Exponential differential equation

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Homework Statement


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?



Homework Equations


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


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.
 

Answers and Replies

  • #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.
 
  • #3
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I've tried that as well but I am still left with no way to relate Y to x.
 
  • #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.
 
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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?
 
  • #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.
 
  • #7
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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?

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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