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!

Initial value problem with y(0)=C

  1. Jun 30, 2012 #1
    Hey everyone, I'm a long-time visitor, it's my first time posting though.


    I have a hw problem that is causing me considerable consternation:

    (y^3)*(dy/dx)=(8y^4+14)*cos(x); y(0)=C

    Oh, and we're supposed to solve the initial-value problem, and then solve for the particular solution in the form (y^4)=...

    I first used separation of variables and then integrated to get:
    (y^4) = (1/4)*Ce^(32sin(x))-(7/4)

    Now that I've solved for the general initial-value equation, I don't know where to go from here. The only way I could think to solve for C is to plug in 0 for x, and C for y--giving you:

    (C^4)=(1/4)*C*(e^0)-(7/4)
    C=((1/4)-(7/4))^(1/3)

    As this is a non-real answer, I cannot figure out where I went wrong. Also, I was told that we would not need to use complex numbers for any of these hw problems.
     
  2. jcsd
  3. Jun 30, 2012 #2
    (y^3)*(dy/dx)=(8y^4+14)*cos(x)
    (y^3)/(8y^4+14)dy=cos(x)dx
    ∫y3/(8y4+14)dy=∫cos(x)dx

    through integrating (did it quickly so you might want to recheck)
    (1/32)ln(8y4+14)+c=sin(x)+c
    ln(8y4+14)=32sin(x)+c
    8y4+14=e32sin(x)+c
    solve from there

    substitute c for y as u did and o for x

    c4=(e32sin(0))(ec)/8 +14/8
    c4=ec/8+14/8
    from there i plugged it into wolframalpha and it provided me c=-1.15, 1.2, 12.0285
     
    Last edited: Jun 30, 2012
  4. Jun 30, 2012 #3

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Hello coolhand. Welcome to PF as a post-er !

    The C in the following solution for y4 the constant of integration. There's no reason to think that this is the same as the C given as the initial value.

    I would call it something else, like k, then using the initial condition, solve for k in terms of C.

    (y4) = (1/4)*k*e^(32sin(x))-(7/4)
     
  5. Jun 30, 2012 #4
    Probably could have been clearer in my description.

    AmritpalS, thanks for breaking it down for me though. My issue arose when I tried to plug in y(0)=C. I wound up creating a new constant (as C was part of the initial conditions) such that:

    ln(8y^4 +14)=32sin(x) +D

    then I solved for D, getting (8C^4 +14).

    Finally I put it all together and got y^4= [(8*(C^4)+14)*e^(32sin(x))]/8

    Thanks again for the help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Initial value problem with y(0)=C
  1. Initial value problem (Replies: 7)

  2. Initial-value problem (Replies: 2)

Loading...