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Second Order Non Homogeneous DE

  1. Nov 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the solution to :y''+2y'+y=t

    2. Relevant equations

    Suppose y(t)=B1t2+B2t+B3

    And I believe, Y(t)=Yh+Yp. That is the solution is equal to the solution to the homogenous equation, plus the particular solution.

    3. The attempt at a solution
    First Solve the homogeneous equation:
    y(t)=B1t2+B2t+B3
    y'(t)=2B1t+B2
    y''(t)=2B1

    Substitute this back into the homogenous problem(y''+2y'+y=0) gives:

    2B1+2(2B1t+B2)+B1t2+B2t+B3=0

    Rearrange:

    B1(t2+4t+2) +B2(t+2)+B3=0

    Solution is t=-2 or B1,B2,B3=0

    Now I'm unsure of what to do?

    As for the particular solution say Yp:

    Yp=ct, where c is a constant so,
    Y'p=c
    Y''p=0

    Substitute this into :y''+2y'+y=t
    0+2c+ct=t

    c=?

    Thanks for any help.
     
  2. jcsd
  3. Nov 23, 2011 #2
    You should try using the annihilator method.

    I got Yp=At+B
    Yp'=A
    Yp"=0 (A and B are constants)

    plugging them back into the ODE :

    2[A] + [At+B] = t

    rearranging:

    (2A+B)[1] + A[t] = t

    so

    A=1

    2A+B=0


    getting A=1 and B= -2

    so then Yp= t-2

    add it to Yh and you should get Y(t)
     
  4. Nov 23, 2011 #3
    oh, and to find Yh just set the ODE = 0 and find the roots
     
  5. Nov 23, 2011 #4

    LCKurtz

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    That is not the correct thing to try for the homogeneous solution. Try [itex]y = e^{rt}[/itex].
     
  6. Nov 23, 2011 #5

    HallsofIvy

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    No, that is NOT how you solve for Yh.

    That's because [itex]B1x^2+ B2x+ B3[/itex] is NOT a solution to the homogeneous equation. Use [itex]y= e^rt[/itex], as LCKurtz suggested, to get the "characteristic equation" to solve for r.

    You cannot just let yp= ct. If the right hand side is a power of t, yp can be a polynomial nof degree at most that power- try yp= ct+ d.
     
  7. Nov 24, 2011 #6
    Actually the annihilator method is correct. In fact I checked with my professor on this. The annihilator method takes out the factor of "guessing" a solution for the particular solution.

    I even had my professor solve this and she got the same answer as I did, especially since we just went over this section in class.

    Annihilator method is actually quite simple for this problem.
     
  8. Nov 24, 2011 #7
    As for Yh, what I mean by set the ODE to 0 is:
    y''+2y'+y=0

    Let y''=r^2, y'=r, y=constant

    ending up with:
    r^2+2r+1=0

    Factoring, you should end up with
    (r+1)(r+1)=0

    where r=-1,-1

    Since this is a repeated root you should use:
    Yh= c1*e^(rt) + c2*t*e^(rt)

    Once you have Yh, then find Yp (particular solution) and add it to Yh for your general solution Y.

    Just remember, since you DON'T have initial conditions

    (for example: y'(0)=0, y(0)=1)

    Your general solution is going to have constants c1 and c2 since there are no initial conditions given
     
  9. Dec 11, 2011 #8
    Sorry for the late reply, yes I made a bad mistake with respect to the characteristic equation . I got it now though thanks!

    Edit: I misunderstood the "hint" on the assignment. Every second order DE of this form, will have that solution given a single root.
     
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