Example 2 Find a series solution around x_(0)=0 for the following differential equation. y''-xy=0

To find a series solution for the differential equation $y'' - xy = 0$ around $x_0 = 0$, we can use the method of power series.<br /><br />Let's assume a power series solution of the form:<br />$y(x) = \sum_{n=0}^{\infty} a_n x^n$<br /><br />Taking the first and second derivatives, we get:<br />$y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}$<br />$y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$<br /><br />Substituting these expressions into the given differential equation, we have:<br />$\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - x \sum_{n=0}^{\infty} a_n x^n = 0$<br /><br />To simplify the equation, let's shift the indices of the series:<br />$\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n - \sum_{n=1}^{\infty} a_n x^{n+1} = 0$<br /><br />Now, we can combine the terms with the same power of $x$:<br />$\sum_{n=0}^{\infty} [(n+2)(n+1) a_{n+2} - a_{n+1}] x^n = 0$<br /><br />For the series to be equal to zero, each coefficient must be equal to zero:<br />$(n+2)(n+1) a_{n+2} - a_{n+1} = 0$<br /><br />Solving for $a_{n+2}$, we get:<br />$a_{n+2} = \frac{a_{n+1}}{n+2}$<br /><br />This recursive relation allows us to find the coefficients of the power series solution. We can start with an initial value, such as $a_0$, and then use the relation to find the subsequent coefficients.<br /><br />For example, if we choose $a_0 = 1$, we can find the next coefficients as follows:<br />$a_1 = 0$<br />$a_2 = \frac{a_1}{3} = 0$<br />$a_3 = \frac{a_2}{4} = 0$<br />and so on.<br /><br />Therefore, the power series solution for the differential equation $y'' - xy = 0$ around $x_0 = 0$ is:<br />$y(x) = \sum_{n=0}^{\infty} a_n x^n$<br /><br />where the coefficients $a_n$ are determined by the recursive relation $a_{n+2} = \frac{a_{n+1}}{n+2}$.