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Find general equation of x′′(t)+5x′(t)+4x(t)=0

  1. Apr 11, 2016 #1
    • Thread moved from the technical forums, so no HH Template is shown.
    Suppose ##x_1(t)## and ##x_2(t)## are two linearly independent solutions of the equations:

    ##x'_1(t) = 3x_1(t) + 2x_2(t)## and ##x'_2(t) = x_1(t) + 2x_2(t)##

    where ##x'_1(t)\text{ and }x'_2(t)## denote the first derivative of functions ##x_1(t)## and ##x_2(t)##

    respectively with respect to ##t##.

    Find the general solution of ##x''(t) + 5x'(t) + 4x(t) = 0## in terms of ##x_1(t)## and ##x_2(t)##.
    ----------------------------------------------------------------------------------------------------------------------------------------

    The general solution of the equation

    ##x''(t) + 5x'(t) + 4x(t) = 0##................................(1)

    is

    ##x(t) = c _1 e^{-4t} + c _2e^{-t}##................................(2)

    Now if i want to express equation (2) in term of ##x_1(t)## and ##x_2(t)## , what exactly i have to do ?
     
    Last edited by a moderator: Apr 11, 2016
  2. jcsd
  3. Apr 11, 2016 #2

    LCKurtz

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    I'm guessing that that equation is supposed to be ##x''(t) \color{red}{\bf -} 5x'(t) + 4x(t) = 0##
    I would first solve the 2x2 matrix system you are given for ##x_1## and ##x_2##.
     
  4. Apr 11, 2016 #3
    The question is correct . X1 = -4 and x2 = -1 . And now what ?
     
  5. Apr 12, 2016 #4

    LCKurtz

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    ##x_1=-4## and ##x_2=-1~##?? Where did those come from? They don't solve either your 2x2 system or your second order system.

    In the last line of post #2 I suggested you solve your 2x2 matrix system for ##x_1## and ##x_2##. Have you done that yet? You should get two linearly independent vectors.

    After that maybe you can make some sense out of the problem.
     
  6. Apr 12, 2016 #5
    sorry my mistake .. $$x _1(t) = 3/2 $$ and $$x _2(t) = -1/2 $$ .
     
  7. Apr 12, 2016 #6

    LCKurtz

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    We are getting nowhere. You aren't showing any work and I don't know what those numbers are supposed to be. They certainly aren't solutions.
     
  8. Apr 12, 2016 #7
    by considering $$x''(t)+5x'(t)+4x(t)=0$$ equation as $$r^2+5r+4=0$$ i get $$r=-4,-1$$ i.e. $$x'(t)=-4,-1$$ which are the two roots of the said equation. from this solution i found the general equation of the said equation i.e. $$x(t) = c_1e^{-4t} + c_2e^{-t}$$ .

    And you are asking to solve $$\left(\matrix{x_1(t)\\x_2(t)}\right)' = \left(\matrix{3 & 2 \\ 1 & 2}\right) \left(\matrix{x_1(t)\\x_2(t)}\right)$$ this ???
     
  9. Apr 12, 2016 #8

    LCKurtz

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    The first line in your first post refers to two linearly independent solutions of that system. So I would expect you would want to find them unless you can find a way to do the problem without finding them explicitly. And when you do that I think you will see why I thought (and still do think) that second order DE has a misprint.
     
  10. Apr 12, 2016 #9

    Mark44

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    This makes no sense. ##x_1(t)## and ##x_2(t)## are functions of t, not constants.
    This makes no sense either
    Isn't this given information?

    I believe that what LCKurtz is suggesting is that you need to solve the system of differential equations as shown in post #1, by diagonalization. To do that, you need to find the eigenvalues and eigenvectors for your matrix, and use them to find a diagonal matrix D to uncouple the system of diff. equations.

    So far in your work I don't see any evidence that you are doing this.
     
  11. Apr 13, 2016 #10

    epenguin

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    LOL it's not transcendentally difficult after all, looks like the only way that might just throw some light on the question.

    You do you see why people are saying it could be a misprint? With a change of sign there is a fairly evident relation between the two parts of the question.

    If at the end it turns out your Prof says he made a mistake in transcription, or that there was a misprint in the book, it wouldn't be the first time that happened here.
     
  12. Apr 13, 2016 #11
    sorry guys . i was wrong about the question .. i found the question and i am attaching the question's image


    m2.jpg
     
  13. Apr 13, 2016 #12

    LCKurtz

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    So why did you tell me the equation was correct, without checking it, when I asked about it? And when are you going to show us how you solve the 2x2 system?
     
  14. Apr 13, 2016 #13

    Mark44

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    If we don't see some effort toward solving the 2x2 system real soon, I'm going to close this thread.
     
  15. Apr 14, 2016 #14
    calculating eigenvalues are 4 and 1 . Eigen vectors are (2,1) and (-1,1)

    and

    $$\left(\matrix{x_1(t) \\ x_2(t)}\right)=c_1e^{4t} \left(\matrix{2 \\ 1 }\right) +c_2e^{t} \left(\matrix{-1 \\ 1 } \right)$$
    or

    $$\left(\matrix{x_1(t) \\ x_2(t)}\right)=\left(\matrix{2 & -1\\ 1 & 1 }\right) \left(\matrix{c_1e^{4t} \\ c_2e^{t} } \right)$$

    so

    $$ x_1(t) = 2c_1e^{4t} - c_2e^{t} $$ and

    $$ x_2(t) = c_1e^{4t} + c_2e^{t} $$

    these are the values of x1 and x2 i found so far .. am i on right direction ?
     
  16. Apr 14, 2016 #15

    LCKurtz

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    This looks correct, and this is where your problem becomes confusing. Your 2x2 system is a vector system of the form$$
    \vec x' = A\vec x$$where$$
    \vec x'=\left(\matrix{x_1(t)\\x_2(t)}\right)' = \left(\matrix{3 & 2 \\ 1 & 2}\right) \left(\matrix{x_1(t)\\x_2(t)}\right)$$and you have a general solution$$
    \vec x = c_1e^{4t} \left(\matrix{2 \\ 1 }\right) +c_2e^{t} \left(\matrix{-1 \\ 1 } \right)$$You could write this as$$
    \vec x = c_1\vec x_1+c_2\vec x_2$$where$$
    \vec x_1 = \left(\matrix{2e^{4t} \\ e^{4t} }\right),~~\vec x_2= \left(\matrix{-e^{t} \\ e^{t} } \right)$$Writing it this way, you have two linearly independent vectors forming a basis for your vector solution. Don't confuse the vectors ##\vec x_1,~\vec x_2## with their components which you are calling ##x_1(t)## and ##x_2(t)##.

    Now the confusion arises in your original post where you state
    When you talk about the two linearly independent solutions of those two equations, you would be talking about the independent vectors we have above. But the general solution of your second order equation would be a scalar, not a vector. So at this point, it isn't clear to me what the question is really asking for. You might have to ask your teacher to clarify this.
     
  17. Apr 15, 2016 #16

    epenguin

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    My understanding is that you are not meant to solve any equation explicitly. (Even though it is not hard to solve any of them.)

    My suggestion is to add the first two equations, and you will then see that you have a simple differential equation in a new variable.

    Maybe you can then see that the third equation is really the same differential equation again but in yet another variable.
     
    Last edited: Apr 16, 2016
  18. Apr 15, 2016 #17

    vela

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    The problem statement seems strangely worded. I'd guess the intent is to solve for ##e^{4t}## and ##e^t## in terms of ##x_1(t)## and ##x_2(t)##.
     
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