Can't solve this differential equation

Question

I am trying to solve this differential equation:

$$xy'' = y$$

But Iam not going anywhere.

I have tried sumation of series but is there some method involving modified Bessel functions?

Thanks

Answer 1

Step 1: Multiply by $x$ to get $x^2D_x^2$ $$ 0=x^2D_x^2y-xy. $$

Step 2: Switch to differential operator of the form $xD_x$: $$ 0=[(xD_x)^2-xD_x]y-xy. $$

Step 3: Recreate a quadratic in front of $y$ by letting $x=cs^2$ noting $xD_x=\frac12sD_s$ (from the power $2$ on $s$): $$ 0=[(\tfrac12sD_s)^2-\tfrac12sD_s]y-cs^2y. $$

Step 4: Let $c=\frac14$ to agree with the coefficient of $s^2D_s^2$ except for differing sign $$ 0=[\tfrac14(sD_s)^2-\tfrac12sD_s]y-\tfrac14s^2y =\tfrac14[s^2D_s^2-sD_s-s^2]y. $$

Step 5: Unfortunately the $sD_s$ has the wrong sign. To fix this, let $y=sz$ and so $$ 0=[sD_s^2-sD_s-s^2]sz=[s^2D_s^2+sD_s-(s^2+1)]z $$ and we recognise this as the modified Bessel's differential equation of order $1$.

Answer 2

I am assuming $y$ is a function of $x$. Assume $y(x)$ can be written as $x^m$. Then the differential equation can be written as $m(m-1)x^{m-2}-x^{m-1}=0$.

Solve it for $m$. and do not forget to add the arbitrary constants in your solution