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Homework Help: Differential equation (substitution)

  1. Apr 18, 2014 #1
    1. The problem statement, all variables and given/known data

    for this question , i 've got my positive 9 but i got -64 , can anyone tell me which part is wrong?

    Question : https://www.flickr.com/photos/123101...3/13907725466/ [Broken]
    Wroking : https://www.flickr.com/photos/123101...n/photostream/ [Broken]

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 18, 2014 #2

    Simon Bridge

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    "flickr: page no found"
    Note: The way to get the most out of this site is to write about how you are thinking about the problem - then we can focus on why you lost so many marks and fix that for the future.
  4. Apr 18, 2014 #3


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    And type the equations here instead of graphics on alternate sites.
  5. Apr 19, 2014 #4
    sorry, i posted at the wrong section, , but i hope you can look at it and help! here's the question and working attached..hope you can help!
    can i do it in this way? please refer to the red circled part

    if i cant do it in such way, can you please show me how do u get the ans please? thanks in advance!

    Attached Files:

    Last edited: Apr 19, 2014
  6. Apr 19, 2014 #5

    Simon Bridge

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    Problem statement reads:
    Using substitution ##y=Vx##, change the DE $$(x^3+xy^2)\frac{dy}{dx}=y^3+yx^2$$... to $$x\frac{dV}{dx}=\frac{-2V}{1+V^2}$$... and show that the solution is $$\ln\left|\frac{xy}{2}\right| = \frac{(x-2y)(x+2y)}{8x^2}$$ is ##x=2## when ##y=1##.

    Good grief!

    Well the first part comes from making the sub and then a bit of algebra.
    You could have cancelled the x^3 out much earlier than you did (like your 3rd line).
    But you made it - well done.

    For the next bit you basically started with separation of variables: $$\int \frac{dx}{x} =\int \frac{1+V^2}{-2V}dV$$... then you wrote:$$-2\int \frac{\frac{1}{2}dx}{\frac{1}{2}x} =\int \frac{1+V^2}{V}dV$$... which is the bit in red. It's a legal move.
    You seem concerned that $$\int \frac{\frac{1}{2}dx}{\frac{1}{2}x}=\ln\left|\frac{1}{2}x\right|+c$$ ##\qquad\qquad## ... instead of ##\ln|x|## but this is fine.

    Recall: ##\ln|\frac{1}{2}x|## differs from ##\ln|x|## by a constant that gets hidden in that "+c" on the end.
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