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In a spatially flat universe containing separately conserved pressureless matter and a coasting fluid with $w=-1/3$, the ratio $\eta ={\rho }_s/{\rho }_m$ is proportional to the scale factor. If the coasting component is modeled by the barotropic fluid perturbation equations with no anisotropic stress, its contribution to the trace gravitational source vanishes because $1+3w=0$. The matter growing mode isA decaying solution is $D_{-}=\sqrt{1+\eta }/{\eta }^{3/2}$. The reduction of order formula with integrating factor ${\eta }^{3/2}\sqrt{1+\eta }$ produces the integral solution. It has $D=\eta [1-4\eta /7+O({\eta }^2)]$ during matter domination and tends to $5/2$ during coasting-fluid domination. Thus growth freezes rather than continuing in proportion to the scale factor. The background geometry is flat despite the same background expansion law as an open dust universe.

For pressureless homologous spherical collapse, let $\varphi$ be its cycloid parameter. The density contrast equation becomes ${\delta }_{\varphi \varphi }-\tan (\varphi /2){\delta }_{\varphi }-3\delta /(1+\cos \varphi )=0$. One solution is ${\delta }_g=\sin \varphi /(1+\cos \varphi {)}^2$. Reduction of order gives an independent solution that decays at collapse,For $ϵ=\pi -\varphi$, the integral is ${ϵ}^5/40+O({ϵ}^7)$, while ${\delta }_g\sim 4{ϵ}^{-3}$. Therefore ${\delta }_d\sim {ϵ}^2/10$. Since $R\sim R_0{ϵ}^2/4$, the displayed powers follow. The growing branch is unique up to scale and addition of the decaying branch; many mixed solutions grow, so it is not literally the only function with growth. Finite perturbations eventually leave the linear regime.

Suppose $y^{\prime \prime }-P(x)y^{\prime }/x-Q(x)y/x=0$ has analytic $P,Q$ near zero, $P(x)=P_0+P_{1}x+\cdots$, and a solution $y_1=1+\beta x+\cdots$, with $P_0$ a positive integer. The indicial roots are $0,P_0+1$. The Abel identity gives $W=C{x}^{P_0}(1+P_{1}x+\cdots )$. Dividing by $y_1^2$ gives $C{x}^{P_0}[1+(P_1-2\beta )x+\cdots ]$. Integrating this reduction of order expression from zero, multiplying by $y_1$, and choosing $C=P_0+1$ proves the displayed normalized expansion. The integral's zero constant eliminates the added multiple of $y_1$. Its nonzero Wronskian proves independence. In this setting the assumed exponent-zero solution prevents a logarithm in the constructed exponent-$P_0+1$ solution.

For the Legendre differential equation $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }+n(n+1)y=0$, put $t=1+x$. The equation becomes $t(2-t)y_{tt}+2(1-t)y_t+n(n+1)y=0$. Its indicial equation is $2r^2=0$. The exponent-zero power series $\sum a_k{t}^k$ obeys the displayed linear recurrence relation. For a nonnegative integer $n$, it terminates at degree $n$, giving a Legendre polynomial with $a_0\ne 0$. Every analytic solution at the endpoint is proportional to this one, since the recurrence determines all coefficients from $a_0$.

The second independent solution has the local form $P_n(x)\log (1+x)+B(x)$, where $B$ is analytic near $-1$. Indeed, reduction of order gives $P_n(x)\int dx/[(1-x^2)P_n(x{)}^2]$, whose integrand is $1/[2P_n(-1{)}^2(1+x)]$ plus an analytic function. Rescale the second solution to obtain the unit coefficient of $P_n\log (1+x)$. Reflection gives the corresponding expansion at $x=1$. This also proves the logarithmic divergence of the independent Legendre function of the second kind at each endpoint.

