Let y_0(x) be a particular solution of the Bessel equation…

Let $y_0(x)$ be a particular solution of the Bessel equation of order zero, $(E_0)\!: \qquad y'' + \frac1x y' + y = 0$, show that $y_0'(x)$ is a particular solution of the Bessel equation of order one, $(E_1)\!: \qquad y'' + \frac1x y' + \left(1-\frac{1}{x^2}\right)y = 0$.

Step 1: Setup

By hypothesis $y_0$ satisfies $(E_0)$ identically for all $x\neq0$: \[y_0'' + \frac1x y_0' + y_0 = 0. \tag{1}\]

The goal is to show that $u:=y_0'$ satisfies $(E_1)$.

Step 2: Differentiating the order-zero equation

Differentiate identity (1) term by term with respect to $x$. The middle term $\dfrac1x y_0'$ requires the product rule, treating $1/x$ and $y_0'$ as a product: \[\frac{d}{dx}\left(\frac1x y_0'\right) = -\frac{1}{x^2}y_0' + \frac1x y_0''.\]

Applying this to (1) term by term gives

\[ y_0''' + \left(\frac1x y_0'' - \frac{1}{x^2}y_0'\right) + y_0' = 0, \]

that is, \[y_0''' + \frac1x y_0'' - \frac{1}{x^2}y_0' + y_0' = 0. \tag{2}\]

Step 3: Rewriting in terms of $u=y_0'$

Set $u=y_0'$, so that $u'=y_0''$ and $u''=y_0'''$. Substituting these into (2): $u'' + \frac1x u' - \frac{1}{x^2}u + u = 0$.

and grouping the two terms in $u$ together, \[u'' + \frac1x u' + \left(1-\frac{1}{x^2}\right)u = 0. \tag{3}\]

Step 4: Conclusion

Equation (3) is exactly $(E_1)$ with $u$ in place of $y$. Since $u=y_0'$ was obtained purely by differentiating the identity (1) that $y_0$ is assumed to satisfy, no additional hypothesis on $y_0$ beyond solving $(E_0)$ was used. Therefore:

the derivative of any solution of the order-zero Bessel equation is a solution of the order-one Bessel equation. $\blacksquare$