Solve $xy'' + y' + xy = 0, y(0)=1, y'(0)=0$ The initial value problem $xy'' + y' + xy = 0; y(0)=1, y'(0) = 0$ has_____
Options:
a) a unique solution.
b) no solution
c) infinitely many solutions 
d) two linearly independent solutions
(Is there any other way around to do it other than power series method?)
Ans) I started with y(x) = $x^r$ but fails to continue.
Then I choose $$y(x) =\sum_{n=0}^{\infty} a_n x^n$$
$$y'(x) =\sum_{n=0}^{\infty} na_nx^{n-1}$$
$$y''(x) = \sum_{n-0}^{\infty} n(n-1)a_nx^{n-2}$$
Substitution leads to 
$$ \sum_{n=0}^{\infty} [n(n-1) + n]a_nx^{n-1} + \sum_{n=0}^{\infty} a_n x^{n+1}=0$$
Change of index leads to
$$ \sum_{n=0}^{\infty} [n(n-1) + n]a_nx^{n-1} + \sum_{n=2}^{\infty} a_{n-2} x^{n-1}=0$$
Rewrite as
$$ a_1 + \sum_{n=2}^{\infty} (n^2a_n + a_{n-2})x^{n-1}=0$$ which implies $a_1=0$.
Recursive relation is
$$n^2a_n = -a_{n-2}$$
Take $a_0 = a_0$ and $a_1 = a_1$
$a_2 = \displaystyle\frac{-a_0}{2^2}$  $a_3 = \displaystyle\frac{-a_1}{3^2}$
$a_4 = \displaystyle\frac{a_0}{2^2.4^2}$  $a_5 = \displaystyle\frac{a_1}{3^2.5^2}$
$a_6 = \displaystyle\frac{-a_0}{2^2.4^2.6^2}$  $a_7 = \displaystyle\frac{-a_1}{3^2.5^2.7^2}$
$$y(x) = a_0 + a_0\sum_{k=1}^{\infty} \frac{(-1)^kx^{2k}}{\Pi_{n=1}^{k} (2n)^2} + a_1\sum_{k=0}^{\infty} \frac{(-1)^kx^{2k+1}}{\Pi_{n=0}^{k} (2n+1)^2}$$
already we got $a_1 = 0$ hence solution will be
$$y(x) = a_0 + a_0\sum_{k=1}^{\infty} \frac{(-1)^kx^{2k}}{\Pi_{n=1}^{k} (2n)^2}$$
Given condition $y(0) = 1$ gives $a_0 = 1$ 
Final solution will be  $$y(x) = 1 + \sum_{k=1}^{\infty} \frac{(-1)^kx^{2k}}{\Pi_{n=1}^{k} (2n)^2}$$ and clearly $y'(0) = 0$ satisfies. So concludes a unique solution.
Is there any other way to do it smartly?
 A: Let $Y(s)$ be the Laplace transform of $y(x)$, then $y' = sY - y(0)$ and $y'' = s^2Y - sy(0) - y'(0) $
Taking the transform of the equation
$$ -(s^2Y - s)' + (sY-1) - Y' = 0 $$
$$ -(2sY + s^2Y' - 1) + sY - 1 - Y' = 0 $$
$$ (1+s^2)Y' + sY = 0 $$
Solving the above first-order ODE gives
$$ Y = \frac{C}{\sqrt{1+s^2}}$$
Using a table lookup, we find that
$$ y(x) = J_0(x) $$
As @Urgje pointed out, you can obtain the same solution by multiplying both sides by $x$
$$ x^2y'' + xy' + x^2y = 0 $$
which is the Bessel equation
A: It should be noted that taking the Laplace Transform of this differential equation as @Dylan did, the solution J0 (The Bessel Function) is a solution, generalized for any initial conditions, as they will cancel algebraically after the Laplace Transform is taken.  For example: $$-(s^2Y-sy(0)-y'(0))'+(sY-y(0))+Y'=0$$ $$-(2sY+s^2Y'-y(0)-0)+(sY-y(0))+Y'=0$$ $$-s^2Y'-sY-Y'=0$$Which then becomes a separable differential equation that can be solved using the same method, and will result in a solution of the form J0 using a lookup table.
Source:
Dylan (https://math.stackexchange.com/users/135643/dylan), Solve $xy'' + y' + xy = 0, y(0)=1, y'(0)=0$, URL (version: 2015-11-11): https://math.stackexchange.com/q/1523454
