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Differential Equation, Substitution/Homogeneous

  1. Jun 30, 2015 #1
    1. The problem statement, all variables and given/known data

    Solve the differential equation:

    question.png

    Here is the books answer to the problem:

    answer.png

    2. Relevant equations

    This is a substitution/homogeneous first order differential equation, which can be converted into a separable differential equation.

    3. The attempt at a solution

    work.jpg

    Where am I going wrong??? (ignore the paragraph breakers in my work, sorry)
     
  2. jcsd
  3. Jun 30, 2015 #2

    BiGyElLoWhAt

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    The expression you have 4 lines from the bottom ##\int\frac{-1}{5m}dV## is rather meaningless. Rather than making the m substitution, try using a trigonometric substitution.
     
  4. Jun 30, 2015 #3

    BiGyElLoWhAt

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    In general, in order for a substitution to be useful, you need to be able to replace all of the old variables with the new variable(s) to get an integral in terms of the new (including the differential)
     
  5. Jun 30, 2015 #4
    How would I go about the trigonometric substitution without the radical? Or can you do it without the radical?
     
  6. Jun 30, 2015 #5

    Mark44

    Staff: Mentor

    The integral in question here, ##\int \frac{2V dV}{1 - 5V^2}## can be done without resorting to a trig substition. As it sits, it's pretty close to ##\int \frac{du}{u}##.
     
  7. Jun 30, 2015 #6
    Yes, I just used m instead of u, as the derivative of the bottom cancels out all the V terms in the equation, and leaves me just with ms to integrate. I don't understand what I have done wrong.
     
  8. Jun 30, 2015 #7

    BiGyElLoWhAt

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    But you haven't gotten rid of all the v's... Ok, yea you did, you just didn't replace the dV with dm. Sorry about that. upon further inspection, you actually have the same answer. At least on the second to last line. Try raising ##e^{\text{Left Hand Side}}=e^{\text{Right Hand Side}}##

    It's a matter of simplification, and I actually had to go through and do it to make sure it was the same. Thanks @Mark44 for straightening me out o0)
     
  9. Jun 30, 2015 #8
    Sorry! I forgot to replace the dV, you are right! I tried raising it to the e, earlier in the day but I suppose I didn't simplify correctly... Thank you for all your help!
     
  10. Jun 30, 2015 #9
    Sorry to re-awaken this problem, but I have gone through the steps of simplifying it as follows...

    work.jpg

    I get this answer instead of the answer to the problem provided up top.
     
  11. Jun 30, 2015 #10

    BiGyElLoWhAt

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    No problem.
    ##e^{2*ln|x|}=e^{ln|x^2|}##
    You have a -1/5 exponent that didn't get handled properly.
     
  12. Jun 30, 2015 #11

    SammyS

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    Your initial result looked better. You did have some errors in the steps, but the result appears to agree with what the book has.
     
  13. Jun 30, 2015 #12

    BiGyElLoWhAt

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    Alternatively, between the 3rd and 4th lines, you put the x^5 in the denominator as if it were negative, but its positive. It should stay in the numerator. I see what you did with the 1/5 now.
     
  14. Jun 30, 2015 #13
    Yeah, I just figured that out and submitted my work. I was so intent on seeing it come out the way the book had it, I completely ignored the logarithmic rules. Thank you for your time on this problem!
     
  15. Jun 30, 2015 #14

    BiGyElLoWhAt

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    No problemo señor
     
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