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Exponential differential equation

  1. Feb 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A curve passes through the point (0,6) and has the property that the slope of the curve at every point P is twice the y-coordinate of P. What is the equation of the curve?



    2. Relevant equations
    Y=Yoe^kx is a solution of (dy/dx)=ky, where k is constant


    3. The attempt at a solution
    I integrated to get the equation for Y above but I am not given any equation to relate x and Y. My initial thought was to set dy/dx=2ky since the slope is always twice the y value. I need to find k but I can't do that without knowing a way to relate x and Y.
     
  2. jcsd
  3. Feb 24, 2009 #2

    Dick

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    Doesn't "slope is always twice the y value" mean dy/dx=2y? Compare that to your relevant equation to figure out what k is.
     
  4. Feb 24, 2009 #3
    I've tried that as well but I am still left with no way to relate Y to x.
     
  5. Feb 24, 2009 #4

    Dick

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    Why don't you show what you tried? You get Y0 by making sure (0,6) is on the curve.
     
  6. Feb 24, 2009 #5
    One thought I had was since I knew the slope was twice the y coordinate of point P, I said that (y-y1)/(x-x1)=2y, using 6 as y1 and 0 as x1. This gives me y=6/(1-2x). When I use this to find matching x and y, I plugged them into the Y=Yoe^kx equation and came out with a negative value for k. This can't be correct since the slope is positive. Right?
     
  7. Feb 24, 2009 #6

    Dick

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    The slope they are talking about is the slope of the tangent line. That's just y'. You are making this too hard. Just write down the solution of y'=2y. What is it? Then try to figure out Y0. Then you are done.
     
  8. Feb 24, 2009 #7

    Mark44

    Staff: Mentor

    Also, the work you did here can be used to give the equation of a straight line between (0, 6) and some point (x, y). It doesn't give you y as a function of x for the curve itself.
     
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