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    Hi guys,

    Is anyone able to help me with this integration question from the Heinemann textbook?

    I've managed to get up to here so far:

    Attachment 513821513823

    But I don't know where to go now

    (The final answer is y^2 = 8x/(x+2))

    Thanks in advance
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    (Original post by jasminetwine)
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    Hi guys,

    Is anyone able to help me with this integration question from the Heinemann textbook?

    I've managed to get up to here so far:

    Attachment 513821513823

    But I don't know where to go now

    (The final answer is y^2 = 8x/(x+2))

    Thanks in advance
    I'm assuming you have \ln y = \frac{1}{2} \ln |x| - \frac{1}{2} \ln |x+2| + c = \frac{1}{2}(\ln |x| - \ln |x+2|) + c

    In which case, this gets you \displaystyle \ln y = \frac{1}{2}\ln \left|\frac{Ax}{x+2} \right|= \ln \sqrt{\frac{Ax}{x+2}} for some arbitrary constant A, using the power rule for logarithms and "cancelling" the logs gets you

    \displaystyle y = \sqrt{\frac{Ax}{x+2}} and can you then use the given information to find A?

    Alternatively, you could have done: \ln y = \frac{1}{2} \left|\frac{Ax}{x+2}right| \Rightarrow 2 \ln y = ln \frac{Ax}{x+2} \Rightarrow \ln y^2 = \ln \frac{Ax}{x+2} and using the power rule on 2\ln y which is more elegant in my opinion.
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    (Original post by Zacken)
    I'm assuming you have \ln y = \frac{1}{2} \ln |x| - \frac{1}{2} \ln |x+2| + c = \frac{1}{2}(\ln |x| - \ln |x+2|) + c

    In which case, this gets you \displaystyle \ln y = \frac{1}{2}\ln \left|\frac{Ax}{x+2} \right|= \ln \sqrt{\frac{Ax}{x+2}} for some arbitrary constant A, using the power rule for logarithms and "cancelling" the logs gets you

    \displaystyle y = \sqrt{\frac{Ax}{x+2}} and can you then use the given information to find A?

    Alternatively, you could have done: \ln y = \frac{1}{2} \left|\frac{Ax}{x+2}right| \Rightarrow 2 \ln y = ln \frac{Ax}{x+2} \Rightarrow \ln y^2 = \ln \frac{Ax}{x+2} and using the power rule on 2\ln y which is more elegant in my opinion.
    How did you get from...

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    To...

    Attachment 513845513847

    Thank you!
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    (Original post by jasminetwine)
    How did you get from...

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    To...

    Attachment 513845513847

    Thank you!
    you can write c as 0.5lnA without any loss of generality...
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    (Original post by jasminetwine)
    How did you get from...

    Name:  image.jpeg
Views: 46
Size:  10.6 KB

    To...

    Attachment 513845513847

    Thank you!
    It's an arbitrary constant, so I can do whatever I'd like with it - but if it makes you happy, we can do this:

    \ln y = \frac{1}{2} \ln x - \frac{1}{2} \ln |x+2| + \frac{1}{2} \ln e^{2c} since c = \ln e^c \Rightarrow c = \frac{1}{2} \ln e^{2c}

    So that I get 2\ln y = \ln \frac{e^{2c}x}{x+2} but then you can just say A=e^{2c} without needing to bother for the sake of simplicity.

    So we get \ln y^2 = \ln \frac{Ax}{x+2} once again, although you could have just as well left it in the form above if you wanted.
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    (Original post by jasminetwine)
    How did you get from...

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    To...

    Attachment 513845513847

    Thank you!
    Call c, 0.5 ln A and then take out the half as a common factor. Inside the bracket use the rules of logs (log a + log b = log ab and log a - log b = log a/b) to collect together the log A, log x and - log |x+2| into one log.


    The trick of changing c to log k is very common in doing the algebra in tidying up solutions of DEs
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    Alternatively, you could plug in your conditions for y, x right away from 2\ln y = \ln x - \ln |x+2| + 2c to find c - if that makes you more comfortable.
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    Thanks for your help your help guys!

    Here are my workings, if anybody is interested!

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    Thanks again!

    Have a lovely weekend everybody!
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    (Original post by jasminetwine)
    Thanks for your help your help guys!
    First class work.

    Which do you think is your favourite method?
 
 
 
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