Question

Using power series method,solve the following differential equation

(x³+3)y"+4xy'+y=0

Answer 1

Let $y = \sum_{n=0}^{\infty}a_nx^n$

Then $y' = \sum_{n=1}^{\infty}na_nx^{n-1}$

And $y'' = \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}$

Then we have that

$(x^3+3)y'' = \sum_{n=2}^{\infty}n(n-1)a_nx^{n+1} + \sum_{n=2}^{\infty}3n(n-1)a_nx^{n-2} ... (*)$

Also

$4xy' = \sum_{n=1}^{\infty}4na_nx^{n} ... (**)$

Adding up and simplifying y, (*) and (**), we have

$\sum_{n=3}^{\infty}(n-1)(n-2)a_{n-1}x^{n} + \sum_{n=0}^{\infty}3(n+1)(n+2)a_{n+2}x^{n}$

+$a_0 + \sum_{n=1}^{\infty}(4n+1)a_nx^n = 0$

Simplifying further, we have

$\sum_{n=3}^{\infty} \big[(n-1)(n-2)a_{n-1} +3(n+1)(n+2)a_{n+2} +(4n+1)a_n \big]x_n$

$+ 6a_2 + 18a_3x+36a_4x^2+a_0+5a_1x+9a_2x^2 = 0$

Considering

$\sum_{n=3}^{\infty} \big[(n-1)(n-2)a_{n-1} +3(n+1)(n+2)a_{n+2} +(4n+1)a_n \big]x_n$

when $n \geq 1,$ we have that

$a_{n+2} = \dfrac{(1-n)(n-2)a_{n-1} - (4n+1)a_n}{3(n+1)(n+2)}$

So the solution to the differential equation is

$y = a_0 + a_1x +a_2x^2$ $+ \sum_{n=1}^{\infty} \big[ \dfrac{(1-n)(n-2)a_{n-1} - (4n+1)a_n}{3(n+1)(n+2)} \big]x^{n+2}$