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Math Help - [SOLVED] Find my mistake...

  1. #1
    Member Altair's Avatar
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    [SOLVED] Find my mistake...

    I have moved to the non-homogeneous equations and was doing the question (2x+y-2)dx + (2y-x+1)dy = 0 and the answer given is,
    <br />
ln (X^2+Y^2) -tan^{-1}(\frac{Y}{X}) = c

    Now, After I got it in the form (after converting it into homogeneous form) \frac{-1 +2V}{-2(V^2+1)} = \frac{dX}{X}, I simplified it till ,
    <br />
ln(({\frac{X^2}{Y^2+X^2})^{1/2}})^2 - tan^{-1}\frac{Y}{X} = ln(X) + C

    Then,

    ln(X) - ln(X^2 + Y^2) - tan^{-1}\frac{Y}{X}  = C

    Where does my mistake lie?
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Altair View Post
    I have moved to the non-homogeneous equations and was doing the question (2x+y-2)dx + (2y-x+1)dy = 0 and the answer given is,
    <br />
ln (X^2+Y^2) -tan^{-1}(\frac{Y}{X}) = c

    Now, After I got it in the form (after converting it into homogeneous form) \frac{-1 +2V}{-2(V^2+1)} = \frac{dX}{X}, [This looks correct!] I simplified it till ,
    <br />
ln(({\frac{X^2}{Y^2+X^2})^{1/2}})^2 - tan^{-1}\frac{Y}{X} = ln(X) + C

    Then,

    ln(X) - ln(X^2 + Y^2) - tan^{-1}\frac{Y}{X}  = C

    Where does my mistake lie?
    \int\frac{-1 +2v}{-2(v^2+1)}dv = {\textstyle\frac12}\!\int\frac1{1+v^2}dv - \int\frac v{1+v^2}dv = {\textstyle\frac12}\tan^{-1}v - {\textstyle\frac12}\ln(1+v^2) (+const.)

    If v = y/x then this is equal to {\textstyle\frac12}\tan^{-1}{\textstyle\frac yx} - {\textstyle\frac12}\ln(x^2+y^2) + {\textstyle\frac12}\ln x^2 = {\textstyle\frac12}\tan^{-1}{\textstyle\frac yx} - {\textstyle\frac12}\ln(x^2+y^2) + \ln x. Then the ln(x) term will cancel with the ln(x) on the right-hand side.
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  3. #3
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    Hello, Altair!

    (2x+y-2)dx + (2y-x+1)dy \:= \:0
    and the answer given is: . \ln(X^2+Y^2) -\tan^{-1}\left(\frac{Y}{X}\right) \;= \;C

    I got it in the form: . \frac{-1 +2V}{-2(V^2+1)}d \;= \;\frac{dX}{X} . . . . Correct!

    I would separate it like this: . \frac{2V-1}{V^2+1}\,dV  \:=\:-2\,\frac{dX}{X}

    Then we have: . \left(\frac{2V}{V^2+1} - \frac{1}{V^2+1}\right)\,dV \:=\:-2\,\frac{dX}{X}

    Integrate: . \ln(V^2+1) - \tan^{-1}(V) \;=\;-2\ln(X)+C

    Back-substitute: . \ln\left(\frac{Y^2+X^2}{X^2}\right) - \tan^{-1}\left(\frac{Y}{X}\right) \;=\;-\ln(X^2) + C

    . . \ln(X^2+Y^2) \:\underbrace{-\: \ln(X^2)}\: - \tan^{-1}\left(\frac{Y}{X}\right) \;=\;\:\underbrace{-\:\ln(X^2)} \:+ C


    Therefore: . \ln(X^2+Y^2) - \tan^{-1}\left(\frac{Y}{X}\right) \:=\:C

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