Question

5. Find at least the first 4 nonzero terms of the solution to (2 + 4x-2x^2)y"-12(z-1)y'-12y = 0, y(1) = 0, y'(1) =-1.

Answer 1

We're set up to find a series solution to the ODE centered around
`x=1`. Let's first replace
`z=x-1`, so that

`y(x)=y(z+1)\implies y'(x)=y'(z+1)\implies y''(x)=y''(z+1)`

and we'll look for a series solution centered at
`z=0`. We have

`y=\displaystyle\sum_(n\ge0)a_nz^n`

`\implies y'=\displaystyle\sum_(n\ge0)(n+1)a_(n+1)z^n`

`\implies y''=\displaystyle\sum_(n\ge0)(n+2)(n+1)a_(n+2)z^n`

The ODE is then

`(2+4x-2x^2)y''-12(x-1)y'-12y=\left(4-2(x-1)^2\right)y''-12(x-1)y'-12y=0`

which in terms of
`z` is

`\left(4-2z^2\right)y''-12zy'-12y=0`

Substituting the series above into the ODE gives

`\displaystyle4\sum_(n\ge0)(n+2)(n+1)a_(n+2)z^n-2\sum_(n\ge0)(n+2)(n+1)a_(n+2)z^(n+2)-12\sum_(n\ge0)(n+1)a_(n+1)z^(n+1)-12\sum_(n\ge0)a_nz^n=0`

`\displaystyle4\sum_(n\ge0)(n+2)(n+1)a_(n+2)z^n-2\sum_(n\ge2)n(n-1)a_nz^n-12\sum_(n\ge1)na_nz^n-12\sum_(n\ge0)a_nz^n=0`

`\displaystyle\left(8a_2+24a_3z+4\sum_(n\ge2)(n+2)(n+1)a_(n+2)z^n\right)-2\sum_(n\ge2)n(n-1)a_nz^n-12\left(a_1z+\sum_(n\ge2)na_nz^n\right)-12\left(a_0+a_1z+\sum_(n\ge2)a_nz^n\right)=0`

`\displaystyle(8a_2-12a_0)+(24a_3-24a_1)z+\sum_(n\ge2)\bigg[(4n^2+3n+2)a_(n+2)-(2n^2+10n+2)a_n\bigg]z^n=0`

so that the coefficients follow the recurrence,

`\begin{cases}a_0=y(0)=0\\\\a_1=y'(0)=-1\\\\a_n=(2(n-2)^2+10(n-2)+2)/(4(n-2)^2+3(n-2)+2)a_(n-2)=(2n^2+2n-10)/(4n^2-13n+12)a_(n-2)&\text{for }n\ge2\end{cases}`

Thanks to the dependency between every other coefficient, we have
`a_n=0` for all even-valued
`n`. Meanwhile,

`a_1=-1`

`a_3=-\frac{14}9`

`a_5=-(700)/(423)`

`a_7=-(23,800)/(16,497)`

so that

`y(z)\approx-z-\frac{14}9z^3-(700)/(423)z^5-(23,800)/(16,497)z^7`

or in terms of
`x`,

`\boxed{y(x)\approx-(x-1)-\frac{14}9(x-1)^3-(700)/(423)(x-1)^5-(23,800)/(16,497)(x-1)^7}`