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: Finding particular integrals

  1. May 5, 2016 #1
    1. The problem statement, all variables and given/known data
    (d^2y/dx^2) + (dy/dx) = cos x

    So you have the trial solution
    y= p*cos(x)+q*sin(x)
    (dy/dx) = -p*sin(x)+q*cos(x)

    The issue Im having is equating the coefficients when after I have subbed them into the initial equation:

    Equating coefficients of sin x gives -q-p=0
    Equating coefficients ofcos x gives -p+q=1
    My question is why is
    = q-p=0

    and same for equating for cos x?

  2. jcsd
  3. May 5, 2016 #2


    Staff: Mentor

    It's much simpler to write the 2nd and 3rd equations as
    y' = -p * sin(x) + q * cos(x)
    y'' = -p * cos(x) - q * sin(x)
    Because the equation ##(-p + q)\cos(x) + (-q - p)\sin(x) = \cos(x)## has to be identically true; i.e., true independent of the values of x.
    So for the equation to be true for all values of x, it must be the case that the coefficient of cos(x) on the left side has to be 1, and the coefficient of sin(x) on the left side has to be 0, since sin(x) doesn't appear on the right side.
    It's also true (and germane to my reasoning here) that the sin and cos functions are linearly independent. That is, neither one is a constant multiple of the other.
  4. May 5, 2016 #3

    Charles Link

    User Avatar
    Homework Helper
    Gold Member

    Don't forget the possibility of solutions to the homogeneous equation, where instead of ## \cos x ##, the right side of the differential equation is equal to zero. The complete solution consists of the particular solution, (which you found), plus any homogeneous solution.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted