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Second Order Linear D.E W/ Constant Coefficients and Zero RHS

  1. Nov 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve the following differential equation and compare computer solutions

    \begin{equation*}
    4y''+12y'+9=0
    \end{equation*}

    2. Relevant equations

    None

    3. The attempt at a solution

    First of all looking at this equation, even though it is in the section where we learned about D.Es with zero right hand side, I don't believe this is the case since the equation is equal to

    \begin{equation*}
    4y''+12y'=-9
    \end{equation*}

    Which is fine, I know how to solve D.E's using the annihilator method (to an extent). I'm sure I am missing something simply in my solution but I just can't get it.

    I start out by writing this out in the operator form

    \begin{equation*}
    4D(D+3)y=-9
    \end{equation*}

    Solving the homogenous equation with roots ##0## and ##-3## which gives me the solutions to the complementary equation

    \begin{equation*}
    y_g=c_1+c_2e^{-3x}
    \end{equation*}

    I then use the operator ##D## to annihilate the constant ##-9## on the RHS to help me solve for the particular solution

    \begin{equation*}
    4D^2(D+3)y=0
    \end{equation*}

    I already know the solution for the term ##(D+3)##, and can find the solution for the term ##d^2y## as follows

    \begin{equation*}
    D^2y=DDy=0 \\
    \end{equation*}

    integrate both sides to solve this D.E

    \begin{equation*}
    DDy=0 \rightarrow \int{\frac{d}{dx} \Big(\frac{dy}{dx}dx\Big)}=\int{0dx} \rightarrow \frac{dy}{dx}=c_3
    \end{equation*}

    Integrate everything again

    \begin{equation*}
    \int{ \frac{d}{dx}(y)dx}=\int{c_1 dx} \rightarrow y= c_3x+c_4
    \end{equation*}

    So from my understanding I take the solutions for the terms ##D^2## and ##D+3##, differentiate them and sub them back into the original D.E ##4y''+12y'+9=0##. I also know that the terms from ##D+3## are part of the general solution and should be annihilated so I don't need to keep them (I did just to make sure)

    \begin{equation*}
    y=c_1+c_2e^{-3x}+c_3+c_4x \\
    y' = -3c_2e^{-3x}+c_4\\
    y''=9c_2e^{-3x}
    \end{equation*}

    Sub into the D.E

    \begin{equation*}
    4(9c_2e^{-3x})+12(-3c_2e^{-3x}+c_4)+9=0\\
    36c_2e^{-ex}-36c_2e^{-3x}+12c_4+9=0\\
    c_4=-\frac{3}{4}
    \end{equation*}

    This is where I get confused, the general solution is the complementary solution plus the particular solution, however in the particular solution I have only been able to solve for ##c_4##, does this mean I can throw away ##c_1##,##c_2## and ##c_3## of the particular solution? This would mean my particular solution is a constant ##-3/4##. In this case my general solution would be

    \begin{equation*}
    y=y_c+y_p \\
    y=c_1+c_2e^{-3x}-\frac{3}{4}x
    \end{equation*}

    The issue is my textbook says the answer is

    \begin{equation*}
    y=(c_1+c_2x)e^{-3x/2}
    \end{equation*}

    I am unsure where I went wrong to get this answer. I suspect the text authors used a different technique than the annihilator method but I can't see how since ##g(x)## (the RHS) does not equal zero. Any help would be greatly appreciated. I am struggling with the D.E portion of my mathematical physics course because we only spent 4 classes on D.Es and I haven't worked with them before in any of my calc classes.

    Thanks again!

