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For dust, curvature and a constant cosmological constant, write $H^2=A{a}^{-3}+B{a}^{-2}+C$. Then $\dot{H}=-3A{a}^{-3}/2-B{a}^{-2}$ and $\ddot{H}=H(9A{a}^{-3}/2+2B{a}^{-2})$, which imply $\ddot{H}+2H\dot{H}-3A{a}^{-3}H/2=0$. Thus $H$ is a solution of the pressureless linear matter perturbation growth equation. In an Einstein-de Sitter universe, $H\propto t^{-1}$ is the decaying mode and the independent growing mode is $t^{2/3}$. Additional dynamical components need not preserve this identity.

To construct the regular open dust linear growth mode, use reduction of order, writing $\delta ={\delta }_{1}u$ and denoting the coefficient of ${\delta }^{\prime }$ by $p(x)$. Since ${\delta }_1$ solves the homogeneous linear differential equation, its terms cancel and leaveThe integrating factor for this first-order equation in $u^{\prime }$ is ${\delta }_1^2{e}^{\int pdx}$, soChoose $C=1$ and the lower integration endpoint zero. This selects the mode regular at early times without adding a multiple of ${\delta }_1$:Its Wronskian with ${\delta }_1$ is ${\delta }_1{\delta }_2^{\prime }-{\delta }_1^{\prime }{\delta }_2=x^{-3/2}(1+x{)}^{-1/2}>0$, proving linear independence.

No explicit antiderivative is needed for the limits. For $x\ll 1$, the integrand is $y^{3/2}[1+O(y)]$, whence its integral is $(2/5)x^{5/2}[1+O(x)]$. For $x\gg 1$, the integrand tends to one, so its integral divided by $x$ tends to one, while $x{\delta }_1\to 1$. ConsequentlyEarly matter domination permits gravitational instability to amplify the density contrast in proportion to the scale factor. Later, the matter density parameter tends to zero and Hubble friction suppresses further amplification. Indeed, for $a\propto t$, neglecting the matter source in the linear matter perturbation growth equation gives $\ddot{\delta }+2\dot{\delta }/t\simeq 0$, whose modes are a constant and $t^{-1}$. These agree with the late limits of the two exact modes.

For pressureless matter with curvature and a possible constant cosmological constant, write the background Friedmann equation asDifferentiating on the expanding branch givesConsequentlyThis proves that the Hubble parameter as a solution of the dust growth equation is valid, including constant-$\Lambda$ and curvature contributions to the background. It does not extend without change to a radiation background or arbitrary dynamical dark energy.

In an Einstein-de Sitter universe, $a\propto t^{2/3}$, $H=2/(3t)$ and $4\pi G\bar{\rho }=2/(3t^2)$. The linear matter perturbation growth equation becomesSubstitute $D=t^p$. The indicial equation is $p(p-1)+4p/3-2/3=0$, with roots $2/3$ and $-1$. ThereforeThe Hubble-parameter solution is the decaying mode in the Einstein-de Sitter universe.

Differentiate the linearized cosmological continuity equation, remembering the derivative of $1/a$:Insert the linear cosmological Euler equation for the peculiar velocity, givingThus the linear matter perturbation growth equation isThe factor $2H$ includes one expansion term from continuity and one from momentum dilution. This equation uses sub-Hubble, pressureless linear perturbations and the stated matter-only source for the gravitational potential.

Change independent variable from proper time to the scale factor. Then $\dot{\delta }=\dot{a}{\delta }_a$ and $\ddot{\delta }={\dot{a}}^2{\delta }_{aa}+\ddot{a}{\delta }_a$. Under radiation domination, $\ddot{a}=-{\dot{a}}^2/a$ and $H^2\simeq 8\pi G{\bar{\rho }}_r/3$. Since ${\bar{\rho }}_m/{\bar{\rho }}_r=a/a_{\mathrm{eq}}$, the linear matter perturbation growth equation becomesPutting $\delta =au$ givesThis keeps the matter self-gravity term in a radiation-dominated background. It is not an exact background equation through radiation-matter equality.

At $a/a_{\mathrm{eq}}\ll 1$, neglecting that small term gives ${\delta }_{aa}+{\delta }_a/a=0$. Integration yieldsThus matter has at most logarithmic growth during this leading radiation approximation, together with a constant independent mode. The constant is non-growing, not a mode that literally falls as $a^{-1}$; that power belongs to $u$ rather than ${\delta }_m$.

For an increasing/decreasing basis of the displayed equation with matter self-gravity retained, put $x=\sqrt{6a/a_{\mathrm{eq}}}$. Its equation becomes $x^2{\delta }_{xx}+x{\delta }_x-x^2\delta =0$. HenceThe Modified Bessel function of the first kind gives $I_0(x)=1+3a/(2a_{\mathrm{eq}})+O((a/a_{\mathrm{eq}}{)}^2)$, which increases slowly. The Modified Bessel function of the second kind gives $K_0(x)=-\frac{1}{2}\ln (a/a_{\mathrm{eq}})+\frac{1}{2}\ln (2/3)-{\gamma }_E+O((a/a_{\mathrm{eq}})|\ln (a/a_{\mathrm{eq}})|)$, where ${\gamma }_E$ is Euler's constant; this mode decreases as $a$ increases. Their leading span is precisely the constant/logarithmic pair above. Mode labels depend on the chosen basis and normalization; no rapid matter-era growth $\delta \propto a$ occurs here. Neglected background corrections can change subleading terms, so the Bessel basis should not be extrapolated through equality.