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ODR: Does this one have a solution?

  1. Dec 16, 2005 #1
    Hi all,

    I'm given this one:

    [tex]
    y' = 10^{x+y}
    [/tex]

    Here's how I went:

    [tex]
    \frac{y'}{10^{y}} = 10^{x}
    [/tex]

    [tex]
    \int\left(\frac{1}{10}\right)^{y}\ dy = \int 10^{x}\ dx
    [/tex]

    [tex]
    \frac{\left(\frac{1}{10}\right)^{y}}{\log \frac{1}{10}}} = \frac{10^{x}}{\log 10} + C
    [/tex]

    [tex]
    \left(\frac{1}{10}\right)^{y} = -10^{x}
    [/tex]

    From which I conclude that it has no solution (on the right side there is negative number whereas on the left side there always will be a positive one).

    Is this right conclusion?

    Thank you.
     
    Last edited: Dec 16, 2005
  2. jcsd
  3. Dec 16, 2005 #2

    StatusX

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    What happened to your constant?
     
  4. Dec 16, 2005 #3
    Well, it just...disappeared [​IMG]
     
  5. Dec 16, 2005 #4

    Tide

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    You're confusing "assumptions" with "conclusions." It is a conclusion to say that there is no solution.

    If the constant of integration just disappeared then you cannot just make a general conclusion. Surely, the problem provided an intial condition or you can infer one. :)
     
  6. Dec 16, 2005 #5
    Of course, I mistakingly got rid of the constant :)
     
  7. Dec 16, 2005 #6

    mezarashi

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    Firstly as mentioned, you cannot just make constants disappear. Secondly, can you elaborate on how you got the negative infront of the 10^x.
     
  8. Dec 16, 2005 #7
    Sorry, there was a typo in the original post. Now I corrected it and hope it's clear.
     
  9. Dec 17, 2005 #8
    Welcoming the constant back, I finished this ODE. Anyway, I got something slightly different than Maples tells me.

    [tex]
    \left(\frac{1}{10}\right)^{y} = -10^{x} + C
    [/tex]

    [tex]
    y \log \frac{1}{10} = \log (C - 10^{x})
    [/tex]

    [tex]
    y = \frac{\log(C-10^{x})}{\log \frac{1}{10}} = -\frac{\log (C-10^{x})}{\log 10}
    [/tex]

    Which implies

    [tex]
    C > 0
    [/tex]

    and for particular C

    [tex]
    10^{x} < C \leftrightarrow x < \frac{\log C}{\log 10}
    [/tex]

    so

    [tex]
    y = -\frac{\log (C-10^{x})}{\log 10},\ x \in \left(-\infty, \frac{\log C}{\log 10}\right), C > 0
    [/tex]

    Proof shows that the equation

    [tex]
    y' = 10^{x+y}
    [/tex]

    so I think it's correct.

    Anyway, Maple tells me that it's

    [tex]
    y(x) = \frac{\log \left(\frac{1}{10^{x} + C \log 10}\right)}{\log 10}
    [/tex]

    but it seems not to satisfy the equation when I differentiated it.

    Where's the truth?

    Thank you.
     
    Last edited: Dec 18, 2005
  10. Dec 18, 2005 #9

    saltydog

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    I'm pretty sure the solution is:

    [tex]y(x)=\text{log}\left[\frac{1}{k-10^x}\right];\quad x<\text{log}(k)[/tex]

    Now . . . does this exists:

    [tex]\int_0^{\text{log(k)}} y(x)dx[/tex]

    ?
     
  11. Dec 18, 2005 #10
    You're right with solution (assuming it is decimal logarithm you use).

    Anyway, why is Maple giving wrong answer? I thought it will be good tool to check my own results but it doesn't seem so...
     
  12. Dec 18, 2005 #11

    saltydog

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    Twoflower, I dont' know about Maple. I use Mathematica and I believe software like it are execllent tools to check results. For example, I just finishing working on:

    [tex]\mathcal{L}^{-1}\left\{\frac{2s-s^2+1}{2(s^3+s-2)}\right\}[/tex]

    Well, that's a mess to do by hand (for me anyway). Have to break it up into several pieces like:

    [tex]\frac{2s}{s^2+s-2},\quad \frac{1}{s-1},\quad \frac{2-s}{2+s+s^2}[/tex]

    and

    [tex]\frac{s^2-1}{s^3+s-2}[/tex]

    Tough to keep track of everything but I can use Mathematica to invert each to verfy each step of my work. I've found this approach, using Mathematica to check steps, very useful as a learning tool. Try and learn to do so with Maple or whatever you use.:smile:
     
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