= Solution
The <matter density parameter> at general <scale factor> is
$$
\Omega_m(a)=\frac{\Omega a^{-3}}{\Omega a^{-3}+(1-\Omega)a^{-2}}=\frac1{1+x}.
$$
The reference to the definition of $x$ in the printed part points to the wrong equation number: the needed definition is the substitution in part (b). Write $B=(1-\Omega)/\Omega$, so $x=Ba$, $d/da=B\,d/dx$, and
$$
H=\frac{H_0(1-\Omega)^{3/2}}{\Omega}\,x^{-3/2}(1+x)^{1/2}.
$$
Expanding the <matter growth equation as a function of scale factor> first gives
$$
\frac{d^2\delta}{da^2}+\left(\frac3a+\frac{d\ln H}{da}\right)\frac{d\delta}{da}-\frac{3\Omega_m(a)}{2a^2}\delta=0.
$$
Using $d\ln H/dx=-3/(2x)+1/[2(1+x)]$ and dividing by $B^2$, we obtain
$$
\boxed{\delta''+\frac12\left(\frac3x+\frac1{1+x}\right)\delta'-\frac{3\delta}{2x^2(1+x)}=0,}
$$
where these primes denote <derivatives> with respect to $x$. For the proposed <linear cosmological density perturbation> $\delta_1=(1+x)^{1/2}x^{-3/2}$,
$$
\frac{\delta_1'}{\delta_1}=-\frac{2x+3}{2x(1+x)},\qquad
\frac{\delta_1''}{\delta_1}=\frac{8x^2+24x+15}{4x^2(1+x)^2}.
$$
The left side of the <differential equation>, divided by $\delta_1$, has numerator
$$
(8x^2+24x+15)-(4x+3)(2x+3)-6(1+x)=0
$$
over the common denominator $4x^2(1+x)^2$. This proves the result directly. Also $\delta_1'/\delta_1<0$ for every $x>0$, so it is a \b[decaying mode]:
$$
\boxed{\delta_1\sim x^{-3/2}\quad(x\to0),\qquad \delta_1\sim x^{-1}\quad(x\to\infty).}
$$
It is proportional to the <Hubble parameter>, an instance of the <Hubble parameter as a solution of the dust growth equation>.
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