Question

Solve and find at least 3 non-zero terms of each infinite series solutions to the differential equation about x=0 for 2x^2y" + xy' +x^2y = 0. Please show steps. Thanks!

Answer 1

`\large \boxed{y=a_0\left( 1-(1)/(6)x^2+(1)/(168)x^4-(1)/(11088)x^6+...\right)}`

Hello, please consider the following.

The equation is

`2x^2y`

Assume that, on a given domain where the sum is defined,

`\displaystyle y=\sum_(n=0)^(\infty) a_nx^n`

is a solution of the equation and we will find a recursion formula for the
`(a_n)_(n\geq 0)`, then

`\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)`

So the equation becomes

`\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)`

And this is equal to 0, so we can say that

`a_1=0\\(2n-1)na_n+a_(n-2)=0 \ \ \text{for n }\geq 2`

It comes

`\boxed{a_n=-(a_(n-2))/(n(2n-1))}`

`a_1=0 \ so \ a_3=0 \ \ and \ \ a_5=0 \ \ so \ \ a_(2n+1)=0\\a_2=(-a_0)/(2(4-1))=(-a_0)/(2*3)=-(a_0)/(6)\\\\a_4=-(a_2)/(4(8-1))=(a_0)/(2*3*4*7)=(a_0)/(168)\\\\a_6=-(a_4)/(6(12-1))=-(a_0)/(2*3*4*7*6*11)=(-a_0)/(11088)`

So

`y=a_0\left( 1-(1)/(6)x^2+(1)/(168)x^4-(1)/(11088)x^6+...\right)`

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