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Homework Help: Solving non Homogeneous second order differential equation

  1. Nov 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the general solution to y'' + y = sec3(x)

    3. The attempt at a solution

    Well I can get the characteristic equation:

    r2 + 1 = 0
    r = +-i

    Then the homogeneous solution is yh = C1excos(x) + C2exsin(x)

    And I know y = yh + yp

    but how do I get yp? Ive never dealt with powers of cos, only polynomials and ex, so Im not sure how to construct the particular solution.

    Thanks for the help!!
  2. jcsd
  3. Nov 5, 2009 #2


    Staff: Mentor

    I tried a few things but didn't get anywhere. You might try yp = Atan x, or yp = Asec x tan x, and see if they get you anywhere. An identity that will probably be useful is sec2x = tan2x + 1.
  4. Nov 5, 2009 #3


    User Avatar
    Science Advisor

    First, yes, the characteristic equation is [itex]r^2+ 1= 0[/itex] which has roots i and -i. That does NOT mean [itex]y(x)= C_1e^x cos(x)+ C_2e^x sin(x)[/itex]. That would correspond to roots 1+ i and 1- i. The general solution to the associated homogeneous equation is [itex]y(x)= C_1 cos(x)+ C_2 sin(x)[/itex]

    Since "sec(x)" is NOT one of the functions that forms solutions for homogeneous linear equations with constant coefficients (polynomials, exponentials, sine and cosine, and combinations of those), there will be no simple way to use "undetermined coefficients".

    But you can use "variation of parameters". Look for a solution of the form y(x)= u(x)cos(x)+ v(x)sin(x) where u(x) and v(x) are unknown functions. Notice that we don't need any special knowledge here. Given any solution, y, we could create many functions "u" and "v" that would work. From y= ucos(x)+ vsin(x), y'= u' cos(x)- u sin(x)+ v' sin(x)+ v cos(x). Because there are, as I said, many different function u and v we can simplify our "search" by also requiring that u' cos(x)+ v' sin(x)= 0. With that condition, y'= -u sin(x)+ v cos(x) and then y"= -u' sin(x)- u cos(x)+ v' cos(x)- v sin(x).

    Putting those into y"+ y= sec3(x) we have (-u' sin(x)- u cos(x)+ v' cos(x)- v sin(x))+ (u cos(x)+ v sin(x))= -u' sin(x)+ v' cos(x)= sec3(x).

    That, -u' sin(x)+ v'cos(x)= sec3(x), and u' cos(x)+ v' sin(x)= 0 give two equations that can be treated as ordinary algebraic equations to be solved for u' and v'. Then integrate to find u and v.
  5. Nov 5, 2009 #4
    Yeah sorry I realize ex shouldn't be there.

    Your stuff was a little hard to follow so I researched the variation of parameters method a little and found basically this:


    since W() = 1 I get

    [tex]yp = -y1 \int sin(x)sec\stackrel{3}{}(x) dx + y2 \int sec\stackrel{2}{} dx[/tex]

    Where y1 = c1cos(x) and y2 = C2sin(x)

    And then after some math I get

    [tex]yp = \frac{C\stackrel{}{2}sin\stackrel{2}{}(x) - C\stackrel{}{1}}{cos(x)}[/tex] ?

    Does this look correct?
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