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Second Order homogeneous linear differential equation solution

  1. Sep 9, 2008 #1
    Hi everyone. I am really confused at the moment learning about Second Order homogeneous linear differential equations. I lay out the background of what I would like to understand. So I understand the actual maths that goes into the diff's, but I do not understand why it should be so, given the general form of a second order homog linear diff

    a\frac{d^{2}y}{dx^{2}} + b\frac{dy}{dx} + cy = 0

    where a, b and c are constants (i deal with just constants for now, in fairness some of this stuff in discussion will translate over when they are considered as functions of x) and y is a function of x.

    I am told that the general solution of this equation is of the form

    y = Au + Bv

    where u and v are different solutions of the differential equation and A and B are non-zero constants. But surly by that that therefor implies that for every second order diff of this form there are at least 3 distinct solutions, y = u, y = v and y = Au + Bv.

    Looking around the net and in books I don't see why this has to be so. I am either given examples and then from that it is implied this must be so, or even worse it is simply stated, I find this extremely frustrating as there is no rigor in any of the stuff I have read. Could anyone provide me with information on either why this is so, or links to more in depth information. Thanks everyone :-)
  2. jcsd
  3. Sep 9, 2008 #2


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    Consider first the plane: since it is two-dimensional, there are two linearly independent vectors v, w (that is, there is no number c such that v = c w). You can write down any vector in this space as A v + B w with A and B numbers.

    A second order differential equation has two linearly independent solutions u and v (linearly independent meaning: there is no number c such that u = c v). Any solution to the equation can be written as A v + B w, with A and B numbers. In fact, you can say the "space of solutions is two-dimensional" (and in fact, this is not even a strange statement as it turns out to be really a vector space in the linear-algebra sense of the word).
  4. Sep 9, 2008 #3


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    No, that implies that there are an infinite number of solutions: a different solution for each choice of A and B. We don't normally specify y= u and y= v separately because they are both variations of Au+ Bv, the first with A=1, B=0 and the second with A=0, B= 1.

    A good differential equations text book should include, perhaps in an appendix, that "The set of all solutions to an nth order, homogeneous, linear differential equation form an n dimensional vector space." If you have studied linear algebra and know what an "nth order, homogeneous, linear differential equation form an n dimensional vector space" that should tell you what you are looking for. If you haven't then you can't complain about lack of "rigor"- you don't have the prerequisites.

    An outline of the proof goes like this:
    1) You should already have seen a theorem that says that, as long as f(x,y) is differentiable in x and y in some neighborhood of (x0, y0), there exist a unique solution fo the initial value problem dy/dx= f(x,y), y(x0)= y0.

    2) That same proof can be extended easily to "vector valued" equations where Y(x) is a vector function with components <y1(x), y2(x), ..., yn(x)>.

    3.) Any nth order, homogeneous, linear differential equation, say
    [tex]a_n(x) \frac{d^n y}{dx^n}+ a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}+ \cdot\cdot\cdot+ a_1\frac{dy}{dx}+ a_0y= 0[/tex]
    can be written as a set of first order differential equations by taking y1/sub]= y, y2/sub]= dy/dx, y3ub]= d2y/dx2, etc. So that your equations become dy1/sub]/dx= y2/sub], etc until the last one is
    [tex]\frac{dyn}{dx}= -\left(a_{n-1}y_{n-1}+ a_{n-2}y_{n-2}+ \cdot\cdot\cdot+ a_0y_1right)/a_n[/tex]

    4.) Thus, there exist a unique solution to the differential equation with Y(x0)= <y1/sub], y2/sub], \cdot\cdot\cdot, yn>
    which corresponds to the nth[/th] order differential equation with initial conditions y(x0)= y1, y'(x0)= y2, y"(x0)= y3, etc.

    5) In particular, there exist a unique solutions with
    a) y1= 1, yi= 0 for [itex]i\ne 1[/itex]
    b) y2= 1, yi= 1 for [itex]i\ne 2[/itex
    Now you can show that those functions are independent and that every solution to the differential equation can be written as a liner combination of those n functions.

    Here's how it would work for the second order differential equation ad2
    y/dx2+ b dy/dx+ cy= 0 that you specify.

    Let y1(x)= y, y2(x)= dy/dx. then we have the two equations
    dy1/dx= y2 and
    dy2/dx= - (by2+ cy1)/a

    Write that as the vector equation dY/dx= F(x,Y) with
    [tex]Y= \left(\begin{array}{c} Y_1x) \\ Y_2(x)\end{array}\right)[/tex]
    Then there exist Y1(x) satifying that differential equation and the intial value conditions
    [tex]Y_1(x_0)= \left(\begin{array}{c} 1 \\ 0\end{array}\right)[/tex]
    and another satisfying that differential equation and the initial value conditions
    [tex]Y_1(x_0)= \left(\begin{array}{c} 0 \\ 1\end{array}\right)[/tex]
    Those two solutions are independent: If AY1+ BY2= 0 (for all x), then taking x= x0 we get A(1)+ B(0)= 0 so A= 0. Then in order to have BY2= 0 for all x, since Y2 is not identically 0 (its second component is 1 at x0) we must have B= 0.

