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Math Help - Missing steps in book

  1. #1
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    Missing steps in book

    Im trying to follow another example in this book, and im not sure how the following is done

    \frac{d^2x}{dr^2} + \frac{1}{r} * \frac{dx}{dr}- \frac{x}{r^2} = \frac{Q}{D}

    apprantley can be rearanged as follows:

    \frac{d}{dr} [ \frac{1}{r} * \frac{d}{dr} (xr) ] =  \frac{Q}{D}

    Thanks in advance

    Calypso
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by calypso View Post
    Im trying to follow another example in this book, and im not sure how the following is done

    \frac{d^2x}{dr^2} + \frac{1}{r} * \frac{dx}{dr}- \frac{x}{r^2} = \frac{Q}{D}

    apprantley can be rearanged as follows:

    \frac{d}{dr} [ \frac{1}{r} * \frac{d}{dr} (xr) ] = \frac{Q}{D}

    Thanks in advance

    Calypso
    When we have proofs or something that simplifies and we don't know how it's done, the best thing to do is work backwards. So let us do that.

    \frac{d}{dr} [ \frac{1}{r} * \frac{d}{dr} (xr) ] = \frac{Q}{D}

    \frac{d}{dr} [ \frac{1}{r} * (x + \frac{dx}{dr} r ) ] = \frac{Q}{D}

    \frac{d}{dr} [ \frac{1}{r}x + \frac{1}{r} \frac{dx}{dr} r ] = \frac{Q}{D}

    \frac{d}{dr} [ \frac{1}{r}x + \frac{dx}{dr} ] = \frac{Q}{D}

    \frac{d}{dr} \frac{1}{r}x + \frac{d}{dr} \frac{dx}{dr} = \frac{Q}{D}

    \frac{dx}{dr} \frac{1}{r} - \frac{1}{r^2}x + \frac{d}{dr} \frac{dx}{dr} = \frac{Q}{D}

    \frac{dx}{dr} \frac{1}{r} - \frac{1}{r^2}x + \frac{d^2x}{d^2r} = \frac{Q}{D}
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  3. #3
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    Quote Originally Posted by calypso View Post
    Im trying to follow another example in this book, and im not sure how the following is done

    \frac{d^2x}{dr^2} + \frac{1}{r} * \frac{dx}{dr}- \frac{x}{r^2} = \frac{Q}{D}

    apprantley can be rearanged as follows:

    \frac{d}{dr} [ \frac{1}{r} * \frac{d}{dr} (xr) ] = \frac{Q}{D}

    Thanks in advance

    Calypso
    Note the last two terms on the LHS

    \frac{1}{r} \frac{dx}{dr}- \frac{x}{r^2} = \frac{d}{dr} \left( \frac{x}{r} \right)

    so all three terms become

    \frac{d^2x}{dr^2} + \frac{d}{dr} \left( \frac{x}{r} \right) = \frac{d}{dr} \left(\frac{dx}{dr} + \frac{x}{r} \right) = \frac{d}{dr} \left(\frac{r\frac{dx}{dr} + x}{r} \right)

    which leads to your expression.
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  4. #4
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    Thanks to both of you for your replies
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