For $a>0$, $b\ne 0$, and a positive integer $\beta$, consider $x^2{y}^{\prime \prime }+x{y}^{\prime }+(-a^2+b{x}^{\beta })y=0$ near $x=0$. Its indicial roots are $a,-a$. The first Frobenius solution, normalized to leading coefficient one, is $y_{+}=x^a{\sum }_{k\ge 0}{c}_{k\beta }{x}^{k\beta }$, with $c_{k\beta }=-b{c}_{(k-1)\beta }/[k\beta (k\beta +2a)]$. If $N=2a$ is a positive integer, a second pure Frobenius solution with leading coefficient one would requireBefore $n=N$, induction makes $d_n=0$ for indices not divisible by $\beta$. If $\beta$ divides $N$, the coefficient $d_{N-\beta }$ is nonzero, and the exceptional equation $0=-b{d}_{N-\beta }$ is impossible. If $\beta$ does not divide $N$, the right-hand side is zero and $d_N$ is free. Thus, in the Frobenius case classification, a positive integer gap gives Case IIIa exactly when $\beta$ divides $2a$, and Case IIIb otherwise.

The coefficient of the necessary logarithm can also be calculated. Let $L=x^2{d}^2/d{x}^2+xd/dx-a^2+b{x}^{\beta }$. The product rule gives $L(y_{+}\log x)=2x{y}_{+}^{\prime }$. Therefore a second solution $x^{-a}\sum d_n{x}^n+K{y}_{+}\log x$, with $d_0=1$, hasFor $N=m\beta$ the preceding recurrence relation gives $d_{(m-1)\beta }=b^{m-1}/[{\beta }^{2m-2}((m-1)!{)}^2]$, soThis proves that the logarithm cannot disappear for a special nonzero value of $b$. For $a=0$, reduction of order instead produces the repeated-root Logarithmic Frobenius solution directly.

For nonnegative integer $n$, a second solution of the Legendre differential equation is obtained by reduction of order from $P_n$ as $P_n(x)\int dx/[(1-x^2)P_n(x{)}^2]$. Normalize it as $Q_0=\frac{1}{2}\log |(1+x)/(1-x)|$ and $Q_1=x{Q}_0-1$, with the usual Legendre polynomial recurrence relation for higher degrees. Inside $(-1,1)$, $Q_n\sim -\frac{1}{2}\log (1-x)$ near $1$ and $Q_n\sim (-1{)}^n\frac{1}{2}\log (1+x)$ near $-1$. These follow from $P_n(1)=1$, $P_n(-1)=(-1{)}^n$ in the integral. Poles in the integral at interior polynomial zeros cancel in the product, by reduction of order across a zero of the known solution. On $x>1$, the integral from infinity defines the decaying normalization $Q_n(x)\sim 2^n(n!{)}^2{x}^{-n-1}/(2n+1)!$. For $n\ge 1$ all bounded exterior solutions decay this way; for $n=0$ an additional constant solution remains bounded.

For pressureless matter in a Friedmann-Lemaître-Robertson-Walker metric with negative spatial curvature of an FLRW universe and no cosmological constant, set $x=(1-{\Omega }_{m0})a/{\Omega }_{m0}>0$. A decaying linear growth factor is $D_{-}(x)=\sqrt{(1+x)/x^3}$. Reduction of order gives the regular growing linear growth factorIts Wronskian with $D_{-}$ is $x^{-3/2}(1+x{)}^{-1/2}$, so the modes are independent. The limits are $D_{+}\sim 2x/5$ during matter domination and $D_{+}\to 1$ in a curvature-dominated universe. Multiplying by $5/2$ normalizes the early mode to $x$. Negative spatial curvature of an FLRW universe thus stops late linear amplification even though the model contains no dark energy.

For $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$, a singular point at zero is regular when $xp(x)$ and $x^{2}q(x)$ extend analytically to zero. Hereso these products are analytic at zero. The equation remains singular there because $q$ has a pole.

Substitute $y={\sum }_{n\ge 0}{a}_n{x}^{n+\sigma }$. The lowest power gives the indicial equation $\sigma (\sigma -1+\lambda )=0$. Comparing the next powers givesFor $0<\lambda <1$ the denominators do not vanish. With $\sigma =0$, $a_1=-a_0/\lambda$ and $a_2=0$, soWith $\sigma =1-\lambda$, the recurrence becomesConsequently $y_2=a_0{x}^{1-\lambda }(1-x{)}^{\lambda }$ on $0<x<1$, with the binomial series for its last factor. The nonzero coefficient ratios tend to one, giving radius one for that power-series factor. The factor $x^{1-\lambda }$ specifies the Frobenius branch; the different leading powers prove independence.

