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Math Help - [SOLVED] Non-homogenous linear ode particular solution help.

  1. #1
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    [SOLVED] Non-homogenous linear ode particular solution help.

    Hey all.

    I am having a lot of trouble finding particular solutions.

    I have three problems:

    A) y'' + y' = 6t^2

    B) y'' - 4y = sin(2t)

    C) y'' + 4y' + 5y = 5t + e^-t

    I am trying to find the particular solutions using the method of undetermined coefficients.

    For A), I found the general solution to be
    Ce^-t + D

    and the particular solution to be

    (2t^4)/(2+t)

    For B), I found the general solution to be
    C + De^(4*t)

    and the particular solution to be

    (-1/8)sin(2t)

    For C), I found the general solution to be
    Ce^(-2*t+i*t) + De^(-2*t-i*t) (everything in the parenthesis after e is what e is raised to.

    and the particular solution to be

    t + (1/2)e^(-t)

    None of these check out. I am flustered and frustrated with this and don't know what else to do. PLEASE HELP!!!

    Thank you
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  2. #2
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    First, you need to solve their homogeneous equivalent so you can come up with the complement function.

    For a, m^2+m=0 is the homogeneous equivalent.

    m(m+1)=0

    y_{c}=C_{1}+C_{2}e^{-t}

    Now what annihilates 6t^{2} and that is D^{3}

    m^{3}*m*(m+1)

    y_{p}=tC_{3}+t^{2}C_{4}+t^{3}C_{5}

    y'_{p}=C_{3}+2tC_{4}+3t^{2}C_{5}

    y''_{p}=2C_{4}+6tC_{5}

    2C_{4}+6tC_{5}+C_{3}+2tC_{4}+3t^{2}C_{5}=6t^{2}

    Now combing and replace Cs with A,Bs.... and solve for the coefficients.

    2B+6tC+A+2tB+3t^{2}C=6t^{2}

    A+2B+t(2B+6C)+3t^{2}C

    \left[ \begin{array}{c c c c} <br />
1 & 2 & 0 & 0 \\<br />
0 & 2 & 6 & 0 \\<br />
0 & 0 & 3 & 6 \end{array} \right]

    Reduced row echelon of our augmented matrix will solve for A B and C
    \left[ \begin{array}{c c c c} <br />
1 & 0 & 0 & 12 \\<br />
0 & 1 & 0 & -6 \\<br />
0 & 0 & 1 & 2 \end{array} \right]

    A=C_{3}=12;<br />
B=C_{4}=-6;<br />
C=C_{5}=2

    Answer is: y=C_{1}+C_{2}e^{-t}+12t+-6t^{2}+2t^{3}
    Last edited by dwsmith; April 7th 2010 at 08:11 PM. Reason: Mistake, wasnt thinking
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    First, you need to solve their homogeneous equivalent so you can come up with the complement function.

    For a, m^2+m=0 is the homogeneous equivalent.

    m(m+1)=0

    y_{c}=C_{1}+C_{2}e^{t}
    Why is it y_{c}=C_{1}+C_{2}e^{t} the general solution to the homogenous, and not
    y_{c}=C_{1}+C_{2}e^{-t}?

    I thought you solved for m (0 and -1 in this case), got two distinct answers a and b and you could plug into
    y_{c}=C_{1}e^{at}+C_{2}e^{bt}



    Have I missed something, or was it a typo?

    Also, if not too much trouble, could you elaborate on how you found <br />
y_{p}=tC_{3}+t^{2}C_{4}+t^{3}C_{5}<br />
would annihilate the 6t^2? I sat and tried a bunch of different things before I gave up and thought it to be hopeless. What is your strategy/technique to attack finding these particular solutions?
    Last edited by kpizle; April 7th 2010 at 07:23 PM. Reason: add more
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  4. #4
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    You are correct it is supposed to be -1.
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  5. #5
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    Quote Originally Posted by dwsmith View Post
    You are correct it is supposed to be -1.
    Radical...okay, so I was on the right track (I realize I butchered the general solution to the homogenous in C pretty badly. Another go with the CORRECT formula will probably yield better results).

    Thank you very much.

    I don't know if you saw the edit to my reply in time to see my last minute question about strategy/technique, so I shall ask again:

    What is your strategy in finding these particular solutions? I tried a bunch of stuff that kept NOT working that I thought I was doing it completely wrong or had the wrong idea or something.

    Thanks again for your reply. It was very helpful!
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  6. #6
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    I edited my original solution because I wasn't paying attention. I didn't finish it but all you need to do is group the like coefficients and factor them out.
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  7. #7
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    D^{3} annihilates square values. What we do to one side we need to do to the other. Since I am using m as my factors, we need to multiple the left which was m(m+1) by m^3.

    Now we have 3 repeated real roots 0. The other 2 are solutions belong to yc.
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  8. #8
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    I finished the solution for you.
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  9. #9
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    Quote Originally Posted by dwsmith View Post
    D^{3} annihilates square values. What we do to one side we need to do to the other. Since I am using m as my factors, we need to multiple the left which was m(m+1) by m^3.

    Now we have 3 repeated real roots 0. The other 2 are solutions belong to yc.
    Ok. I was able to get that the particular solution for a is

    12t - 6t^2 + 2t^3 which checks out. Thank you kindly.

    But I'm having trouble seeing how you went from D^3
    m^3*m*(m+1) to At + Bt^2 + Ct^3

    Well...I hadn't seen that you had finished it for me until I had posted this. Heh. Thank you.

    I don't see what the ms did to get to the generic particular equation...
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  10. #10
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    For B, I have taken steps back.

    General to the homogenous:

    r^2 - 4r = 0
    r=0, 4

    so... Yc = C_1 + C_2e^{4t}

    I took
    Yp = Asin(2t)
    Y'p = 2Acos(2t)
    Y''p = -4Asin(2t)

    Substitution:

    ( -4Asin(2t)) - 4 ( Asin(2t)) = -8Asin(2t)

    then, I did (more substitution)

    -8Asin(2t) = sin(2t)

    and found A = -1/8

    so, Yp = -(1/8)sin(2t)

    But when I try to plug the general solution

    y = C_1 + C_2e^{4t} - (sin(2t)/8)

    It doesn't work. Where did I go wrong?
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  11. #11
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    Found the error. Wrong general solution...needed

    r^2 - 4 = 0

    instead of

    r^2 - 4r = 0

    Yp was correct.

    Correct general equation

    y = C_1e^{2t} + C_2e^{-2t} - (sin(2t)/8)

    and everything works.
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  12. #12
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    Yup B is correct.
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