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Differential Equations: Inhomogenous case

  1. Mar 16, 2009 #1
    This is a question aimed at the understanding of a step by step method for solving DEs.

    consider for instance;

    [tex]
    \begin{equation}
    \frac{d^{2} x}{d t^{2}} + 3 \frac{dt}{dt} - 4 = Ce^{itn}
    \end{equation}
    [/tex]

    Where i is the complex notation, C some constant, and n some argument of that angle.

    Now I fully understand how to solve the homogenious part of this equation (I think);

    [tex]
    \begin{equation}
    \frac{d^{2} x}{d t^{2}} + 3 \frac{dt}{dt} - 4 = 0
    \end{equation}
    [/tex]

    Let [tex] $ x(t) = e^{zt} $[/tex]
    [tex]\therefore[/tex]

    [tex] $ x'(t) = ze^{zt} $[/tex]
    [tex] $ x''(t) = z^{2} e^{zt} $[/tex]
    [tex]\therefore[/tex] We can write down;

    [tex] $ z^{2} + 3z - 4 = 0 $[/tex]

    [tex]\therefore[/tex]
    z = 1
    or z = -4
    [tex]\therefore[/tex]

    [tex]
    \begin{equation}
    x = Ae^{t} + Be^{-4t}
    \end{equation}
    [/tex]

    But then I don't understand either my Lecturer or my course textbook for solving the equation for the inhomogenious case. I understand the proof for the adding together of the Particular Solution and the Complimentary Function, However how do you find a Particular Solution?

    Do I do:
    [tex]
    \begin{equation}
    x = Ae^{t} + Be^{-4t} + e^{itn}
    \end{equation}
    [/tex]

    Or go back and write;

    [tex]
    \begin{equation}
    z^{2} e^{zt} + 3ze^{zt} - 4e^{zt} = Ce^{itn}
    \end{equation}
    [/tex]
    ...and then what would I do?


    So the question is, how do I. In a step by step method, solve for the inhomogenious case?

    Cheers,
    Haths
     
  2. jcsd
  3. Mar 16, 2009 #2

    Mark44

    Staff: Mentor

    You need to include a constant multiplier in your particular solution: xp = Deitn. Substitute xp in the left side of your differential equation to solve for D. Aside: is there a specific value given for C in your diff. eqn?

    The general solution x(t) will be the solution to the homogeneous equation + the particular solution.
     
  4. Mar 16, 2009 #3
    Sorry that doesn't help me at all.

    [tex]$ x_{p} = De^{itn}$[/tex]

    from what you've said just sounds like the;

    [tex]$ x_{p} = Ce^{itn}$[/tex]

    I already have. C is just some random constant, which is why I think what you've just said is a restatement of what I've identified already.

    [hr]

    However do you mean, rearrange the formula to;

    [tex]

    \begin{equation}
    z^{2} e^{zt} + 3ze^{zt} - 4e^{zt} + Ce^{itn} = 0
    \end{equation}

    [/tex]

    But again how does that help? I don't understand how to solve this.

    Cheers,
    Haths
     
  5. Mar 16, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Judging by your solution, it looks like the homogeneous equation you are actually solving is x''+3x'-4x=0. Not x''+3x'-4=0. If that's the correct equation then it's very easy to get an inhomogeneous solution to x''+3x'-4x=C*exp(int). Just substitute A*exp(int) for x and solve for A in terms of C. If it's what you actually wrote x''+3x'-4=C*exp(int) then it's a bit more complicated. The homogeneous part of that is x''+3x'=0 and the '4' needs to come from the particular solution.
     
  6. Mar 17, 2009 #5

    Mark44

    Staff: Mentor

    No you don't. The solution to the homogeneous equation that you have is x(t) = Aet + Be-4t. The particular solution I offered is different from your solution to the homogeneous equation.

    As Dick already pointed out, there is some confusion about exactly what your differential equation is, which you gave in the first post in this thread as
    [tex]\begin{equation}\frac{d^{2} x}{d t^{2}} + 3 \frac{dt}{dt} - 4 = Ce^{itn}\end{equation}[/tex]
    Your later work indicates that the equation is probably this:
    [tex]\begin{equation}\frac{d^{2} x}{d t^{2}} + 3 \frac{dx}{dt} - 4x = Ce^{itn}\end{equation}[/tex]
    It would be nice for us to know exactly what the problem you're working is.



     
  7. Mar 18, 2009 #6
    No no. I have no idea where you've thought the equation had a -4x term in it.

    It shouldn't matter what I write down as working, because I'm asking you what I should be doing at each step. If I've written down somthing wrong, just say; "No that's wrong, this is what you should do next."

    I was asking you guys if that's what I was meant to be doing. The underrulling line command didn't appear to work.

    Anyhow from a very old libary book it says I should let the R.H.S. of the equation be my dy/dx function, then do exactly the same thing, as for the homogenious case. To get that part of the equation.

    Is this true...?

