Question

Solve in series 2x(1-x)d^2y/dx^2+(1-x)dy/dx+3y=0

Answer 1

Solution

Let

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

$\frac{dy}{dx}=\sum_{n=1}^{\infty}{na_nx^{n-1}}=\sum_{n=0}^{\infty}{(n+1)a_{n+1}x^n}$

$\frac{d^2y}{dx^2}=\sum_{n=0}^{\infty}{n(n+1)a_{n+1}x^{n-1}}=\sum_{n=0}^{\infty}{(n+1)(n+2)a_{n+2}x^n}$

$(1-x)\frac{dy}{dx}=a_1+\sum_{n=1}^{\infty}[(n+1)a_{n+1}-na_n]x^n$

$2x(1-x)\frac{d^2y}{dx^2}=4a_2x+2\sum_{n=0}^{\infty}{[(n+1)na_{n+1}-(n-1)na_n]x^n}$

Substitution into equation:

$4a_2x+2\sum_{n=2}^{\infty}{\left[n\left(n+1\right)a_{n+1}-\left(n-1\right)na_n\right]x^n}+a_1+\left[2a_2-a_1\right]x+\sum_{n=2}^{\infty}{\left[\left(n+1\right)a_{n+1}-na_n\right]x^n}+3a_0+3a_1x+3\sum_{n=2}^{\infty}{a_nx^n}=0$

Coefficient near xⁿ are equal to zero.

x⁰ : 3a₀+a₁=0 => a₁=-3a₀ 

x¹ :4a₂+2a₂-a₁+3a₁=0 => 6a₂+2a₁=0 => a₂=-a₁/3= a₀ 

xⁿ (n>1):2n(n+1)a_n+1-2(n-1)na_n+(n+1)a_n+1-na_n+3a_n=0 => (2n+1)(n+1)a_n+1 +(3+n-2n²)a_n=0 =>

a_n+1 =(2n²-n-3)a_n/[(2n+1)(n+1)]=(2n-3)a_n/(2n+1)

a₃ =a₂/5= a₀/5, a₄ =3a₃/7= 3a₀/(5*7), a₅ =5a₄/9=5*3a₀/(5*7*9), a₆ =7a₅/11=7*5*3a₀/(5*7*9*11), …,

a_n+1 =3a₀/[(2n+1)(2n-1)]

General formula for a_n: a_n =3a₀/[(2n-1)(2n-3)], n>0

Therefore solution is

$y(x)=3a_0\sum_{n=0}^{\infty}\frac{x^n}{(2n-1)(2n-3)}$