Answer
(a) $J'(0) = 0$
(b) $J''(0) = -\frac{1}{2}$
Work Step by Step
(a) It is given that $y = J(x)$ and $J(0) = 1$
$xy''+y'+xy = 0$
$xJ''(x)+J'(x)+xJ(x) = 0$
We can find $J'(0)$:
$(0)J''(0)+J'(0)+(0)J(0) = 0$
$0+J'(0)+0 = 0$
$J'(0) = 0$
(b) We can find $J''(0)$:
$xJ''(x)+J'(x)+xJ(x) = 0$
$J''(x)+xJ'''(x)+J''(x)+J(x)+xJ'(x) = 0$
$J''(0)+(0)J'''(0)+J''(0)+J(0)+(0)J'(0) = 0$
$J''(0)+0+J''(0)+J(0)+0 = 0$
$2J''(0)+1 = 0$
$2J''(0) = -1$
$J''(0) = -\frac{1}{2}$
