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Thread: [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 $\displaystyle (2x+y-2)dx + (2y-x+1)dy = 0$ and the answer given is,
    $\displaystyle
    ln (X^2+Y^2) -tan^{-1}(\frac{Y}{X}) = c$

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

    Then,

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

    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 $\displaystyle (2x+y-2)dx + (2y-x+1)dy = 0$ and the answer given is,
    $\displaystyle
    ln (X^2+Y^2) -tan^{-1}(\frac{Y}{X}) = c$

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

    Then,

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

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

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

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    Hello, Altair!

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

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

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

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

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

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

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


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

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