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General Solution for (x^2)*y +x*y'-y=1/(x+1)

  1. Jul 21, 2012 #1
    General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    Where do I start?
     
  2. jcsd
  3. Jul 21, 2012 #2
    Re: General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    Find a solution to the homogeneous equation first i.e. RHS = 0
    Hint: Try x^n

    Then find a solution to the forced equation i.e. RHS = 1/(x+1)
    Hint: Try something along the lines of a/(x+1)
     
  4. Jul 21, 2012 #3
    Re: General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    by substition I get:
    (x^2)(sum(n=2))a(n)x^(n-2)+x(sum(n=1))a(n)x^(n-1)-(sum(n=0))a(n)x^(n)
    Not sure what to do for n=0 for terms with (x^2) and x so I can end up with the (recurance relation)*x^n.
     
  5. Jul 21, 2012 #4
    Re: General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    You're not using recurrence relations here, just try x^n and see what values of n will give you zero for the homogeneous equation.
     
  6. Jul 21, 2012 #5
    Re: General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    Is this correct? (for the nonhomogeneous)
    n(n-1)x^n+nx^n-x^n=0
    (n^2-n+n-1)x^n=0
    (n-1)(n+1)=0 n=+/- 1
    y=Ax+Bx=>y=x(A+B)
     
  7. Jul 21, 2012 #6
    Re: General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    You have the right values of n for the homogeneous equation i.e. n=+/- 1

    So that means x^1 is a solution and x^(-1) is a solution i.e. y = Ax + B/x solves the homogeneous equation.

    Now all you need is a particular integral which solves the forced (non homogeneous) equation :)
     
  8. Jul 21, 2012 #7
    Re: General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    I don't understand why we seem to be ignoring the (x^2) and x from the original formula?
    And why they are not involved in the general soluton?
     
  9. Jul 21, 2012 #8
    Re: General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    Have we ignored the x^2 and the x?

    Clearly y = x or y = 1/x are solutions to (x^2)*y'' + x*y' - y = 0

    Sorry, you'll have to explain carefully where it is that you're getting confused...
     
  10. Jul 22, 2012 #9
    Re: General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    Are these fundamental solutons? Or do I need initial values for a fundamental solution? Does this mean there are infinite amount of solutions, if so, is there a core of solutions? Or a center or solutions that the rest radiate out from?
     
    Last edited: Jul 22, 2012
  11. Jul 22, 2012 #10
    Re: General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    I'd say before I start explaining things, it would help if you tell me what you already know about differential equations and how to solve them. It seems to me like you might want to review a lot of the material as there are some fundamental concepts you seem to have missed out
     
  12. Jul 22, 2012 #11
    Re: General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    Yes, I understand. I thought there might be some quick answers. The book we are using doesn't seem to be the greatest for beginners. I've seen what looks to me better books from the library that are ten years old. I appreciate your help. Maybe you can suggest some books or resources. Any help would be appreciated.
     
  13. Jul 22, 2012 #12
    Re: General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    Here are the lecture notes that I used in first year undergrad. They should cover pretty much everything and more!
     

    Attached Files:

  14. Jul 22, 2012 #13
    Re: General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    Maybe this is an easier question. What is the difference between x^n and a(n)x^n. The same, but only different? Is a(n)x^n used for power series? I'm sure these are one of the many processes to solving ODE. Maybe you can also point me in another direction. I understand the concept of ODE's and PDE's, but is there a group that I can only call NON ODE's (for lack of a better term(s))? Again, any help is appreciated?
     
  15. Jul 22, 2012 #14
    Re: General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    Right, so basically, here are the general ideas:

    ODE's are when you have a differential equation where the function you're dealing with only depends on one variable e.g. y(x).
    The other types of differential equations are PDE's where you deal with functions of more than one variable e.g. y(x,t)

    Now, when you want to solve an ODE, you have to look at what order it is. So if the highest derivative in the ODE is y'(x) then it's of first order. If the highest derivative in the equation is y''(x) then it's of second order, etc.

    The point here is you should think back to standard integration,
    if you have dy/dx = f'(x), then y = f(x) + c, where c is an arbitrary constant.
    This is important because dy/dx = f'(x) has infinitely many solutions and it's your boundary condition that determines a unique solution to your problem by allowing you to evaluate c. So if you're just finding the general solution, you have y = f(x) + c, where c is a degree of freedom.

    This tells us that whenever we find the general solution to a first order ODE, we should always have one degree of freedom i.e. a constant that can be determined by some boundary condition e.g. if we are given y(0).

    Similarly if we're trying to solve a second order ODE, we would expect two degrees of freedom as you're effectively integrating twice and obtaining two constants as a result.

    The other important thing to think about is whether the ODE is linear i.e. if y1 and y2 are solutions of the ODE, is Ay1 + By2 also a solution for any constants A and B?
    If so, the ODE is linear and more importantly this means that if you can find a y1 and a y2 which solve the ODE such that y1 is not a scalar multiple of y2, then you've found your general solution i.e. y = Ay1 + By2 where A and B are arbitrary constants (your degrees of freedom).
    For a first order linear ODE, if you find a solution y1, then your general solution will be Ay1 where A is some arbitrary constant.

    This is why in the example I said try x^n, because if x^n solves your equation then Ax^n solves it and A is your constant i.e. degree of freedom.

    When you're doing series solutions, plug in y = Ʃa(n)*x^n
    and then try to find a recurrence relation for the a(n) where usually a(0) and a(1) end up being arbitrary constants.

    I'd say read the notes for further details and some worked examples, but above are all the basic general ideas for solving ODE's and how it works. The best way to get a feel for it is by doing loads of practice questions.
     
  16. Jul 22, 2012 #15
    Re: General Solution for (x^2)*y"+x*y'-y=1/(x+1)

    Yes, I'll read the notes. Thank you.
     
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