Question

y"+(x-1)y'+y=0

Answer 1

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

Applying Power series solution.

Let

$y=\sum_{n=0}^\infty a_nx^n$

So

$y'=\sum_{n=1}^\infty na_nx^{n-1}\\ y''=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}$

Substitute these values for $y,y'$ and $y''$ into the DE.

$\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+(x-1)\sum_{n=1}^\infty na_nx^{n-1}+\sum_{n=0}^\infty a_nx^n=0\\ \sum_{n=2}^\infty n(n-1)a_nx^{n-2}+x\sum_{n=1}^\infty na_nx^{n-1}-\sum_{n=1}^\infty na_nx^{n-1}+\sum_{n=0}^\infty a_nx^n=0\\ \sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^n-\sum_{n=1}^\infty na_nx^{n-1}+\sum_{n=0}^\infty a_nx^n=0\\ \sum_{n=0}^\infty (n+1)(n+2)a_{n+2}x^n+\sum_{n=0}^\infty na_nx^n-\sum_{n=0}^\infty (n+1)a_{n+1}x^{n+1}+\sum_{n=0}^\infty a_nx^n=0\\$

Now, equating the coefficients of $x^n$ to 0

$(n+1)(n+2)a_{n+2}+na_n-(n+1)a_{n+1}+a_n=0\\ (n+1)(n+2)a_{n+2}=(n+1)a_{n+1}-(n+1)a_n\\ a_{n+2}=\frac{a_{n+1}-a_n}{n+2}$

Now putting $n=0,1,2,\cdots$ . We have

$a_2=-\frac{1}{2}(a_0-a_1)\\ a_3=\frac{1}{3}(a_2-a_1)=-\frac{1}{6}(a_0+a_1)\\ a_4=\frac{1}{4}(a_3-a_2)=\frac{1}{12}(a_0-2a_1)\\ \cdots$

Recall that ;

$y(x)=\sum_{n=0}^\infty a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots$

We will finally have y(x) in terms of $a_0,a_1.$

$y(x)=a_0+a_1x-\frac{1}{2}(a_0-a_1)x^2-\frac{1}{6}(a_0+a_1)x^3+\frac{1}{12}(a_0-2a_1)x^4+\cdots\\ y(x)=a_0(1-\frac{1}{2}x^2-\frac{1}{6}x^3+\frac{1}{12}x^4+\cdots)+a_1(x+\frac{1}{2}x^2-\frac{1}{6}x^3-\frac{1}{6}x^4+\cdots).$