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Homework Help: 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:


    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

    through integrating (did it quickly so you might want to recheck)
    solve from there

    substitute c for y as u did and o for x

    c4=(e32sin(0))(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


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    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
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