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I First Order Differential Equation

  1. Jan 2, 2016 #1
    Basically, I am confused by one question in a practice paper in which the equation is given as follows:

    dy/dx = e^-2y

    and I know the general solution is equal to : y = -0.5e^-2y + C

    which would make sense if it was direct integration however it seems to me it is in fact separable integration as there is a lone function of y. Below is a picture of my attempt at a solution through separable integration and would appreciate if anyone has any input? Am I wrong in thinking it's seperable?
     

    Attached Files:

  2. jcsd
  3. Jan 2, 2016 #2

    Mark44

    Staff: Mentor

    That's not the solution, as it does not give y as a function of x.
    The image you posted is terrible! It goes from barely legible to completely unreadable because of stuff you have crossed out. There are several errors in the lines at the end.

    Please include your work inline, not as an image.
     
  4. Jan 2, 2016 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I hope you don't 'know' that because it is not true!
    You are not "wrong in thinking it's separable" but you are "separating" it wrong.
    dy/dx= e^(-2y) can be written in "differential" form as e^(2y)dy= dx and then integrating to get e^(2y)/2= x+ C which could then be solved for y as a logarithm of x.
    But you seem to be "separating" it as dy= e^(-2y)dx. You can't integrate that because on the right you have a function of y that you cannot integrate with respect to x.
     
  5. Jan 2, 2016 #4

    Krylov

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    Education Advisor

    Please stop polluting the technical math subforums with these kinds of very badly asked homework questions. I encounter them too often, they are not good for the quality of this particular subforum and, in fact, not good for the quality of PF as a whole.
     
  6. Jan 2, 2016 #5

    Mark44

    Staff: Mentor

    @santeria13, please repost your question in the Homework section (under Calculus & Beyond). Note that you must use the homework template, and show your work inline in the post, not as an attachment.

    Thread closed.
     
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