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Variation of Parameters Differential Eq.

  1. Apr 9, 2006 #1
    Use the method of variation of parameters to find a particular solution of [​IMG][/PLAIN] [Broken]
    ok i know first i do [​IMG][/PLAIN] [Broken] then r = +/- 1i then i make that into the equation [​IMG][/PLAIN] [Broken] so cosx is [​IMG][/PLAIN] [Broken] and sinx is [​IMG][/PLAIN] [Broken] so now the wronskian is w= [​IMG][/PLAIN] [Broken] which would be (cosx)(cosx) - (sinx)(-sinx). Now what do I do from here? I remeber something about integrals being used next, such as [​IMG][/PLAIN] [Broken] and a u2 integral
    Last edited by a moderator: Apr 22, 2017 at 9:28 AM
  2. jcsd
  3. Apr 10, 2006 #2


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    You would be better off to actually LEARN "variation of parameters rather than just trying to memorize formulas!

    Since sin(x) and cos(x) are solutions to the homogeneous equation, try a solution to the entire equation of the form y(x)= u(x) sin(x)+ v(x) cos(x). (There are, in fact, many possible such solutions: for example for any solution, y, choose u= 0 v= u/cos(x).) Differentiating y'= u'sin(x)+ u cos(x)+ v' cos(x)- u sin(x). Now, require that u' sin(x)+ v' cos(x)= 0. We can do this precisely because there are many possible solutions.

    Since u' sin(x)+ v' cos(x)= 0, y'= u cos(x)- v sin(x) and, differentiating again, y"= u' cos(x)- u sin(x)- v' sin(x)- v cos(x). Putting that back into the original equation, y"+ y= (u' cos(x)- u sin(x)- v' sin(x)- v cos(x))+ u sin(x)+ v cos(x)= u' cos(x)- v' sin(x)= csc2(y). We have, also, u' sin(x)+ v' cos(x)= 0. Solve those two equations, algebraically, for u' and v' and then integrate. Warning: typically, the integrals cannot be done in closed form- you may need to write the solution to the differential equation in terms of those integrals.
  4. Apr 10, 2006 #3
    Hmm that is the way our professor told us how to do it. Also another way someone had told me to do it was doing the wronskian twice.. there seems to be so many ways to do this type of problem
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