    Edit: Error in my second last equation ##y=c_1+c_2e^{-3x}-\frac{3}{4}##, I have edited it to be ##y=c_1+c_2e^{-3x}-\frac{3}{4}x##
     
    Last edited: Nov 27, 2015
  2. jcsd
  3. Nov 27, 2015 #2

    BvU

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  4. Nov 27, 2015 #3
    Thanks for the reply BvU,

    I am aware of both. From my understanding the characteristic equation is just another way of looking at the operator form of a question? For example the characteristic equation of the D.E ##4y''+12y'+9=0## is just ##4m^2+12m+9=0## and in operator form is ##(4D^2+12D+9)y=0##. From my understanding the operator form and characteristic equation give you the same information (eg. the roots of the characteristic equation would be the same as the roots for the operator form) just in a different way.

    Also I have seen the case of repeated roots (along with unique roots and imaginary roots) for example ##(D+3)^2## gives the solution ##y=(c_1+c_2x)e^-3x##. I will admit that I get confused when the repeated roots are not in the form ##(D-a)^2##, for example (like in the problem above ##D^2## or if we were to get if our operator form is in a form ##(D^2+D+a)^2## (repeated form of a degree 2 polynomial). Hence why I tried deriving it in my solution. Repeated roots are covered very briefly in my text and only cover forms of ##(D-a)^2## however some of my homework problems are not of this form.

    Also to confuse me even further a classmate suggested I use wolfram to check the solution of the D.E above and it gives the same answer as my answer ##y=c_1+c_2e^{-3x}-\frac{3}{4}##. It doesn't seem like my text and the wolfram answers are equivalent so I am pretty confused now.
     
  5. Nov 27, 2015 #4

    BvU

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    You are perfeecetly correct. Perhaps the textbook has a misprint and meant the equation $$
    \begin{equation*}
    4y''+12y'+9y=0
    \end{equation*}$$.

    Which is what I read (by mistake, having looked at the book answer first o:) ) and worked out

    Well done !
     
  6. Nov 27, 2015 #5
    You're absolutely right, it is a misprint. I just looked at the errata and the equation is ##4y''+12y'+9y=0##, all that headache for nothing haha. Thanks again, I appreciate it!
     
  7. Nov 27, 2015 #6

    SteamKing

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    In spite of what the title of this thread states, this is a second-order linear ODE with constant coefficients, 'cuz you have y".
     
  8. Nov 27, 2015 #7
    I would never do the annihilator method, unless instructed to.

    I would have solved this problem, using the method of undetermined coefficients. I would have even done variation of parameters before annihilator method.
     
  9. Nov 27, 2015 #8
    You're right SteamKing, that was my bad.

    MidgetDwarf, my prof told us that annihilator method was the one he preferred since for him it used less memorization than the method of undetermined coefficients. Just as an aside from my understanding the annihilator method is essentially the same as the method of undetermined coefficients, just a different way of getting to your expression with the undetermined coefficients. I haven't learned about variation of parameters yet, probably will soon. To each their own I guess.
     
  10. Nov 27, 2015 #9

    LCKurtz

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    Yes, you are correct. The annihilator method is the method of undetermined coefficients without the guesswork. Also, given a problem where you could use either variation of parameters or the annihilator method, the latter would almost always be preferred.
     
  11. Nov 27, 2015 #10

    Ray Vickson

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    Another way is to note that
    [tex] 12y' + 9 =12 \left(y' + \frac{9}{12} \right) = 12 \frac{d}{dx} \left(y + \frac{9}{12}x \right), [/tex]
    so writing ##y + (9/12)x = z## the DE becomes, ##z''+ 12 z' = 0##.
     
  12. Nov 27, 2015 #11

    However, there really is not any guesswork for typical problems in a book. Always go up a degree in your guess, and a solution can be found.

    The annilahor method, is usually longer to do manually, however, it is the same method, as you pointed out.
     
  13. Nov 29, 2015 #12

    epenguin

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    Your would have so little difficulty in e.g. factorising (4x2 + 12x + 9) that IMO this is just crying out to be dealt with by factorisation of the differential operator.
     
  14. Nov 30, 2015 #13

    Mark44

    Staff: Mentor

    Changed the title to reflect the actual order of this problem.
     
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