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Differential equation, confused on directions, F(x,y)?

  1. Jan 18, 2006 #1
    Hello everyone, I am confused on what they are wanting me to do here on this problem:
    has an implicit general solution of the form
    In fact, because the differential equation is separable, we can define the solution curve implicitly by a function in the form
    Find such a solution and then give the related functions requested.
    here is my work, i think they wanted me to do, they gave no intial condition so i solved for C:

    but it was incorrect :cry:
  2. jcsd
  3. Jan 18, 2006 #2
    Well what is supposed to be the correct answer? I can't see much wrong with what you've done.

    Edit: You seem to have replaced the y with x after the integration of the exponential.
    Last edited: Jan 18, 2006
  4. Jan 18, 2006 #3
    thats a good question, it doesn't tell me. it will tell me if i get it right, not if i get it wrong. Whats this implicit general solution stuff do i have to take partial derivatives or somthing?
  5. Jan 18, 2006 #4


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    Staff Emeritus
    Science Advisor

    WHY do you keep writing [itex]\frac{1}{e^{-2x}}[/itex]? Surely you know that [itex]\frac{1}{e^{-2x}}= e^{2x}[/itex]

    Your instructions were to write the answer as F(x,y)= K and then write it as F(x,y)= G(x,y)+ H(x,y)= K

    You already have 5e2x- arcsin(x/2)= K. Can't you just look at that and pick G and H?
  6. Jan 18, 2006 #5


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    Homework Helper

    Going from the second line to the third line your "y" became an "x". As far as what they want:

    They asked for [tex]F(x,y)=G(x)+H(y)=K[/tex]

    So "Find such a solution" is satisfied by [tex]F(x,y)=-\mbox{arcsin}\left( \frac{x}{2}\right) + 5e^{2y} = K[/tex]

    but the answer they want is "give the related functions requested"

    so try [tex]G(x)= -\mbox{arcsin}\left( \frac{x}{2}\right) \mbox{ and }H(y)= 5e^{2y} [/tex]
  7. Jan 18, 2006 #6
    Ahhh, thanks so much everyone it was correct. weee
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