1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: IVP problem

  1. Apr 14, 2007 #1


    User Avatar

    1. The problem statement, all variables and given/known data
    Please take a look at my work and help me figure out where I went wrong. Thanks!
    Use separation of variables to solve the IVP
    [tex]\[ty'=\sqrt{1-y^2},\quad y(1)=1,\quad t>0 [/tex]

    2. Relevant equations

    3. The attempt at a solution
    Use separation of variables to solve the IVP
    [tex] \[ty'=\sqrt{1-y^2},\quad y(1)=1,\quad t>0 [/tex]

    [tex]dy/dt = \sqrt{1-y^2}/t [/tex]

    [tex]1/\sqrt{1-y^2} dy = 1/t dt[/tex]

    [tex] \int 1/\sqrt{1-y^2} dy = \int 1/t dt[/tex]

    [tex] arcsin(y) = ln (abs (t)) + C [/tex]

    [tex] C = 1 [/tex]

    therefore the formula should be

    [tex]y = sin(ln(abs(t))+1) [/tex]

    Last edited: Apr 14, 2007
  2. jcsd
  3. Apr 15, 2007 #2

    Gib Z

    User Avatar
    Homework Helper

    I don't understand what you think is the problem.
  4. Apr 15, 2007 #3
    I don't see a problem except that your value of C is incorrect, because with that choice y(1) is not equal to 1.
  5. Apr 15, 2007 #4


    User Avatar
    Science Advisor

    arcsin(1) is [itex]\pi/2[/itex], not 1.

    I will also point out that this is of the form [itex]y'= \sqrt{1- y^2}/t[/itex] and the function is NOT Lipshchitz in y on any interval containing y= 1. Therefore, the solution is not unique. y= 1 for all t is another obvious solution and, in fact, there are an infinite number of solutions.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook