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Homework Help: Complementary function and particular integral

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data
    The quantities x and y are related by the differential equation :

    [tex]\frac{dy}{dx}+3y=6x+5[/tex]

    (i) Find the complementary function of th above differential equation
    (ii) Find also a particular integral for this differential equation, and hence write down the general solution


    2. Relevant equations
    Integrating factor


    3. The attempt at a solution
    I know how to find the general solution, but I don't know what complementary function and particular integral are....

    Thanks
     
  2. jcsd
  3. Jan 25, 2010 #2

    tiny-tim

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    Hi songoku! :smile:
    Complementary function (or complementary solution) is the general solution to dy/dx + 3y = 0.

    Particular integral (I prefer "particular solution") is any solution you can find to the whole equation.

    (basically, you make a wild but intelligent :rolleyes: guess … in this case, I'd go for a polynomial :wink:)

    Since the difference between any two particular solutions is a complementary function, that means that you can choose just one particular solution, add all the complementary functions to it, and you get all the solutions.​
     
  4. Jan 25, 2010 #3

    vela

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    If you use an integrating factor to solve this problem, you end up with an integral of the integrating factor multiplied by (6x+5). The portion of the solution which comes from this term, sans the integration constant, might be what's being called the particular integral.
     
  5. Jan 25, 2010 #4
    Hi tiny-tim and vela :smile:
    So, differential equation will have complementary solution only if the form :
    dy/dx + (a)y = r(x) ?

    particular integral = particular solution?

    This is what I read on my book :
    "the general solution of a differential equation, involving an arbitrary constant, represents more than one, even infinitely many, solutions. the solutions obtained by assigning definite value to the constant is called a particular solution. typically, the value is determined by making the solution satisfy a given condition, traditionally referred to as initial condition."

    So I think particular integral is not the same as particular solution. To find particular solution, we have to find the general solution first.

    how to make wild but intelligent :rolleyes: , such that you can guess polynomial? and how to apply it to solve the question?

    my book never mentioned about complementary function and particular integral :grumpy:


    I found the general solution : y = (6x+5)/3 - 2/3 + Ce^(-3x). You mean the particular integral is (6x+5)/3 - 2/3 ?

    Thanks ^^
     
  6. Jan 25, 2010 #5

    vela

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    The general solution to a linear differential equation consists of two pieces. One piece is the solution to the homogeneous equation; the other is the solution to the inhomogeneous equation. The latter piece is typically referred to as the particular solution. Your book isn't using the term in this way; perhaps that is why the problem refers to a particular integral instead, to avoid confusion.

    Yes, but keep in mind I'm just making an educated guess. If u(x) is the integrating factor, the solution is given by

    [tex]y=\frac{1}{u(x)}\int u(x)(6x+5)dx + \frac{C}{u(x)}[/tex]

    The term due to the arbitrary constant is the solution to the homogeneous equation. Since
    the first term contains an integral and yields what's typically called the particular solution (though not by your book), I figured that's what particular integral means. So tiny-tim and I are saying the same thing: the particular integral is what most people call the particular solution.
     
  7. Jan 25, 2010 #6

    tiny-tim

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    Hi songoku! :smile:
    No, the complementary solution is where the RHS is zero (the "homogenoeus equation").
    No, I think they're the same …

    "the solution obtained by assigning a definite value to the constant" is a particular solution (in this case, it has the constant = 0).

    Remember, a "particular solution" is any, repeat ANY, solution to the complete equation.​
    No NO NOOOO!

    You can find the particualr solution first.

    It usually has nothing to do with the general complementary solution (in this case, it was linear, and the complementary solution was exponential).
    I believe there are some rules somewhere, but I don't remember them. :redface:

    Usually, it's fairly obvious … you look at the RHS, and imitate it …

    in this case, the RHS is a polynomial, so you try a polynomial.
    Yes, except you'll lose marks in the exam if you don't simplify that, to 2x +1. :wink:
     
  8. Jan 25, 2010 #7
    Hi vela and tiny-tim :wink:
    If so, what is the term for "particular solution", which is the solution we get using the initial value to find the constant? In my book, it is called particular solution, but if particular solution = particular integral, maybe you know other term for calling my definition of "particular solution"

    ( I hope I don't cause confusion...)

    dy/dx + 3y = 6x + 5

    dy/dx - 6x = 5 - 3y

    Can I say complementary function is solution to :
    dy/dx - 6x = 0 ?

    But my book said the value of the constant is obtained using the initial condition given from the question and the constant is not always zero.


    So if the RHS is exponential, I try exponential, and if it is trigonometry, I also try trigonometry?

    Thanks a lot :smile:
     
  9. Jan 26, 2010 #8

    tiny-tim

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    Hi songoku! :smile:
    Yup! :biggrin:
    oops! :redface: i should have been more specific …

    the complementary solution is the general solution when the LHS has only functions of y (so not including functions of x, or constants)

    to put it another way, y is there as (d/dx)0y, the 0th derivative of y :wink:
    I suspect your book is asking eg "Find the particular solution (or integral) of dy/dx + 3y = 6x + 5 for which y = 7 when x = 12" …

    this is using "particular" in its non-technical sense, just as one might ask "find the equation of the particular tangent to this circle that passes through (12,7)".
     
  10. Jan 26, 2010 #9
    Hi tiny-tim :wink:
    Actually for this question, it doesn't ask about particular solution. But the others are just like what you wrote (there is initial value to find the particular solution).

    Let take y = 2 and x = 0 for this question, then the particular solution will be :

    y = 2x + 1 + 1/(e^3x) and this is not the same as particular integral, which is y = 2x + 1

    I think I understand how to find the complementary solution and particular integral. I just a little bit confused about the terms "particular solution" and "particular integral"

    Thanks
     
  11. Jan 26, 2010 #10

    vela

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    I searched the web a bit, and it seems there are two senses in how the term particular solution applies to differential equations. The more common one is in referring to the solution to the non-homogeneous equation. This solution is sometimes called the particular integral, as in your homework problem. But the term particular solution is also apparently used to draw a distinction between a general solution with still arbitrary constants and the solution in which those constants have specific values to satisfy the initial conditions. This latter usage is what your book uses.

    To summarize: Your problem uses "particular" in the more common sense, and the particular integral is the same thing as the particular solution. You can ignore what your book says.
     
  12. Jan 26, 2010 #11

    tiny-tim

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    ah! as in the (non-scientific) expression "From the general to the particular". :smile:
     
  13. Jan 26, 2010 #12
    ah! That's what I mean :smile:


    Thanks a lot to you both !!

    *NB
    tiny-tim, can you help me on other thread about increasing function (you've answered the question earlier). Mark has different opinion..https://www.physicsforums.com/showthread.php?t=372401

    tiny-tim and vela : please help me too at this : https://www.physicsforums.com/showthread.php?t=372400

    And I am very sorry if this is inappropriate or I break the rules...Thanks in advanced
     
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