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Non-homogeneous (Annihilator method?)

  1. Feb 29, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]y'=(2x+y+1)^{-1}[/tex]


    2. Relevant equations
    Find the complimentary Yc and then use annihilator to find Yp (<-- at least that is what I am trying)


    3. The attempt at a solution

    [tex]y'=(2x+y+1)^{-1}[/tex]

    [tex]\Rightarrow \frac{1}{y'}=2x+y+1[/tex]

    [tex]\Rightarrow \frac{1}{y'}-y=2x+1[/tex]

    In operator form:

    [tex]\Rightarrow (\frac{1}{D}-1)y=2x+1[/tex]

    Finding Yc by characteristic equation:

    [tex]\Rightarrow \frac{1}{m}-1=0[/tex]

    [tex]\Rightarrow m=1[/tex]

    Thus, [tex]Y_c=c_1e^x[/tex]

    To find Yp, multiplying both sides of [itex](\frac{1}{D}-1)y=2x+1[/itex] by the operator [itex]D^2[/itex]:

    [tex]\Rightarrow D-D^2=D(1-D)=0[/tex]

    [tex]\Rightarrow D=0,1[/tex]

    [tex]\Rightarrow Y_p=A+Bxe^x[/tex] due to the repeated root.

    Does this look okay so far?

    If so, I just find Y'p and Y''p and plug them back into [tex]\Rightarrow \frac{1}{y'}=2x+y+1[/tex]

    and compare coefficients of like powers of x right?

    Is there a better way to do this? (keep in mind it is still early on in the class, so we may not have learned the easier way if so)
     
  2. jcsd
  3. Feb 29, 2008 #2
    You can make this equation exact by calculating an integrating factor. To see this, rewrite it as:

    [tex]dx-(2x+y+1)dy=0[/tex]

    Now you can obtain an integrating factor (containing only y) to make it exact. The way with the differential operator seems a bit odd to me for this kind of equation.
     
  4. Feb 29, 2008 #3
    I am not sure that I am at all familiar with that method. The only thing I have done with an integrating factor is when I have an equation of the form

    [tex]\frac{dy}{dx}+P(x)y=g(x)[/tex]

    and then I have always just used [tex]\mu =e^{\int P(x)dx}[/tex] as that what you mean?

    I don't see how this equation is exact?
     
  5. Feb 29, 2008 #4

    arildno

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    Well, the simplest is to set: u=2x+1+y.

    Then we get:
    [tex]\frac{du}{dx}=2+\frac{dy}{dx}=2+\frac{1}{u}=\frac{2u+1}{u}[/tex]

    This separates nicely as:
    [tex]\frac{udu}{2u+1}=dx[/tex]
    or:
    [tex]\frac{1}{2}(\int(1-\frac{1}{2u+1})du=x+C\to{u}-\ln(\sqrt{2u+1})=2x+C[/tex]
    Thus, we get:
    [tex]y+1-\ln(\sqrt{4x+2y+3})=C[/tex]
     
  6. Feb 29, 2008 #5
    It is not exact. The integrating factor makes it exact. The equation

    [tex]\frac{dy}{dx}+P(x)y=g(x)[/tex]

    is an application of the use of integrating factors. You can make the following differential equation:

    [tex]P(x,y)dx+Q(x,y)dy=0[/tex]

    exact by multiplying it with an integrating factor. Two cases exist, first, the following does not contain y explicitly:

    [tex]\frac{1}{Q}\cdot \left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x} \right)[/tex]

    gives the integrating factor:

    [tex]\lambda(x)=exp\left(\int \frac{1}{Q}\cdot \left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x} \right) dx \right)[/tex]

    or if the following does not contain x explicitly:

    [tex]\frac{-1}{P}\cdot \left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x} \right)[/tex]

    gives the integrating factor:

    [tex]\lambda(y)=exp\left(\int \frac{-1}{P}\cdot \left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x} \right) dy \right)[/tex]

    Multiplying this with the differential equation makes it an exact one. Look this up in your textbook, it has to be in there. And thus the way to solve the original problem you posted. You need to use the second type, this is [tex]\lambda(y)[/tex]. Can you proceed from here? You will need the proof of these formulas in order to solve this equation. In case you have to use the differential operator, I can't help you.
     
  7. Feb 29, 2008 #6
    It seems someone was faster :smile:. The result is indeed the one you get when you use the integrating factor. I got:

    [tex]e^{-2y} \cdot \left(x+\frac{3}{4}+\frac{y}{2}\right)=K[/tex]
     
  8. Feb 29, 2008 #7
    Stupid question, but what allows me to make this substitution. I have seen so many different substitutions in this course and they are just too many to keep track of. I have seen, let y=ux let x=vy let u=Ax+By+C<---hey, is that the one you used?

    I have no problem solving skills when it comes to this course. With the exception of a Bernoulli....I have no good way of "spotting" what kind of equation I am dealing with.

    Are there "dead giveaways" that I should be seeing...or is it just trial and error all day?
     
  9. Feb 29, 2008 #8
    I would say it is mainly practice. If you understand the theory then it is "just" a matter of making as much exercises as you can find. In the end you will learn to "see" what is the preferred method in more and more cases. However there is no definite way of solving differential equations and very often I have to try several methods before I find the solution. This is something I am not blaming myself, it is the absence of a method that makes it difficult. There is really only one thing that can be solved systematically and that is the calculation of a derivative. There is no method for doing integrals or solving differential equations, only practice can make you better by applying the known (or unknown) "tricks".
     
  10. Feb 29, 2008 #9
    I see what you mean. I guess then the real problem is that this "theory" that you speak of that I need to understand. I don't know what it is! All we are learning are these cookbook style methods for solving very specific kinds of diff eqs.

    But when he takes all of these different kinds of diff eqs and throws them on a quiz, and they're not in standard form or whatever you call it, they all look the same. . . and they all look different all at once. Then it's trial and error.

    Every one of these cookbook style methods has involved some very lofty assumptions that have yet to be explained to us.

    So I do not have much of the theory of general approach to solving a diff eq.

    Know what I mean.

    Edit: Also, this expression:[tex]exp\left(\int \frac{1}{Q}\cdot \left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x} \right) dx \right)[/tex]

    Does that just mean "e to the" all of that?
     
    Last edited: Feb 29, 2008
  11. Feb 29, 2008 #10
    I know exactly what you mean. When I was studying differential equations, I also learned the methods step by step and after every different method there were a number of exercises which were a direct application of the theory. That was not so difficult. However at the end of a few chapters there were a large set of problems that didn't state what method to use and solving those was the real challenge. Then you need to know all the different methods and learn how to "see" the correct solution. That was the perfect preparation for the exams. It takes time no doubt about that.

    Yes it does. A very important class of equations is the one that is exact. Whether a differential equation is exact or not is easily checked. Write the equation as:

    [tex]P(x,y)dx+Q(x,y)dy=0[/tex]

    It is exact if:

    [tex]\frac{\partial P(x,y)}{\partial y} = \frac{\partial Q(x,y)}{\partial x} [/tex]

    If it is not, then you (sometimes) can make it exact by multiplying it with a function which is called the integrating factor. This factor can be found as I wrote a little earlier.

    In the case of the equation you gave, it is:

    [tex]\lambda(y)=exp\left(\int \frac{-1}{1}(0+2)dy \right)=exp(-2y)=e^{-2y}[/tex]

    If you multiply this with the original equation it becomes exact:

    [tex]e^{-2y}dx-e^{-2y}(2x+y+1)dy=0[/tex]

    is exact and can be solved using the method of an exact differential equation.
     
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