= Solution
Integrating the given expansion equation gives $R^{3/2}=t/t_0$, with the integration constant fixed by $R(t_0)=1$. Thus the <scale factor> and <Hubble parameter> are $R=(t/t_0)^{2/3}$ and $H=2/(3t)$. The <Newtonian derivation of the flat Friedmann equation> gives $H^2=8\pi G\rho/3$, so
$$
\rho=\frac1{6\pi Gt^2},\qquad c^2=\gamma K\rho^{\gamma-1}=c^2(t_0)(t/t_0)^{-2(\gamma-1)}.
$$
Use the <comoving coordinate> $\boldsymbol q=\boldsymbol x/R$ and the <density contrast> $\delta=\rho_1/\rho$. Linearized <mass conservation>, the peculiar-velocity <Euler equation for fluid motion> and the <Poisson equation> for the <peculiar gravitational potential> are
$$
\dot\delta=-R^{-1}\nabla_q\cdot\boldsymbol v,\qquad
\dot{\boldsymbol v}+H\boldsymbol v=-R^{-1}\nabla_q(c^2\delta+\Phi),\qquad
\nabla_q^2\Phi=4\pi GR^2\rho\delta.
$$
Differentiate the first equation, substitute the second and use the third. A Fourier component of comoving <wavenumber> $k$ obeys
$$
\ddot\varepsilon+2H\dot\varepsilon+\left(\frac{c^2k^2}{R^2}-4\pi G\rho\right)\varepsilon=0.
$$
Since $c^2k^2/R^2=k^2c^2(t_0)t_0^{2\gamma-2/3}t^{-2\gamma+2/3}$, the normalization is $\Lambda=kc(t_0)t_0^{\gamma-1/3}$, as printed in the PDF. The resulting <linear cosmological density perturbation> equation is
$$
\boxed{\ddot\varepsilon+\frac4{3t}\dot\varepsilon+\left(\frac{\Lambda^2}{t^{2\gamma-2/3}}-\frac2{3t^2}\right)\varepsilon=0.}
$$
To reduce it to a <Bessel equation>, put $\nu=\gamma-4/3$, $z=\Lambda/(\nu t^\nu)$ and $\varepsilon=t^a y(z)$. As $\dot z=-\nu z/t$, multiplying the substituted equation by $t^{2-a}$ gives
$$
\nu^2z^2y''+\nu(\nu-2a-1/3)zy'+\left[\nu^2z^2+a^2+a/3-2/3\right]y=0.
$$
Taking $a=-1/6$ makes the first two coefficients proportional to those in the <Bessel equation>, and the remaining constant is $-25/36$. Hence
$$
\boxed{\nu=\gamma-\frac43\in(0,1/3),\qquad\lambda=\frac5{6\nu}=\frac5{6\gamma-8}.}
$$
For noninteger $\lambda$, the <Bessel density modes of a polytropic expanding universe> are
$$
\boxed{\varepsilon=t^{-1/6}\left[A J_{-\lambda}(z)+B J_\lambda(z)\right].}
$$
There is an exceptional case within the allowed interval: if $\lambda=n$ is an integer, $J_{-n}=(-1)^nJ_n$, so these are not two independent solutions. The full solution then uses $t^{-1/6}[A Y_n(z)+B J_n(z)]$, with the <Bessel function of the second kind> providing the missing branch. For example $\gamma=29/18$ gives $\lambda=3$. A basis using $J_\lambda,Y_\lambda$ is valid for every allowed exponent.
For early times, $z\to\infty$. The <large-argument asymptotic expansion of the Bessel function of the first kind> has oscillatory phase $z-\pi a/2-\pi/4$ and envelope $(2/(\pi z))^{1/2}$ for order $a$; the <Bessel function of the second kind> has the same envelope. Thus
$$
|\varepsilon|_{\rm envelope}\propto t^{-1/6}z^{-1/2}\propto\boxed{t^{\nu/2-1/6}=t^{(3\gamma-5)/6}.}
$$
This decreases because $3\gamma<5$. At late times $z\to0$, the leading powers are $J_{\pm\lambda}(z)\propto z^{\pm\lambda}$ for noninteger order, giving $t^{-1}$ and $t^{2/3}$. The $Y_n$ branch gives the same growing power in the integer case. Therefore \b[a generic perturbation grows as $t^{2/3}$]; an initial condition with zero growing-mode coefficient instead has the decaying $t^{-1}$ mode.
To find the growth scale including its dependence on $\nu$, set $\varepsilon=t^{2/3}h(t)$. The equation becomes
$$
h''+\frac8{3t}h'+\Lambda^2t^{-2\nu-2}h=0.
$$
Substitute $h=1+a\Lambda^2t^{-2\nu}+\cdots$. The leading coefficient equation is $-2\nu(5/3-2\nu)a+1=0$. The <pressure correction to a growing polytropic cosmological mode> is consequently
$$
\varepsilon_+=C t^{2/3}\left[1+\frac{\Lambda^2t^{-2\nu}}{2\nu(5/3-2\nu)}+\cdots\right].
$$
Since $ck/R=\Lambda t^{-\nu-1}$ and $t^{-2}=6\pi G\rho$, the correction becomes small when $(ck/R)^2t^2$ is small compared with $2\nu(5/3-2\nu)$. The factor $2(5/3-2\nu)$ is between two and $10/3$, giving the stated order-of-magnitude onset condition
$$
\boxed{\frac{ck}{R}\lesssim\sqrt{6\pi\nu G\rho}.}
$$
For a well-developed asymptotic growing mode, replace $\lesssim$ by $\ll$. This is a <Jeans instability> interpretation: the physical acoustic frequency must be sufficiently low compared with the gravitational expansion scale, or equivalently the physical wavelength must be sufficiently long for self-gravity to overcome pressure support. It is an approximate condition for the dust-like growing branch, not an exact initial-condition-independent instant at which any perturbation begins increasing. The instantaneous sign change of the coefficient in the original equation instead occurs at $c^2k^2/R^2=4\pi G\rho$.
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