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Math Help - Integrating Factor

  1. #1
    Member Altair's Avatar
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    Integrating Factor

    The question is :-

    y -xy' = x + yy'

    The question asks to find the integrating factor and to solve it further.

    As this is a homogeneous equation I tried the formula of \frac{1}{Mx + Ny} and both partial fraction formulas but the answer which is \frac{1}{x^2 + y^2} wouldn't come. I get \frac{1}{-x^2-y^2}.

    Secondly even by using the answer of I.F I get stuck at final integration. Please do it for me. I dunno that how does the terms tan^{-1} (\frac{x}{y}) and -ln\sqrt{x^2 + y^2} come ?
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  2. #2
    Senior Member Peritus's Avatar
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    first divide the ODE by x


    \begin{gathered}<br />
  \frac{y}<br />
{x} - y' = 1 + \frac{y}<br />
{x}y' \hfill \\<br />
   \hfill \\<br />
   \Leftrightarrow y' = \frac{{\frac{y}<br />
{x} - 1}}<br />
{{\frac{y}<br />
{x} + 1}} \hfill \\ <br />
\end{gathered}

    now let us use the following substitution:

    <br />
v = \frac{y}<br />
{x} \Rightarrow y' = v + xv'<br />

    after applying this substitution to our ODE we get:

    <br />
xv' =  - \frac{{1 + v^2 }}<br />
{{1 + v}}<br />

    this is a separable equation:

    <br />
\int {\frac{1}<br />
{x}dx}  =  - \int {\frac{{1 + v}}<br />
{{1 + v^2 }}} dv<br />

    <br /> <br />
\ln \left| x \right| + C =  - \arctan (v) - 0.5\ln \left( {1 + v^2 } \right)<br />

    you can now back substitute to obtain the solution in an implicit form.
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  3. #3
    Member Altair's Avatar
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    I want to do it by using the I.F.
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  4. #4
    Senior Member Peritus's Avatar
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    Quote Originally Posted by Altair View Post
    I want to do it by using the I.F.
    be my guest...
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