    Any solution can be written as a linear combination: let y be any solution to that differential equation. Let A= y(0), B= y'(0). Then it is easy to show that the vector
    [tex]Y(x)= \left(\begin{array}{c}y(x) \\ y'(x)\end{array}\right)[/tex]
    satisfies the vect90 differential equation and with intial conditions
    [tex]Y(x_0)= \left(\begin{array}{c}A \\ B\end{array}\right)[/tex]
    and, in fact, so does AY1+ BY2. By the "uniqueness" of solutions, then, we must have Y(x)= AY1+ BY2 which, comparing first components meand y(x)= Ays[bu]1
    (x)+ By2[/sub(x).
  5. Sep 9, 2008 #4
    CompuChip and HallsofIvy thanks so much, btoh very helpfull. HallsofIvy, you know I did actually believe that the way everything looked it seems that there should be an infinite number of solutions, which makes perfect sense.
    Honestly, I have never seen that statement anywhere. It is very funny you should say that as on my way home on the bus today I was thinking about what CompuChip had said (before you had posted) and thought, surly then this must form some sort of functional space, which I was sort of rite. You see the thing is all of my school texts (Im doing A-level) and all the texts ive found on the internet, nowhere has a statment resembling that ever poped up. You are perfectly rite I dont have the prerequisites, clearly I dont, but for some resason it is not deemed nesersary to know the fundamental stuff. I would ask, is there a text book or anything you can suggest that would not skip over fundamental stuff for calculus, I would have loved if any article or text I came across described the second order diffs as you did, however they havnt. I suppose it is because I am only doing A-levels at the moment, mabye this information would be discovered later on at uni?

    Thanks guys has helped alot
  6. Sep 9, 2008 #5


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    I couldn't really make out from your reply whether you've just not seen that particular statement, or you didn't do any linear algebra at all. But the next paragraph will assume the former.

    As I said, functional spaces are just like vector spaces. For example, you know that a polynomial looks like
    [tex]a_0 + a_1 x + \cdots + a_n x^n[/tex]
    but I bet you never thought of the set of [tex]\{ 1, x, x^2, \cdots \}[/tex] as forming a basis for the vector space of polynomials (and yes, it is a vector space: 0 is a polynomial, the sum of two polynomials is a polynomial, as is a scalar multiple of one). Usually linear algebra is developed with Euclidean spaces ([itex]\mathbb R^n[/itex]) in mind, and functions are something completely different.

    Usually, different subjects (differential equations, calculus, linear algebra, ...) are taught and viewed separately and it is only after studying mathematics for somewhat longer that one starts to discover and appreciate the interesting connections between them. I am pleased that you thought about it a lot and came to the same conclusion, in so far as your current knowledge allows.

    So if you don't have a solid foundation in linear algebra yet, I would recommend doing that first. The subject may seem a bit boring and simple at first but things like bases, vector spaces, dimension of a space (even a one-dimensional vector space usually has infinitely many elements :smile:), orthogonality, etc. turn up in a lot of different subjects.
    Since I don't really know of any good books on either linear algebra, or differential equations, I can't really advise you on that. Judging by your posts, you like differential equations but you may also find an area like functional analysis interesting (which deals with function spaces much more). So probably I should leave it to the experts to point you to good references in an orderly manner (instead of mentioning all the wonderful parts of mathematics you might like to study, ever) :smile:
  7. Sep 9, 2008 #6
    thanks CompuChip, in actuall fact i do know [tex] \{ 1, x, x^2, \cdots \} [/tex] forms the basis for the vector space of polynomials. However I have knowledge of quite a lot of this, but as is evident of this thread, not that in depth, again the issue being that this information in this sort of form isnt available to me except my hopeful searches around the net. Im not sure as to the level of maths you have learnt up to, I presume higher than A-level, the issue I find id that all course content seems to be designed to be User friendly, in the process omitting important information, the underlying principals. Thanks though CompuChip
  8. Sep 9, 2008 #7


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    Sorry I'm not from the UK so I don't really know what is covered in "A-level mathematics" and what's not. I am a mathematics university graduate, so presumable that is higher than you are, yes :smile:
    The good part is, that I got the subjects presented in a somewhat logical order, with one semester for each of them, and that I had 3 years to go through them. But all the pieces only fell together in the second year and later, in secondary school I didn't even learn about vectors and matrices :(

    Anyway, when answering posts on this forums it's always tricky to find out at what level people are, especially since education quality, rigor and order of presentation varies strongly between countries and institutions. But they are a very good way to increase your knowledge, and get the depth and rigor you are looking for. You'll just have to forgive me if I tell you something you already knew :wink:
  9. Sep 9, 2008 #8
    Hey CompuChip, thanks, yeh your about 4 years ahead of my learning level lol. I presume when you say secondary you mean 16-18 education (or 16-17 something in that area :-) which is where I am at the moment). Thats no problem at all as well, most of the stuff youve said I havnt actually known anways, just know what its aboutish. I suppose I can take happiness in the fact that Im not supposed to really know this stuff yet anyway, ill be patient then Ill get there eventually :smile: Thanks again really extremely helpful to both you CompuChip and HallsofIvy
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