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Simple Differential equation with reduction of order

  1. Nov 6, 2012 #1
    use method of reduction of order to find second solution:

    t2y''-4ty+6y = 0 , y1(t)= t2


    Attempt:

    So I've done all the steps, up to the substitution, but I'm having problems with what appears to be a simple linear equation but I can't solve it: Any ways, with w = v' I arrive at:

    w't4 = 0

    Everything else canceled out. But trying to get an integrating factor with this would lead me to 0.
     
  2. jcsd
  3. Nov 6, 2012 #2

    LCKurtz

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    Actually, you are OK so far. Divide out the ##t^4## and you have ##w'=0## or ##v''=0##. Integrate it twice and remember when you are done that ##y=t^2v##. You will see what a second independent solution is.
     
    Last edited: Nov 6, 2012
  4. Nov 6, 2012 #3
    Your right. This working with zero in this scenario is so weird....

    I had another quick question with regards to undetermined coefficients:

    y''+2y' = 3 + 4sin(2t)

    So working out everything I get everything correct, but I'm missing a term in the particular solution. my solution is:

    Yp(t) = -1/2cos(2t) - 1/2sin(2t)

    I guess the issue is when I'm setting up my guess and setting the expressions equal. so my initial "guess" is:

    Yp(t) = A + B cos(2t) + C sin(2t)

    obviously after diffeentiating this twice the constant A term disappeared and as well there is no "y" in the differential equation. yet in the solution they got:

    3/2 -1/2cos(2t) - 1/2sin(2t) plus the complimentary equation. How do I get the 3/2?
     
  5. Nov 6, 2012 #4

    LCKurtz

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    You didn't show your ##y_c## complementary solution. But ##y=A## obviously is a solution of the homogeneous equation, so it can't work for ##y_p##. Try ##y_p=At## plus the sine an cosine terms.
     
  6. Nov 6, 2012 #5

    Sorry for the delayed response. had class. my complementary solution was:

    yc(t) = c1e0t + c2e-2t

    So where does the A term appear in the complemetary solution? is it because c1 is only the constant without e0t in it?
     
  7. Nov 6, 2012 #6

    LCKurtz

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    If you simplify ##c_1e^{0t}## what do you get?
     
  8. Nov 7, 2012 #7


    Yes. I get the constant alone. Is it because it's the constant or should I look at it as it's because they are the same degree polynomial?
     
  9. Nov 7, 2012 #8

    LCKurtz

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    Too many pronoun "its" in that question. I have no idea what you are asking me.
     
  10. Nov 7, 2012 #9

    Sorry. What I was asking is that since in the complementary solution I have y(t) = c1 + c2e-2 does the constant c1 match with the "A" that I have as my "guess" for the polynomial portion of the particular solution? and if so, then that's my I add a "t" to it? i.e: At + (the rest)?
     
  11. Nov 7, 2012 #10

    LCKurtz

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    Yes. The fact that a constant ##c_1## is a solution of the homogeneous equation means that a constant, no matter whether you call it ##A## or something else can't be a solution of the NH equation. So no point in trying ##y_p = A##.
     
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