When $\lambda =1$, both indicial roots become zero and the analytic Frobenius solution in both cases is a multiple of $1-x$. To obtain the missing solution use reduction of order. Away from zero and one, $p=1/x$, henceThus the real general solution on any interval avoiding the singular points isIts Wronskian with $1-x$ is $1/x\ne 0$, proving independence. For complex solutions choose a logarithm branch on a suitable simply connected region.

For conserved nonrelativistic dark matter and radiation, ${\rho }_{\mathrm{DM}}\propto a^{-3}$ and ${\rho }_{\gamma }\propto a^{-4}$, so $y={\rho }_{\mathrm{DM}}/{\rho }_{\gamma }=a/a_{\mathrm{eq}}$ obeys $\dot{y}=yH$. Denote differentiation with respect to $y$ by a prime. ThenThe supplied Friedmann equations implySubstitution in the linear growth equation and division by $y^2{H}^2$ give the Mészáros equationFor $D_1=1+3y/2$, $D_1^{\prime }=3/2$ and $D_1^{\prime \prime }=0$, making the two remaining terms cancel exactly. For reduction of order, put $\delta =f{D}_1$ and $P=(2+3y)/[2y(1+y)]=1/y+1/[2(1+y)]$. The equation reduces to $f^{\prime \prime }+(2D_1^{\prime }/D_1+P)f^{\prime }=0$, whenceUsing the supplied antiderivative and reversing an irrelevant overall sign gives the independent solutionIts Wronskian is $D_1{D}_2^{\prime }-D_1^{\prime }{D}_2=-1/[y\sqrt{1+y}]\ne 0$. The asymptotic expansions areThus subhorizon dark matter during radiation domination has constant and logarithmic modes, rather than the superhorizon $a^2$ growth of b. A combination with negative $A_2$ supplies positive logarithmic growth. During matter domination, the growing mode is proportional to $a$, agreeing with b, while the second mode decays as $a^{-3/2}$. These results neglect subhorizon radiation clustering; the Mészáros equation is the approximation stated in the question, not a full photon–matter perturbation system through horizon entry.

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 a putative second Frobenius solution with the smaller indicial root $-a$, set $y_{-}=x^{-a}{\sum }_{n\ge 0}{d}_n{x}^n$, with $d_0=1$ and $d_j=0$ for $j<0$. Its recurrence relation isThe Frobenius case classification concerns the gap $2a$ between the indicial roots. In the course convention, Case IIIa is the obstructed recurrence requiring a Logarithmic Frobenius solution; Case IIIb permits two pure Frobenius series.

If $a>0$ and $2a\notin \mathbb{Z}$, this is Case I: no denominator vanishes. There are two linearly independent pure Frobenius solutions, withTheir different leading powers establish linear independence.

If $a=0$, this is Case II, with a repeated indicial root. There is only one pure Frobenius series with a nonzero leading coefficient. To see the second solution's form, reduction of order givesHere $y_{+}(t)=1+O(t^{\beta })$, so the integrand is $t^{-1}+O(t^{\beta -1})$. ConsequentlyThe logarithm has a nonzero coefficient and cannot be removed by adding a multiple of $y_{+}$.

If $N=2a$ is a positive integer, the only potentially obstructed coefficient is $n=N$. Below that index, $d_n=0$ unless $\beta$ divides $n$. At the exceptional index the equation readsIf $N=m\beta$ for some positive integer $m$, all coefficients $d_0,d_{\beta },\dots ,d_{(m-1)\beta }$ are nonzero because $b\ne 0$. The exceptional equation is inconsistent. This is Case IIIa, and the independent second solution has the formThe Frobenius resonance for an Euler equation with monomial potential explains why divisibility, rather than the integer gap alone, decides this outcome. For example, with $d_0=1$, applying the original differential operator to $y_{+}\log x$ gives $2x{y}_{+}^{\prime }$. Cancellation of the coefficient of $x^a$ requires $K=-b{d}_{N-\beta }/(2a)\ne 0$.

If $\beta$ does not divide $N$, then $d_{N-\beta }=0$; this includes $N<\beta$ through the negative-index convention. The exceptional equation is $0=0$, so $d_N$ is free. This is Case IIIb: set $d_N=0$ to select an independent pure Frobenius series starting with $x^{-a}$. Subsequent coefficients again follow the recurrence relation. Changing $d_N$ simply adds a multiple of $y_{+}$, the undetermined coefficient at Frobenius resonance.

