Question

(x^2+1) y''+xy'-xy=0 in series solution

Answer 1

Given

$(x^2+1) y''+xy'-xy=0$

Let

$\begin{aligned} y&=\sum_{n=0}^{\infty} a_{n} x^{n}\\ y'&=\sum_{n=1}^{\infty} na_{n} x^{n-1}\\ y''&=\sum_{n=2}^{\infty} n(n-1)a_{n} x^{n-2} \end{aligned}$

Then

$\begin{aligned} (x^2+1)\sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2}+x \sum_{n=1}^{\infty} na_{n} x^{n-1}-x\sum_{n=0}^{\infty} a_{n} x^{n}&=0\\ \sum_{n=2}^{\infty} n(n-1) a_{n} x^{n}+\sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2} + \sum_{n=1}^{\infty} na_{n} x^{n}-\sum_{n=0}^{\infty} a_{n} x^{n+1}&=0\\ \sum_{k=2}^{\infty} k(k-1) a_{k} x^{k}+\sum_{k=4}^{\infty}( k-2)(k-3) a_{k-2} x^{k} + \sum_{k=1}^{\infty} ka_{k} x^{k}-\sum_{k=1}^{\infty} a_{k-1} x^{k}&=0\\ 2a_2x^2+6a_3x^3+\sum_{k=4}^{\infty} k(k-1) a_{k} x^{k}+ \sum_{k=4}^{\infty}( k-2)(k-3) a_{k-2} x^{k} \\ +a_1x+2a_2x^2+3a_3x^3 + \sum_{k=4}^{\infty} ka_{k} x^{k}-a_0x-a_1x^2-a_2x^3-\sum_{k=4}^{\infty} a_{k-1} x^{k}&=0\\ (a_1-a_0)x+(4a_2-a_1)x^2+(9a_3-a_2)x^3\\ +\sum_{k=4}^{\infty}[ k(k-1) a_{k}+(k-2)(k-3)a_{k-2}+ka_k-a_{k-1} ]x^k&=0 \end{aligned}$

Then we get

$\begin{aligned} a_1-a_0&=0\ \ \to a_1=a_0\\ 4a_2-a_1&=0\ \ \to a_2= \frac{1}{4}a_0\\ 9a_3-a_2&=0\ \ \to a_3=\frac{1}{36}a_0\\ k^2a_k&=(k-2)(k-3)a_{k-2}-a_{k-1}\\ a_{k}&=\frac{(k-2)(k-3)a_{k-2}-a_{k-1}}{k^2},\ \ \ \ \ k\geq 4 \end{aligned}$

Hence

$\begin{aligned} \end{aligned}$ $\begin{aligned} a_4&=\frac{2a_2-a_3}{16}=\frac{17}{576}a_0\\ a_5&=\frac{2a_2-a_3}{16}=\frac{79}{14400}a_0\\ \vdots \end{aligned}$

Then we can write the solution as

$\begin{aligned} y&=\sum_{n=0}^{\infty}a_nx^n\\ &=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots \\ &=a_0+a_0x+\frac{1}{4}a_0x^2 +\frac{1}{36}a_0x^3+\frac{17}{576}a_0x^4+\frac{79}{14400}a_0x^5+\cdots \end{aligned}$