    Haths
     
  8. Mar 18, 2009 #7

    Mark44

    Staff: Mentor

    From your work. You showed one of the terms as dt/dt, so it wasn't a stretch for us to infer that you might have omitted the x from your -4 term. You also showed what appears to be a characteristic equation of z^2 + 3z - 4 = 0. This equation comes from this homogeneous differential equation: y'' + 3y' - 4y = 0. The roots of this latter equation are y = -4 and y = 1. The solutions to the homogeneous differential equation are y = Ae^(-4t) and y = Be^t, which you show later in your work.
    It shouldn't matter but it does. Very often people post questions that are different from what they are given in their problems, make typos in what they post, and otherwise manage to stray from the exact formulations of their problems. From what I've seen, people who respond on this forum try to take into account what the poster means, and not just what he or she actually wrote. In some cases, a well-developed sense of clairvoyance is helpful.
    If the very first thing you write down is wrong (meaning the problem posted is different from the actual problem--and this happens fairly often), it's difficult or impossible for us to determine this.
    Your differential equation is apparently x'' + 3x' = 4 + Ce^(i*nt). Dick has suggested how you can approach this problem in his earlier response.
     
  9. Mar 18, 2009 #8

    lurflurf

    User Avatar
    Homework Helper

    so you have an equation of the form
    Ly=f
    in this case
    L=D^2+3D-4
    f=C*exp(i*t*n)
    we know how to solve problems of the form
    Ly=0
    and we know f is the solution to such a problem
    in this case L=D-i*n or if a real basis is prefered L=D^2+n^2
    thus we can reduce one problem into the other
    Ly=f
    Tf=0
    so
    TLy=0
    we solve this problem
    y=yp+yc
    where
    Lyp=0
    so the problem is complete when we solve
    Lyc=f
    this is simple algebra as we know the form of yc
    if we knew L and T had only 0 as a common solution we could solve
    Ty=0
    to find yc (really we are finding a space containing yc, that is the form of yc)
    but if T and L have nontrivial solutions in common the common solutions are replaced in yc by solutions of TL that are not solutions to Ly=0 or Ty=0
     
  10. Mar 18, 2009 #9

    Dick

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    Science Advisor
    Homework Helper

    Right. So solve the homogeneous equation x''+3x'=0 and apply the technique of 'variation of parameters' to find a particular solution. You can also do it by splitting the problem into finding a particular solution of x''+3x'=4 and x''+3x'=C*exp(i*n*t) and then adding them. Both are pretty easy.
     
    Last edited: Mar 18, 2009
  11. Mar 18, 2009 #10
    Fair point Mark, you can assume it's a dx on the differential, but that's the only typo.

    I'm sorry, all of this is gobbledigook to me. It might be explaining it very well in your mind. But I don't know what "variation of parameters" is and a quick google didn't explain.

    Dick just says the same words as my textbook, but didn't explain what he means by it, hence why I said I don't know how to solve hence could someone show what is meant by those words.


    Haths
     
  12. Mar 18, 2009 #11

    lurflurf

    User Avatar
    Homework Helper

    The general case may be easier to understand given an example
    (D^2-3D+2)y=C*exp(a*x)
    (D-a)(D-1)(D-2)y
    assume a is not 1 or 2
    y=A*exp(x)+B*exp(2x)+E*exp(a*x)
    y=yp+yc
    where
    yp=A*exp(x)+B*exp(2x)+
    yc=E*exp(a*x)
    (D-1)(D-2)y=(D-1)(D-2)(yp+yc)=(D-1)(D-2)yc=E(a-1)(b-2)exp(a*x)
    so E=C/(a-1)/(b-2)
    now we have the cases where a=1 or 2
    these can be handled by solving TLy or by taking limits
     
  13. Mar 18, 2009 #12

    Dick

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    Science Advisor
    Homework Helper

    The problem is simple enough that you don't have to use variation of parameters. A time honored way of finding a particular solution is to simply guess a solution. You only need to find one. x''+3*x'=4 is pretty easy to guess. So is x''+3*x'=C*exp(int) if you guess that x(t) might be proportional to exp(int). Just try. You haven't done that for a while. And then solve the homogeneous ODE x''+3x=0.
     
  14. Mar 23, 2009 #13
    See this is why for the first time I've been disappointed with the help I've received here, because simply saying 'solve' isn't a help when I'm asking what do you mean by 'solve'. If you ask me to solve a quadratic equation for its roots I know what your on about. If you ask me to find the points of inflection on f(x) I know what your asking me to solve. If I didn't I'd ask you what do you mean by that, and that's what I'm asking. Now in the bold example above I know though my studies what you mean by that.

    Guessing is not a step by step method.

    What upsets me the most is that nobody has shown me what they meant by their explanation. lurflurf's posts helped and so did Dick's first one after I'd gone away and received some in person help, but they were cryptic posts.

    [hr]

    Is this what you were describing all along...?

