Q.$y=[\log(x+\sqrt{x^2+a^2})]^m$, show that $(x^2+a^2)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}=0$

Given $y=[\log(x+\sqrt{x^2+a^2})]^m$. Let $u=\log(x+\sqrt{x^2+a^2})$ and $s=\sqrt{x^2+a^2}$, so $y=u^m$.

Step 1 — a standard simplification: $u'=\dfrac{1+\frac{x}{s}}{x+s}=\dfrac{(s+x)/s}{x+s}=\dfrac1s$. Differentiating again, $u''=-\dfrac{s'}{s^2}=-\dfrac{x/s}{s^2}=-\dfrac{x}{s^3}$.

Step 2 — differentiate $y=u^m$: $y_1=mu^{m-1}u'=\dfrac{mu^{m-1}}{s}$, so $s\,y_1=mu^{m-1}$.

Step 3 — differentiate $y_1$ again: $y_2=m(m-1)u^{m-2}(u')^2+mu^{m-1}u''=\dfrac{m(m-1)u^{m-2}}{s^2}-\dfrac{mxu^{m-1}}{s^3}$.

Step 4 — multiply by $s^2=(x^2+a^2)$: $(x^2+a^2)y_2=m(m-1)u^{m-2}-\dfrac{mxu^{m-1}}{s}$.

Step 5 — and $x\,y_1=x\cdot\dfrac{mu^{m-1}}{s}=\dfrac{mxu^{m-1}}{s}$.

Step 6 — add: $(x^2+a^2)y_2+xy_1=m(m-1)u^{m-2}-\dfrac{mxu^{m-1}}{s}+\dfrac{mxu^{m-1}}{s}=m(m-1)u^{m-2}$. …

Differentiate $y=u^m$ with $u=\log(x+\sqrt{x^2+a^2})$, using the neat facts $u'=1/\sqrt{x^2+a^2}$ and $u''=-x/(x^2+a^2)^{3/2}$; the genera …

Assuming the target identity is $=0$ for every $m$ without checking — a direct computation (or a quick numerical check, e.g. $a=1,m=3,x=\sqrt3$) shows the true right-hand side is $m(m-1)[\log(x+\sqrt{x^2+a^2})]^{m-2}$, which is …