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Homework Help: Need help proving solution of a differential equation

  1. Apr 2, 2010 #1
    I need help proving the general soultion to this equation, dy/dx=(y-(y^2))/x, is x/(x+C)

    where C cannot equal -x. When I separate the variables and integrate I get

    ln|y|-ln|1-y|=ln|x|+C, and I cannot make this look like the general solution. I'm not sure if

    I did the integration wrong but I think it is right. Im not too sure what to do after I

    integrate. Any help would be greatly appreciated. Thanks
     
  2. jcsd
  3. Apr 2, 2010 #2
    Your integration is correct. Try combining the ln y functions and then raising both sides to the e power. From there, it should just be a simplification problem to solve.
     
  4. Apr 2, 2010 #3
    Once you simplify, you may not see it right away I noticed when doing the simplification myself.

    You should end up with y=-xc/(1-xc) but if we multiply the top and the bottom by (-1/c)/(-1/c), we would get x/[(-1/c)+x]. -1/c=c2 or C depending on how you want to notate it. Thus, yielding x/(C+x)
     
  5. Apr 4, 2010 #4
    Im not sure how you ended up with y=-xc/(1-xc) when I combine the ln functions and raise to the e power I get y/(1-y)=x+C then you can change that into (1-y)/y=1/(x+C), then (1/y)-1=(1/x+C) im not sure if that is the way you did it so can you please elaborate more on how you got y=-xc/(1-xc). Sorry if I am being difficult i usually dont have this much trouble with these kind of problems. thanks
     
    Last edited: Apr 4, 2010
  6. Apr 4, 2010 #5
    First of all, elnx + c= Cx not x+c.

    That means you will have y/(1-y)=Cx

    Next, you need to multiple both sides by (1-y); thus, obtaining y=Cx(1-y)=Cx-ycx.

    Now just add yCx to both sides. y+yCx=Cx.

    Then factor: y(1+Cx)=Cx.

    y=Cx/(1+Cx)

    Multiple by (1/c)/(1/c) which is equal to 1.

    y=x/((1/c)+x) and 1/c is just another constant.

    y=x/(C2+x)
     
  7. Apr 4, 2010 #6
    A HA!!!!!! That makes perfect sense now you have to raise the entire right side of the equation to the e not just ln|x| to the e power plus c to the e power, because you raised ln|y/(1-y)| to the e power, wow, thanks you definitely saved me from a lot of frustration
     
    Last edited: Apr 4, 2010
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