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

1. Mar 11, 2013

SithsNGiggles

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. Mar 11, 2013

HallsofIvy

Staff Emeritus
Let u= 1- x. Then $dy/du= dy/dx (dx/du)= -dy/dx$ and $d^2y/du^2= -(d/dx)(-dy/dx)= d^2y/dx^2$. The differential equation becomes $ud^2y/^u+ y= 0 with u= 0. So take your solution to [itex](1- x)y''+ y= 0$, and substitute u= 1-x in place of x.

3. Mar 11, 2013

SithsNGiggles

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.

4. Mar 12, 2013

SithsNGiggles

I take it back! I understand why $u_0$ is 0. Thanks again!