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Basic differential equations, homogeneous method

  1. Jul 7, 2008 #1
    xy'-y =sqrt((x^2)+y(^2))
    so by method of homogeneous equations I get arcsinh(y/x) = ln|x|+c but don't know what to do from there. the book suggests using sinhx = 1/2 (e^x - e^-x) but I don't know how to incorporate this. The answer should be y= (1/2) * (cx^2 - 1/c). How do you get this????

    This is my work so far:
    y'-(y/x) = sqrt (1+(y/x)^2)
    since y=ux...
    du/(sqrt(1+u^2) = dx/x
    taking the antiderivatives gives
    arcsinh(u) = ln|x|+c
    where
    arcsinh|y/x| = ln|x|+c
     
    Last edited: Jul 7, 2008
  2. jcsd
  3. Jul 7, 2008 #2

    rock.freak667

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    Homework Helper

    [itex]sinh^{-1}(\frac{y}{x}) = lnx + ln A[/itex] (I put c=lnA to make things simpler)
    [itex]\Rightarrow sinh^{-1}(\frac{y}{x}) = lnAx[/itex]


    [tex]\Rightarrow \frac{y}{x} = sinh(lnAx)[/tex]


    and they said that


    [tex]sinhX=\frac{e^X - e^{-X}}{2}[/tex]


    So [itex]sinh(lnAx)[/itex] should make the equation

    [tex]\frac{y}{x}=\frac{e^{lnAx}-e^{-lnAx)}}{2}[/tex]

    Do you see how to continue from here?
     
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