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Serious problem about Differential equation

  1. Mar 2, 2008 #1
    Hi! I have 3 really problem here that i beg anyone who knows more to help me:
    1. How to solve the equation of the kind xy'-3y=3 using power series.
    I usually solve kind pf homogeneous equations. Here the problem rises when i want to equalise all the coefficient of x^n to 0. Because of that 3 there i don't now how to procede.
    2. I don't know what is exactly an ordinary point. They say that an ordianry point is a point at which an ordinary differential equation is "analytic",but i don't understand that term "analytic".
    3. Is it important to look for the radius of convergence first, when you want to solve an ODE using power series? I always consider around x=0.
  2. jcsd
  3. Mar 2, 2008 #2


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    That is exactly why memorizing formulas without recognizing them is a mistake!

    If this were xy'- 3y= 0,you would know what to do? Then you would write [itex]y(x)= \sum_{n=0}^\infty a_n x^n[/itex] so that [itex]y'(x)= \sum_{n=1}^\infty n a_n x^{n-1}[/itex] and your equation would be
    [itex]xy'- 3y= \sum_{n=1}^\infty n a_n x^n- \sum_{n=0}^\infty 3a_n x^n= 0[/itex]
    Notice that the first sum is from 1 to infinity while the second sum is from 0 to infinity. That happens because when you differentiate a power series, you lose the first term. Of course, since you say you can solve homogenous equations, you know that you need to treat that case separately: If n= 0, 3a0= 0 so a0= 0. If n> 0 then
    nan- 3an= (n- 3)an= 0 which makes the solution kind of trivial! Either an= 0 or n= ? (Of course, it's a separable equation and easy to solve that way.)

    The only difference now is that you have
    [itex]xy'- 3y= \sum_{n=1}^\infty n a_n x^n- \sum_{n=0}^\infty 3a_n x^n= 3[/itex]
    You can think of that right hand side as 3x0 in order to match it to the left hand side. Again, do the n=0 case separately: 3a0= 3 so a0= 1 now. Do you see what to do if you had x2 or 3+ x2 on the right side? For some more complicated function, like ex or cos(x), expand them in their Taylor series and then equate corresponding coefficients.

    A real valued function is analytic at a point if its Taylor's series exists at that point (so it must be infinitely differentiable in order to have a Taylor's series) and converges to the function on some neighborhood of the point. Caution- there exist functions that "have" Taylor's series but the radius of convergence is 0. There exist functions that have Taylor's series that converge for all x but not to the function! Fortunately for you almost all, if not all, of the examples you will see are "not analytic" because they have a factor in the denominator that goes to 0 at that point.

    I wouldn't consider it "necessary", although it might save you some work if you found the radius of convergence was 0!

    You shouldn't consider only "around x= 0". You might well have a problem where the "initial condition" is given at some point other than x= 0. If you were told that y(a)= 1, satisfying that condition will be much easier if you wrote [itex]y= \Sum a_n (x- a)^n[/itex]. Or you could "shift" your variable: write t= x- a and rewrite the equation in terms of t rather than a.
  4. Mar 3, 2008 #3
    And can someone show me (give me hint on) how to find legendre polynomials on 1st and 2nd degrees using (1-2xh+h^2)^(-1/2)=∑h^n P_n(x) .
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