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Equation of a curve using a given point and slope

  1. Sep 16, 2011 #1
    Find a curve whose slope at any point (x,y) is equal to 5y, and which passes through the point (1,-2)

    All previous examples which i have seen of this type of question give a much more complex slope and always in terms of x.
    This is causing me a bit of confusion with my answer. Should I work though it by integrating the slope and putting this equal to x instead of y i.e

    x=5/2y^2 + C

    then plug in my given x and y to obtain C=-9
    giving a final equation of

    x= 5/2y^2 -9

    or should I work through using a general solution using arbitrary constant?

    All help will be very much appreciated
  2. jcsd
  3. Sep 16, 2011 #2


    Staff: Mentor

    This problem implicitly asks you to solve the initial condition differential equation, dy/dx = 5y, y(1) = -2.

    Your solution doesn't work. For your equation, dy/dx = 1/(5y).
  4. Sep 16, 2011 #3
    From the initail conditions given I then re-arrange to give dy/y = 5dx?

    and integrate this to obtain lny = 5x +c
  5. Sep 16, 2011 #4


    User Avatar
    Science Advisor

    dy/dx= 5y so dy/y= 5dx has nothing to do with the initial condition, that y(1)= -2.

    Be careful! Integrating dy/y does NOT give ln y. It gives ln|y|. Often, we are working with positive numbers, but the condition that y(1)= -2 makes the difference important! Now, what value of c gives y satisfying y(1)= -2?
  6. Sep 17, 2011 #5
    Working through i get


    so A=-2e^-5

    so y=-2e^-5e^5x ??
  7. Sep 17, 2011 #6


    Staff: Mentor

    Sure, but you can check this yourself. Once you have found the solution, most of the hard work is done, and it's a simple matter to verify that your solution works.
    Any solution you get must satisfy two conditions:
    1) y' = 5y
    2) y(1) = -2

    If both of these are true for your solution, then your function is a solution to the initial condition problem.

    Last, how you wrote your solution is not technically correct and it's more difficult to read than it needs to be.

    y = -2e^(-5) * e^(5x)

    or (using the X2 button, in the expanded menu after you click Go Advanced), you can write it like this:
    y = -2e-5e5x

    or you can learn a bit of LaTeX to make it look like this:
    [tex]y = -2e^{-5}e^{5x}[/tex]
  8. Sep 17, 2011 #7
    Thanks a lot mark44 and halls of ivy. Appreciate your help
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