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It looks simple, but I can’t do it

  1. Dec 2, 2007 #1
    Can anyone help me how to make the following equation each variable x and y together with dx and dy respectively?

    [tex]y'=\frac {(4x^2 + y^2)} {xy}[/tex]

    Thanks in advance
     
  2. jcsd
  3. Dec 2, 2007 #2
    Try letting [tex]z=y/x[/tex] and creating a differential equation for [tex]z[/tex] using the one for [tex]y[/tex].
     
  4. Dec 3, 2007 #3

    HallsofIvy

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    y'= dy/dx so this is
    [tex]\frac{dy}{dx}= \frac{4x^2+ y^2}{xy}[/tex]
    [tex]xydy= (4x^2+ y^2)dx[/tex]
    [tex]-(4x^2+y^2)dx+ xydy= 0[/tex]
    Is that what you meant?

    You could also let v= y/x and make the substitution y= xv. In that case, dy= xdv+ vdx and the equation becomes
    [tex](-4x^2+ x^2v^2)dx+ x(vx)(xdv+ vdx)= 0[/itex]
    [tex]x^2(v^2- 4)dx+ x^2vdv+ x^2v^2dx= 0[/itex]
    [tex]x^2(2v^2-4)dx+ x^2vdv= 0[/itex]
    As long as x is not 0 we can divide by [itex]x^2[/itex]
    [tex](2v^2-4)dx+ vdv= 0[/tex]
    [tex]dx+ \frac{v}{2v^2- 4}dv= 0[/tex]
    Is that what you mean?
     
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