Answer to Question #275550 in Differential Equations for Aman

Question

Solve the following differential equations in series.

x²d²y/dx + xdy/dx +(x²-4)y=0

Answer 1

$y(x)=\displaystyle\sum_{n=0}^{\infin}a_nx^n$

$y'(x)=\displaystyle\sum_{n=1}^{\infin}na_nx^{n-1}$

$y''(x)=\displaystyle\sum_{n=2}^{\infin}na_n(n-1)x^{n-2}$

$x^2\displaystyle\sum_{n=2}^{\infin}na_n(n-1)x^{n-2}+x\displaystyle\sum_{n=1}^{\infin}na_nx^{n-1}+(x^2-4)\displaystyle\sum_{n=0}^{\infin}a_nx^n=0$

$(x^2-4)(a_0+a_1x)+a_1x+\displaystyle\sum_{n=2}^{\infin}(na_n(n-1)+na_n+(x^2-4)a_n)x^n=0$

$\displaystyle\sum_{n=2}^{\infin}na_n(n-1)x^{n}+\displaystyle\sum_{n=1}^{\infin}na_nx^{n}+\displaystyle\sum_{n=0}^{\infin}a_nx^{n+2}-4\displaystyle\sum_{n=0}^{\infin}a_nx^n=0$

$\displaystyle\sum_{n=2}^{\infin}na_n(n-1)x^{n}+\displaystyle\sum_{n=1}^{\infin}na_nx^{n}+\displaystyle\sum_{n=2}^{\infin}a_{n-2}x^{n}-4\displaystyle\sum_{n=0}^{\infin}a_nx^n=0$

$-4a_0-4a_1x+a_1x+\displaystyle\sum_{n=2}^{\infin}(na_n(n-1)+na_n+a_{n-2}-4a_n)x^{n}=0$

$a_0=a_1=0$

$na_n(n-1)+na_n+a_{n-2}-4a_n=0,n\ge 2$

$a_n=-\frac{a_{n-2}}{n(n-1)+n-4}=-\frac{a_{n-2}}{n^2-4},n\neq 2$

$y(x)=-\displaystyle\sum_{n=3}^{\infin}\frac{a_{n-2}}{n^2-4}x^n$