[tex]\large \boxed{y=a_0\left( 1-\dfrac{1}{6}x^2+\dfrac{1}{168}x^4-\dfrac{1}{11088}x^6+...\right)}[/tex]
Hello, please consider the following.
The equation is
[tex]2x^2y" + xy' +x^2y = 0[/tex]
Assume that, on a given domain where the sum is defined,
[tex]\displaystyle y=\sum_{n=0}^{\infty} a_nx^n[/tex]
is a solution of the equation and we will find a recursion formula for the [tex](a_n)_{n\geq 0}[/tex], then
[tex]\displaystyle y=\sum_{n=0}^{\infty} a_nx^n\\\\y'=\sum_{n=1}^{\infty} na_nx^{n-1}\\\\y''=\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}[/tex]
So the equation becomes
[tex]\displaystyle 2x^2y'+xy'+x^2y=\sum_{n=2}^{\infty} 2n(n-1)a_nx^2x^{n-2}+\sum_{n=1}^{\infty} na_nxx^{n-1}+\sum_{n=0}^{\infty} a_nx^2x^n\\\\=\sum_{n=2}^{\infty} 2n(n-1)a_nx^{n}+\sum_{n=1}^{\infty} na_nx^n+\sum_{n=0}^{\infty} a_nx^{n+2}\\\\=\sum_{n=2}^{\infty} 2n(n-1)a_nx^{n}+\sum_{n=1}^{\infty} na_nx^n+\sum_{n=2}^{\infty} a_{n-2}x^{n}\\\\=a_1x+\sum_{n=2}^{\infty} 2n(n-1)a_nx^{n}+ na_nx^n+a_{n-2}x^{n}\\\\=a_1x+\sum_{n=2}^{\infty} \left((2n-2+1)na_n+a_{n-2}\right)x^{n}[/tex]
And this is equal to 0, so we can say that
[tex]a_1=0\\(2n-1)na_n+a_{n-2}=0 \ \ \text{for n }\geq 2[/tex]
It comes
[tex]\boxed{a_n=-\dfrac{a_{n-2}}{n(2n-1)}}[/tex]
[tex]a_1=0 \ so \ a_3=0 \ \ and \ \ a_5=0 \ \ so \ \ a_{2n+1}=0\\a_2=\dfrac{-a_0}{2(4-1)}=\dfrac{-a_0}{2*3}=-\dfrac{a_0}{6}\\\\a_4=-\dfrac{a_2}{4(8-1)}=\dfrac{a_0}{2*3*4*7}=\dfrac{a_0}{168}\\\\a_6=-\dfrac{a_4}{6(12-1)}=-\dfrac{a_0}{2*3*4*7*6*11}=\dfrac{-a_0}{11088}[/tex]
So
[tex]y=a_0\left( 1-\dfrac{1}{6}x^2+\dfrac{1}{168}x^4-\dfrac{1}{11088}x^6+...\right)[/tex]
We can go further to find a generic expression but only 3 non-zero terms were requested.
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
