Question

Solve 𝑦 ′′ − 𝑥𝑦 ′ + 𝑦 = 0 assuming a power series. Find the range of 𝑥 for which the solution is valid

Answer 1

$y''-xy'+y=0$

It is clear from the equation above that it has no singular point. Therefore the range of x for which the solution is valid is $(-\infty,\infty)$

Let the solution of our equation be of the form $y=\displaystyle\sum_{n=0}^{\infty}a_nx^n$ .....(1)

From which,

$y'=\displaystyle\sum_{n=0}^{\infty}na_nx^{n-1}=\displaystyle\sum_{n=1}^{\infty}na_nx^{n-1}$

$y''=\displaystyle\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}=\displaystyle\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}$

We then substitute the above expressions in our equation

$\displaystyle\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}-\displaystyle\sum_{n=1}^{\infty}na_nx^{n}+\displaystyle\sum_{n=0}^{\infty}a_nx^n=0$

$\displaystyle\sum_{n=0}^{\infty}(n+1)(n+2)a_{n+2}x^{n}-\displaystyle\sum_{n=1}^{\infty}na_nx^n+\displaystyle\sum_{n=1}^{\infty}a_nx^n+a_0=0$

$2a_2+6a_3x+\displaystyle\sum_{n=1}^{\infty}[(n+1)(n+2)a_{n+2}-na_n+a_n]x^n+a_0=0$

Comparing both sides of equal powers of x we have,

$2a_2+a_0=0\implies a_2=-\frac{1}{2}a_0$

$6a_3=0\implies a_3=0$

$a_{n+2}=\frac{n-1}{(n+1)(n+2)}a_n$ , $n\ge 0$ .....(1)

From the recurrence relation (2) above we get,

$a_{4}=\frac{1}{(2+1)(2+2)}a_2=-\frac{1}{24}a_0$

$a_5=0=a_7=a_9=0...$

$a_{6}=\frac{4-1}{(4+1)(4+2)}a_4=\frac{1}{10}\cdot (-\frac{1}{24}a_0)=-\frac{1}{240}a_0$

Substitute the values of $a_2,a_3,a_4,a_5,a_6$ in the relation (1) to get:

$y=a_1x+a_0(1-\frac{1}{2}x^2-\frac{1}{24}x^4-\frac{1}{240}x^6...)$

$\therefore$ the solution to our equation is

$y=Ax+B(1-\frac{1}{2}x^2-\frac{1}{24}x^4-\frac{1}{240}x^6...)$

Where A and B are arbitrary constants.