By means of a power series (centred at the ordinary point xo = 0), find the two linearly independent solutions (y1 (x) and y2(x)) of the differential equation: y" + 4.x? y = 0

Assume the power series solution for the given differential equation to be as follows:

$y=\sum _{k=0}^{\infty }{a}_k{\left(x-x_0\right)}^k$

In our problem, it is given that $x_0=0$. 

$$\begin{gathered} \left\{y=\sum _{k=0}^{\infty }{a}_k{x}^k \\ y\text{'}=\sum _{k=1}^{\infty }k{a}_k{x}^{k-1} \\ y\text{'}\text{'}=\sum _{k=2}^{\infty }k\left(k-1\right)a_k{x}^{k-2}\right\} \end{gathered}$$..........................................(1)

Substitute equation (1) in the differential equation. 

$$\begin{gathered} y\text{'}\text{'}+4x^{2}y=0 \\ \sum _{k=2}^{\infty }k\left(k-1\right)a_k{x}^{k-2}+4x^2\sum _{k=0}^{\infty }{a}_k{x}^k=0 \end{gathered}$$

$\sum _{k=2}^{\infty }k\left(k-1\right)a_k{x}^{k-2}+4\sum _{k=0}^{\infty }{a}_k{x}^{k+2}=0$...........................(2)

Substitute $k-2=n$ for the first term and $k+2=n$ for the 2nd term in equation (1).

$\sum _{n=0}^{\infty }\left(n+2\right)\left(n+1\right)a_{n+2}{x}^n+4\sum _{n=2}^{\infty }{a}_{n-2}{x}^n=0$

Taking the first two coefficients of the 1^st summation and equating it to zero, we get, 

$$\begin{gathered} 2a_2+6a_3+\sum _{n=2}^{\infty }\left[\left(n+2\right)\left(n+1\right)a_{n+2}+4a_{n-2}\right]x^n=0 \\ \Rightarrow a_2=0, a_3=0, \left(n+2\right)\left(n+1\right)a_{n+2}+4a_{n-2}=0 \end{gathered}$$

$\Rightarrow a_{n+2}=-\frac{4a_{n-2}}{\left(n+2\right)\left(n+1\right)}$..............................................(3)