    Define the function y.
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  2. #2
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    Quote Originally Posted by aujt74 View Post
    Let x+y=z

    a) d/dx [ (1/3)x^3+x+2 ] = 0, where x is an element of the set C
    Solve for x.


    Here's a kick start.
    \frac{d}{dx}\left[\frac{1}{3}x^3+x+2\right] = x^2+1

    Now solve

     x^2+1 = 0


    Quote Originally Posted by aujt74 View Post
    Let x+y=z

    b) dy/du = ln(e), y(1)=1
    Define the function y.
    also consider that \ln(e)=1
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  3. #3
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    Thanks for that kickstart!

     x^2+1 = 0
    so x= i

    \frac{dy}{du}=\ln(e) = 1
    so integrating wrt to u y = u ?

    That's right isn't it?
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  4. #4
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    Quote Originally Posted by aujt74 View Post
    \frac{dy}{du}=\ln(e) = 1
    so integrating wrt to u y = u ?

    That's right isn't it?
    Yes. In general, y = u + C, but you have y(1) = 1, so 1 = 1 + C and C = 0.
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  5. #5
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    Quote Originally Posted by aujt74 View Post

     x^2+1 = 0
    so x= i
    x = \pm i
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  6. #6
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    Thanks for all the help so far!

    Quote Originally Posted by aujt74 View Post
    Let x+y=z
    x = \pm i

    y = u

    so z = ?

    I don't understand how to evaluate z? I've not been given any values for u.
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  7. #7
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     <br />
z= \pm i +u<br />
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