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Math Help - [SOLVED] Question 3

  1. #1
    Member Altair's Avatar
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    [SOLVED] Question 3

    The answer for <br />
x sin (\frac {y}{x})y' = xcos(\frac {y}{x}) + ysin(\frac {y}{x})<br />
is given as,


    <br />
xcos(\frac {y}{x}) = c.<br />

    Where as my answer after simplifying the following step,

    <br />
\frac {dv}{cot(v)} = \frac {dx}{x}<br />

    was,

    <br />
-cos (\frac {y}{x}) = xc<br />

    I have tried my best to get the required form but I couldn't. Where does the problem lie?
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  2. #2
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    Quote Originally Posted by Altair View Post
    The answer for <br />
x sin (\frac {y}{x})y' = xcos(\frac {y}{x}) + ysin(\frac {y}{x})<br />
is given as,


    <br />
xcos(\frac {y}{x}) = c.<br />

    Where as my answer after simplifying the following step,

    <br />
\frac {dv}{cot(v)} = \frac {dx}{x}<br />

    was,

    <br />
-cos (\frac {y}{x}) = xc<br />

    I have tried my best to get the required form but I couldn't. Where does the problem lie?
    \frac {dv}{cot(v)} = \frac {dx}{x} \Rightarrow \tan v \, dv = \frac{1}{x} \, dx \Rightarrow \int \tan v \, dv = \int \frac{1}{x} \, dx

    \Rightarrow -\ln |\cos v| = \ln |x| + A \Rightarrow \ln |1/\cos v| = \ln |x| + A \Rightarrow \frac{1}{\cos v} = Bx where B = e^A

    \Rightarrow \frac{1}{B} = x \cos v \Rightarrow C = x \cos \left( \frac{y}{x} \right), where C = 1/B.
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  3. #3
    Flow Master
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    Quote Originally Posted by Altair View Post
    The answer for <br />
x sin (\frac {y}{x})y' = xcos(\frac {y}{x}) + ysin(\frac {y}{x})<br />
is given as,


    <br />
xcos(\frac {y}{x}) = c.<br />

    Where as my answer after simplifying the following step,

    <br />
\frac {dv}{cot(v)} = \frac {dx}{x}<br />

    was,

    <br />
-cos (\frac {y}{x}) = xc<br />

    I have tried my best to get the required form but I couldn't. Where does the problem lie?
    The mistake you probably made was in not realising that - \ln a = \ln a^{-1} = \ln \frac{1}{a} ......
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  4. #4
    Member Altair's Avatar
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    Exactly.
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