1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Initial-value problem

  1. Oct 23, 2006 #1
    i have no idea on how to start this problem.


    Find a solution of the initial-value problem

    dy/dx= -y^2 , y(0)=.0625


    so..

    (1/-y^2)dy = dx

    int(1/-y^2)dy=int(dx)

    1/y=x+c
    please if someone can help me start it or explain it to me
     
  2. jcsd
  3. Oct 23, 2006 #2

    radou

    User Avatar
    Homework Helper

    Note that you have [tex]-\frac{1}{y} = x + C[/tex]. Now all you have to do is write [tex]y = -\frac{1}{x+C}[/tex] and use the initial value.
     
    Last edited: Oct 23, 2006
  4. Oct 23, 2006 #3
    so am i solving for C?
     
  5. Oct 23, 2006 #4

    radou

    User Avatar
    Homework Helper

    Yes, you are.
     
  6. Oct 23, 2006 #5
    so
    C= (1/y)-x ??

    so C=(1/.0625) if y(0)=.0625 ????
    is this correct?
     
  7. Oct 23, 2006 #6

    radou

    User Avatar
    Homework Helper

    Yes, C = -16.
     
  8. Oct 23, 2006 #7
    negative???
     
  9. Oct 23, 2006 #8

    radou

    User Avatar
    Homework Helper

    Yes, since [tex]y(x) = -\frac{1}{x+C} \Rightarrow y(0) = -\frac{1}{C} = 0.0625[/tex]
     
  10. Oct 23, 2006 #9
    but the integral of -1/y^2 dy is 1/y
     
  11. Oct 23, 2006 #10

    radou

    User Avatar
    Homework Helper

    Yes, you're right. I apologize, that minus sign slipped through somehow. :smile: So, your answer is correct.
     
  12. Oct 23, 2006 #11
    ok but i put the answer c=16 and it was wrong. is there another step?
     
  13. Oct 23, 2006 #12
    ok i got it the answe is y=1/(x+16)

    thank you for your help
     
  14. Oct 23, 2006 #13

    radou

    User Avatar
    Homework Helper

    Yes, I just wanted to write that the answer is a function, not a constant.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Initial-value problem
  1. Initial value problem (Replies: 7)

  2. Initial-value problem (Replies: 2)

Loading...