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Find the solution?

  1. Aug 17, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the solution of y"+2y'+3y=sin(t)+δ(t-3∏); y(0)=0, y'(0)=0.


    2. Relevant equations
    Here's the work:
    s^2*Y(s)-s*y(0)-y'(0)+2(s*Y(s)-y(0))+3Y(s)=1/(s^2+1)+e^(-3pi*s)
    s^2*Y(s)+2sY(s)+3Y(s)=1/(s^2+1)+e^(-3pi*s)
    Y(s)(s^2+2s+3)=1/(s^2+1)+e^(-3pi*s)



    3. The attempt at a solution
    The answer is y=(1/4)sin(t)-(1/4)cos(t)+(1/4)(e^(-t))(cos(sqrt(2)t))+(1/sqrt(2))(u3pi(t))(e^(-(t-3pi))(sin(sqrt(2))(t-3pi)
     
  2. jcsd
  3. Aug 17, 2013 #2

    LCKurtz

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    Why stop there? What do you get for ##Y(s)?## Then where are you having trouble finding the inverse?
     
  4. Aug 18, 2013 #3
    I know that you have to divide (s^2+3s+2) to the other side but I don't know what I'll get. Can you show me what I get for Y(s)?
     
  5. Aug 18, 2013 #4
    Divide with (s^2+3s+2) then you might have to do a partial fraction decomposition so you can find the inverse.
     
  6. Aug 18, 2013 #5
    But what's Y(s)?
     
  7. Aug 18, 2013 #6
    Divide both sides with (s^2+3s+2)

    Y(s)=1/(s^2+2s+3)(s^2+1)+e^(-3pi*s)/(s^2+2s+3)
     
  8. Aug 18, 2013 #7
    But what's the inverse laplace transform of 1/((s^2+1)(s^2+2s+3))?
     
  9. Aug 18, 2013 #8

    LCKurtz

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    Do you remember partial fractions from calculus? Figure out ##A,B,C,D## to write$$
    \frac 1 {(s^2+1)(s^2+2s+3)}=\frac {As+B}{s^2+1}+\frac {Cs+D}{s^2+2s+3}$$Then we can talk about the inverse.
     
  10. Aug 18, 2013 #9

    LCKurtz

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    Are you saying the algebra is too complicated for you?
     
  11. Aug 19, 2013 #10

    HallsofIvy

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    Personally, I have always disliked the "Laplace Transform Method" (which is what you are using although you don't say that). Problems like this can always be done more simply just using the "characterstic equation" method (the "advantage" for engineers is that with the Laplace Transform, you can just apply formulas without having to think.)

    Here, the corresponding homogeneous equation is y''+ 3y'+ 2y= 0 which has characteristic equation [itex]r^2+ 3r+ 2= (r+ 2)(r+ 1)= 0[/itex] and so has general solution [itex]y(x)= C_1e^{-t}+ C_2e^{-2t}[/itex].

    Now we need only find a single function that satisfies the entire equation to add to that general solution to the homogeneous equation. I would be inclined to use "undetermined coefficients" but that requires knowing the general form of the function and you might not be familiar with the derivatives of that delta function.

    So instead, I will use "variation of parameters". We seek a solution of the form [itex]y(t)= u(t)e^{-t}+ v(t)e^{-2t}[/itex]. There are, of course, many different choices for functions u(t) and v(t) that will work. Differentiating, [itex]y'(t)= u'(t)e^{-t}- u(t)e^{-t}+ v'(t)e^{-2t}- 2v(t)e^{-2t}[/itex]. We can narrow our search among those "many different choices" and simplify the equation by requiring that [itex]u'(t)e^{-t}+ v'(t)e^{-2t}= 0[/itex]. That leaves [itex]y'(t)= -u(t)e^{-t}- 2v(t)e^{-2t}[/itex].

    Differentiating again, [itex]y''(t)= -u'(t)e^{-t}+ u(t)e^{-t}- 2v'(t)e^{-2t}+ 4v(t)e^{-2t}[/itex]. And putting those into the equation,
    [tex]y''+ 3y'+ 2y= -u'(t)e^{-t}+ u(t)e^{-t}- 2v'(t)e^{-2t}+ 4v(t)e^{-2t}+ 3(u(t)e^{-t}- 2v(t)e^{-2t})+ 2(u(t)e^{-t}+ v(t)e^{-2t})[/tex]
    [tex]= (-u'(t)e{-t}- 2v'(t)e^{-2t})+ u(t)(e^{-t}- 3e^{-t}+ 2e^{-t})+ v(4e^{-2t}- 6e^{-2t}+2)[/tex]
    [tex]=-u'(t)e^{-t}- 2v'(t)e^{-2t}= sin(t)+ \delta(t- 3\pi)[/tex]
    An equation involving only the two first derivatives of u and v (there are no second derivatives because of our requirement that [itex]u'(t)e^{-t}+ v'(t)e^{-2t}= 0[/itex] and no u and v themselves because [itex]e^{-t}[/itex] and [itex]e^{-2t}[/itex] satisfy the homogeneous equation.)

    Together with the required [itex]u'(t)e^{-t}+ v'(t)e^{-2t}= 0[/itex], that gives two equations we can solve algebraically for u' and v':

    [itex]u'(t)e^{-t}+ v'(t)e^{-2t}= 0[/itex]
    [itex]-u'(t)e^{-t}- 2v'(t)e^{-2t}= sin(t)+ \delta(t- 3\pi)[/itex]

    An obvious first step is to add the two equations, eliminating u':
    [itex]-v'e^{-2t}= sin(t)+ \delta(t- 3\pi)[/itex] so that [itex]v'= -e^{2t}sin(t)- e^{2t}\delta(t- 3\pi)[/itex]

    "[/itex]-e^{2t}sin(t)[/itex] can be integrated "by parts" while the integral of [itex]-e^{2t}\delta(t- 3\pi)[/itex] is, of course, just [itex]-e^{6\pi}[/itex].

    To find u(t), multiply the first equation by two and then add:
    [itex]u'(t)e^{-t}= sin(t)+ \delta(t- 3\pi)[/itex] so that
    [itex]u'= e^t sin(t)+ e^t \delta(t- 3\pi)[/itex]
    and that can be integrated easily.
     
  12. Aug 19, 2013 #11
    LCKurtz, how would you solve for A, B, C, D? I know it's (As+B)(s^2+2s+3)+(Cs+D)(s^2+1)=1, then you get As^3+2As^2+3As+Bs^2+2Bs+3B+Cs^3+Cs+Ds^2+D=1. But what's after that?
     
  13. Aug 19, 2013 #12
    Never mind. I get it now.
     
  14. Aug 20, 2013 #13
  15. Aug 20, 2013 #14

    LCKurtz

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    @success: Before continuing, please show us what you got for your partial fraction decomposition.
     
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