    Say we already have the homogenious case function I will call this H(t). I want to find the solution x(t) given the equation;

    [tex]

    \begin{equation}
    \frac{d^{2} x}{d t^{2}} + 3 \frac{dx}{dt} - 4 = Ce^{itn}
    \end{equation}

    [/tex]

    I know that;

    [tex]

    \begin{equation}
    x(t) = H(t) + P(t)
    \end{equation}

    [/tex]

    Where P(t) is the particular solution. H(t) is;

    [tex]

    \begin{equation}
    H(t) = Ae^{t} + Be^{-4t}
    \end{equation}

    [/tex]

    To find the particular solution we go back to the differential equation and let x(t) equal the R.H.S. (Right hand side) of equation 1 (numbering them in the order they appear in this post). The we replace the dx/dt and [tex]$ d_{2}x / dt^{2}$[/tex] terms with the differentials x'(t) and x''(t). Hence;

    [tex]

    \begin{equation}
    x'(t) = inCe^{int}
    \end{equation}

    [/tex]

    [tex]

    \begin{equation}
    x''(t) = -n^{2}Ce^{int}
    \end{equation}
    [/tex]

    Subsituting in equation 1;

    [tex]

    \begin{equation}
    -n^{2}Ce^{int} + 3inCe^{int} - 4 = Ce^{itn}
    \end{equation}

    [/tex]

    Rearranging to isolate C;


    [tex]

    \begin{equation}
    -n^{2}Ce^{int} + 3inCe^{int} - Ce^{itn} = 4
    \end{equation}

    [/tex]


    [tex]

    \begin{equation}
    C(-n^{2}e^{int} + 3ine^{int} - e^{itn}) = 4
    \end{equation}

    [/tex]

    [tex]

    \begin{equation}
    C = \frac{4}{( (-n^{2} - 1 ) e^{int} + 3ine^{int} )}
    \end{equation}

    [/tex]

    Knowing that this is C and that it satisfies the DE we had above, P(t) is;

    [tex]

    \begin{equation}
    P(t) = \frac{4}{( (-n^{2} - 1 ) e^{int} + 3ine^{int} )}e^{itn}
    \end{equation}

    [/tex]

    Hence a solution to the DE is;

    [tex]

    \begin{equation}
    x(t) = Ae^{t} + Be^{-4t} + \frac{4}{( (-n^{2} - 1 ) e^{int} + 3ine^{int} )}e^{itn}
    \end{equation}

    [/tex]

    Haths
     
  15. Mar 23, 2009 #14

    Mark44

    Staff: Mentor

    Dick omitted a prime in the differential equation that you quoted in your last post. In the other two places in that same paragraph, the differential equation is correct.
    This should be :
    And then solve the homogeneous ODE x''+3x'=0.

    Your solutions to this homogeneous differential equation are incorrect.
    Neither function is a solution to x'' + 3x' = 0, nor is the sum of these functions a solution.

    Given that you are trying to solve a nonhomogeneous differential equation, it is not at all unreasonable, IMO, for us to advise you to solve the related homogeneous differential equation, because I'm reasonably sure that you were exposed to this type of differential equation before learning about solving nonhomogeneous differential equations.

    Surely you understand that being asked to solve a differential equation means to find a function or a collection of functions that satisfy the differential equation.

    When someone asks us how to work a particular type of problem, most of us giving help and advice have expectations of the abilities of the person asking, and those expectations are higher for differential equation-type problems than for something at a lower level, such as solving a quadratic equation. We tend to omit some details at this level that we would be more likely to provide for someone asking about solving a quadratic equation.

    To solve your nonhomogeneous DE, first find the solutions to the homogeneous version. Hint: characteristic equation. Second, find particular solutions to the nonhomogeneous DE. Reread Dick's suggestions a couple of posts back.
     
  16. Mar 23, 2009 #15
    Is that a yes or a no to my question in the last post?

    ___________________________________________________________

    Why do I want to solve;

    x''+3x=0

    ??? It's not the oridginal example, there is a -4 missing from it if it meant to be the oridginal Homogenious DE we are trying to solve, and I never even tried solving x''+3x=0 for the final solution. Hence my comment about his posts being cryptic as there is no explaination of where this is coming from or why it's even involved.

    ___________________________________________________________

    Yes, but the finding part is what I'm asking how to do. I thought I had made that clear with the my posts.

    Haths
     
  17. Mar 23, 2009 #16

    Mark44

    Staff: Mentor

    You don't. You want to solve x'' + 3x' = 0.


    Your original differential equation, with two small changes, is this:
    x'' + 3x' = 4 + Ceitn

    Notice the prime in 3x' and that the constant is now on the right side.

    Your homogeneous DE is x'' + 3x' = 0. Part of your general solution to the equation above is the solution to the homogeneous problem.

    First, find the solution to the equation x'' + 3x' = 0. The characteristic equation for this DE will be helpful. This is the solution to the homogeneous DE. To check, make sure that the 2nd derivative + 3 times the first derivative adds to 0.

    Second, find solutions to x'' + 3x' = 4 and to x'' + 3x' = Ceitn, following the suggestions Dick gave a few posts back. It's probably easier to do these separately. These solutions combined will be your particular solution to the nonhomogeneous DE. To check, make sure that the 2nd derivative + 3 times the first derivative adds to 4 + Ceitn.

    Third, combine the solution to the homogeneous equation with the solution to the nonhomogeneous equation. This will be your general solution.



     
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