# DiffEQ 2nd order series sol'n problem: (1 - x)y'' + y = 0, x0 = 0

1. Jun 5, 2005

### VinnyCee

This problem is from section 5.2 in Boyce, DiPrima's Differential Equations 8th edition.

$$(1 - x)\,y''\,+\,y\,=\,0$$

I get:

$$2\,a_2\,+\,a_0\,+\,\sum_{n\,=\,1}^{\infty}\,\left[(n\,+\,2)\,(n\,+\,1)\,a_{n\,+\,2}\,-\,n\,(n\,+\,1)\,a_{n\,+\,1}\,+\,a_n\right]\,x_n\,=\,0$$

$$2\,a_2\,+\,a_0\,=\,0$$

$$a_2\,=\,-\frac{1}{2}\,a_0$$

and the recursion formula:

$$a_{n\,+\,2}\,=\,\frac{n\,(n\,+\,1)\,a_{n\,+\,1}\,-\,a_n}{(n\,+\,2)\,(n\,+\,1)}$$

Now I am totally lost, just like in this other thread, because I don't know how to move on from this step. Please help

2. Jun 6, 2005

### VinnyCee

What now?

Using the recursion formula above, i get the following:

\begin{flalign*} a_{(0)+2}& =\frac{[(0)\,+\,1]\,(0)\,a_{(0)\,+\,1}\,-\,a_0}{[(0)\,+\,2]\,[(0)\,+\,1]}&\:\:\:\:\:\:\: a_2& =\frac{-\,a_0}{2*1}\\ a_{(1)+2}& =\frac{[(1)\,+\,1]\,(1)\,a_{(1)\,+\,1}\,-\,a_{(1)}}{[(1)\,+\,2]\,[(1)\,+\,1]}& a_3& =\frac{-a_0\,-\,a_1}{3*2}\\ a_{(2)+2}& =\frac{[(2)\,+\,1]\,(2)\,a_{(2)\,+\,1}\,-\,a_{(2)}}{[(2)\,+\,2]\,[(2)\,+\,1]}& a_4& =\frac{-a_1\,-\,\frac{1}{2}\,a_0}{4*3}\\ \end{flalign*}

Last edited: Jun 6, 2005
3. Jun 6, 2005

### VinnyCee

Almost done...

After fixing the eroor in the first term of the recursion series above, I now set a0 = 1 and a1 = 0 to seperate the two-types of terms left in the formula:

$$y(x)\,=\,\sum_{n\,=\,0}^{\infty}\,a_n\,x^n\,=\,a_0\,+\,a_1\,x\,-\,\frac{a_0}{2\,*\,1}\,x^2\,-\,\frac{a_0\,+\,a_1}{3\,*\,2}\,x^3\,-\,\frac{a_1\,+\,\frac{1}{2}\,a_0}{4\,*\,3}\,x^4$$

$$y_{a_0\,=\,1,\,a_1\,=\,0}(x)\,=\,(1)\,+\,(0)\,x\,-\,\frac{(1)}{2\,*\,1}\,x^2\,-\,\frac{(1)\,+\,(0)}{3\,*\,2}\,x^3\,-\,\frac{(0)\,+\,\frac{1}{2}\,(1)}{4\,*\,3}\,x^4\,=\,1\,-\,\frac{1}{2}\,x^2\,-\,\frac{1}{6}\,x^3\,-\,\frac{1}{24}\,x^4\,+\,...$$

Now with a0 = 0 and a1 = 1:

$$y_{a_0\,=\,0,\,a_1\,=\,1}(x)\,=\,(0)\,+\,(1)\,x\,-\,\frac{(0)}{2\,*\,1}\,x^2\,-\,\frac{(0)\,+\,(1)}{3\,*\,2}\,x^3\,-\,\frac{(1)\,+\,\frac{1}{2}\,(0)}{4\,*\,3}\,x^4\,=\,x\,-\,\frac{1}{6}\,x^3\,-\,\frac{1}{12}\,x^4\,-\,\frac{1}{24}\,x^5\,+\,...$$

Now, factoring out the a0 and a1 terms from each:

$$y(x)\,=\,a_0\,\left[1\,-\,\frac{1}{2}\,x^2\,-\,\frac{1}{6}\,x^3\,-\,\frac{1}{24}\,x^4\,+\,...\right]\,+\,a_1\,\left[x\,-\,\frac{1}{6}\,x^3\,-\,\frac{1}{12}\,x^4\,-\,\frac{1}{24}\,x^5\,+\,...\right]$$

The book says this is the answer, but is there any way to generalize this equation more? Possibly by using a summation?

4. Jun 6, 2005

### dextercioby

U could use factorials,but what would be the purpose?You found the solution,i say that's enough.

Daniel.