Question

Solve the boundary value problem

Answer 1

Another Cauchy-Euler ODE. Take $y=x^r$ and you get

$r(r-1)x^r-16rx^r+64x^r=0$
$r^2-17r+64=0$
$\implies r=\dfrac{17}2\pm\dfrac{\sqrt{33}}2$

so that the general solution is

$y=C_1x^{(17+\sqrt{33})/2}+C_2x^{(17-\sqrt{33})/2}$

With the given boundary conditions, you have

$y(1)=0\implies 0=C_1+C_2$
$y(2)=1\implies 1=C_12^{(17+\sqrt{33})/2}+C_22^{(17-\sqrt{33})/2}$
$\implies C_1=\dfrac{2^{(-17+\sqrt{33})/2}}{2^{\sqrt{33}}-1},C_2=-\dfrac{2^{(-17+\sqrt{33})/2}}{2^{\sqrt{33}}-1}$