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Math Help - Undertermined Coeffecients.

  1. #1
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    Undertermined Coeffecients.

    I can get this about 3/4ths of the way done, but i'm missing something and i don't know where it comes from.

    Solve the initial value problem
    y'' +y=cos(x) y(0)=1 y'(0)=-1

    My method was to first find the roots of the equation r^2 +1. That gives +-i
    Yp= c1cos(x) + c2sin(x)
    And the characteristic equation i believe would be (i may be confusing characteristic and particular, but either way, that doesn't impact my problem, i don't believe)
    Yc=Acosx + Bsinx

    then
    Y'c= -Asinx +Bcosx (just so i don't skip a step)
    Y''c= -Acosx -Bcosx

    If I plug this back into the originial equation of y''+y=cosx, all the terms on the left side cancel each other out, and i get cosx=0. I'm not sure what to do with that. A and B are both 0?

    Anyway, I then solve for the initial conditions, and I get c1=1, and c2=-1.

    Final solution cosx-sinx. But my answer tells me there is a +1/2xsinx that i am missing. Where does that come from? I tried another problem and had the ssame issue, i got the first two terms right, but then there was a -1/5 sinx. I'm not seeing how these are gotten.
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  2. #2
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    Re: Undertermined Coeffecients.

    Quote Originally Posted by isuckatcalc View Post
    I can get this about 3/4ths of the way done, but i'm missing something and i don't know where it comes from.

    Solve the initial value problem
    y'' +y=cos(x) y(0)=1 y'(0)=-1

    My method was to first find the roots of the equation r^2 +1. That gives +-i
    Yp= c1cos(x) + c2sin(x)
    And the characteristic equation i believe would be (i may be confusing characteristic and particular, but either way, that doesn't impact my problem, i don't believe)
    Yc=Acosx + Bsinx

    then
    Y'c= -Asinx +Bcosx (just so i don't skip a step)
    Y''c= -Acosx -Bcosx

    If I plug this back into the originial equation of y''+y=cosx, all the terms on the left side cancel each other out, and i get cosx=0. I'm not sure what to do with that. A and B are both 0?

    Anyway, I then solve for the initial conditions, and I get c1=1, and c2=-1.

    Final solution cosx-sinx. But my answer tells me there is a +1/2xsinx that i am missing. Where does that come from? I tried another problem and had the ssame issue, i got the first two terms right, but then there was a -1/5 sinx. I'm not seeing how these are gotten.
    The homogenous solution is, as you found, Yh= c1cos(x) + c2sin(x).

    So a 'particular solution' of the form Yp=Acosx + Bsinx you are attempting to find obviously will not work.

    I suggest you now consult your class notes or textbook for what 'particular solution' to construct when this happens .... What do they say?

    The final solution will then be y = Yh + Yp, and you then use the given boundary conditions to solve for the arbitrary constants c1 and c2. You would probably benefit from reviewing examples from your class notes and textbook. You might benefit from reading this: http://www.mathhelpforum.com/math-he...ns-177497.html
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