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Solving an ODE about a point using a solution about another point?

  1. Mar 11, 2013 #1
    1. The problem statement, all variables and given/known data

    The first task was to solve ##(1-x)y''+y=0## about x = 0, which I've already found.

    Now I have to use this solution to solve ##\color{red}{xy''+y=0}## about x = 1.

    2. Relevant equations

    3. The attempt at a solution

    I found the solution about x = 0 (after a lot of rewriting/simplifying) to be

    ##y=a_0\left(1-\dfrac{1}{2!}x^2-\dfrac{1}{3!}x^3-\dfrac{1}{4!}x^4-\dfrac{2}{5!}x^5-\dfrac{7}{6!}x^6+\cdots\right)+a_1\left(x-\dfrac{1}{3!}x^3-\dfrac{2}{4!}x^4-\dfrac{5}{5!}x^5-\dfrac{18}{6!}x^6+\cdots\right)##

    Supposing that's accurate, how can I use it find the next solution?
     
  2. jcsd
  3. Mar 11, 2013 #2

    HallsofIvy

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    Science Advisor

    Let u= 1- x. Then [itex]dy/du= dy/dx (dx/du)= -dy/dx[/itex] and [itex]d^2y/du^2= -(d/dx)(-dy/dx)= d^2y/dx^2[/itex]. The differential equation becomes [itex]ud^2y/^u+ y= 0 with u= 0. So take your solution to [itex](1- x)y''+ y= 0[/itex], and substitute u= 1-x in place of x.
     
  4. Mar 11, 2013 #3
    Just for clarification, why is this new ODE about u = 0, and not u = 1? Since we let ##u=1-x##, why don't we have ##u_0=1-x_0\Rightarrow u_0=1-0=1?##

    Thanks for the tip!

    EDIT: I see that having ##u_0=1## doesn't change the question... I guess I'd just like to know why ##u_0## doesn't change.
     
  5. Mar 12, 2013 #4
    I take it back! I understand why ##u_0## is 0. Thanks again!
     
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