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Math Help - solving ODEs

  1. #1
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    solving ODEs

    I've just started this unit at university.

    A question was to solve this ODE:

    y' = y

    Why is it that the answer is not y= y^2/2 + c,
    but instead y= ce^x

    also should it actually be written as y(x)=ce^x, or does that not matter?

    Is there an algebraic approach to do this question? I put it in wolfram, and the final two steps were:
    y(x) = e^(x+c)
    Simplify
    => y(x) = ce^x

    I don't quite understand how that simplifies.

    Any help is appreciated, thanks!
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  2. #2
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    Re: solving ODEs

    y'=y
    You must integrate with respect to the same variable on both sides of the equation.
    One integral of y' is = Integral y'(x)*dx = y(x) +c
    One integral of y with respect to y is = integral y*dy = y/2 +c
    But it is of no use here, because you should integrate relatively to x , that is : Integral y*dx
    Do not confuse with Integral y*dy
    Since you don't know yet y(x), you cannot compute Integral y*dx. So, you need another method. The other method leads to y= c*exp(x)
    Last edited by JJacquelin; August 7th 2013 at 03:32 AM.
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  3. #3
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    Re: solving ODEs

    Another example :
    y(x) = x
    Integral y*dy = y/2 +c = (x^4)/2 +c
    Integral y*dx = Integral xdx = (x^3)/3 +c
    Of couse, both are different.
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  4. #4
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    Re: solving ODEs

    Quote Originally Posted by JJacquelin View Post
    So, you need another method. The other method leads to y= c*exp(x)
    I am still somewhat confused.
    Also, are we supposed to know this 'other method' (i'm doing a second year university math subject, we use a book called Advanced engineering mathematics),
    or are we supposed to be able to recognise the solution
    for y'=y is y= c*exp(c)
    by looking at it?
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  5. #5
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    Re: solving ODEs

    dy/dx = y' = y is a so called "separable" ODE. This is one of the simplest kind of ODE, as explained in all books.
    Ordinary Differential Equation -- from Wolfram MathWorld (See eq. 17)
    dy = y*dx
    dy/y = dx
    ln(y) = x +C
    y = exp(x+C)
    y = exp(C)*exp(x)
    exp(C) = c
    y = c*exp(x)
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  6. #6
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    Re: solving ODEs

    excellent, thank you so much!

    I don't know why the textbook simply didn't show us that.
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  7. #7
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    Re: solving ODEs

    I have never seen a Differential Equations textbook that didn't show that! What textbook are you using?
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  8. #8
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    Re: solving ODEs

    Advanced Engineering Mathematics
    by Erwin Kreyszig, 10th ed.
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