The complete answer is thereforeIn particular, for $\beta =1$ every positive integer gap gives Case IIIa. For $\beta =2$, positive integer $a$ gives Case IIIa, while $a=1/2,3/2,5/2,\dots$ gives Case IIIb; the remaining positive $a$ give Case I. The condition $b\ne 0$ matters: at $b=0$ the two Cauchy-Euler equation powers already give independent solutions whenever $a>0$.

Use $t=x+1$. The Legendre differential equation becomesSubstituting a Frobenius series $y=t^r{\sum }_{k\ge 0}{a}_k{t}^k$, $a_0\ne 0$, gives the indicial equation $2r^2=0$: the indicial root zero is repeated. For $r=0$, the coefficient of $t^k$ yieldsorFor $0\le k<n$ the multiplying factor is nonzero. At $k=n$ it vanishes, so $a_{n+1}$ and all later coefficients are zero. Thus one solution is a polynomial of exact polynomial degree $n$, including a nonzero constant when $n=0$. Denote a chosen normalization by $P_n$.

The linear recurrence relation determines every analytic coefficient from $a_0$, so there cannot be a second independent analytic solution at $t=0$. To see the required logarithm explicitly, reduction of order gives the second solution locally asNear $t=0$, the integrand is $1/(2a_0^{2}t)$ plus an analytic function. After rescaling the second solution, its form is thereforeThe analytic remainder converges locally; its constant term may be set to zero by adding a multiple of $P_n$. The nonzero logarithmic coefficient makes $y_2$ linearly independent of $P_n$. This Frobenius expansion of the Legendre equation at an endpoint gives a Logarithmic Frobenius solution; the same endpoint calculation at $x=1$ explains why the independent Legendre function of the second kind is excluded when boundedness at both ends is required.

Where $y_1$ is nonzero, the Wronskian identity $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$ becomes the first-order equationIts integrating factor is $1/y_1$, giving $(y_2/y_1{)}^{\prime }=W/y_1^2$. Integration proves the reduction of order formulaTaking $C_0=0$ gives the displayed representative in the question, with $y_2(x_0)=0$ when $y_1(x_0)\ne 0$. Changing the lower limit adds a multiple of $y_1$ and does not change the linearly independent solution modulo the known one. The scale of $W$ separately fixes the normalization of $y_2$.

Differentiate the determinant and use the original second-order equation for each solution:This proves Abel's identity $W^{\prime }+pW=0$ rather than merely quoting it.

For the specific equation, substituting $y_1=1-x$, $y_1^{\prime }=-1$, $y_1^{\prime \prime }=0$ leaves $(1-x^2)-(1+x)(1-x)=0$. Away from zero its normalized first-derivative coefficient is $p(x)=x-1/x$, soAlthough the equation is singular at zero, this expression extends analytically there. Near zero, $y_1$ is nonzero and the reduction of order integrand is regular. With the prescribed lower limit,The integrand through degree three isAfter integrating and multiplying by $1-x$,The condition $y_2^{\prime \prime }(0)=1$ fixes $C=1$, and the lower limit already gives $y_2(0)=0$. Therefore the first three nonzero terms areThe representation genuinely solves the second-order equation: writing $L[y]=y^{\prime \prime }+p{y}^{\prime }+qy$, differentiating its Wronskian gives $W^{\prime }+pW=y_{1}L[y_2]-y_{2}L[y_1]=y_{1}L[y_2]$. Its left-hand side is zero and $y_1\ne 0$ near zero. Analyticity then extends the original unnormalized equation to zero itself. Its Wronskian with $y_1$ is nonzero for sufficiently small $x\ne 0$, proving linear independence. The vanishing Wronskian at the singular point zero does not contradict linear independence on a regular interval. The integral representation is used locally near zero, before the zero of $y_1$ at one.

Write $H=\dot{R}/R$. A fluid element in the homologous spherical flow obeys $\dot{\mathbf{r}}=H\mathbf{r}$, so $\mathbf{r}/R$ remains constant. Also $\nabla \cdot {\mathbf{u}}_0=3H$. Along every element, the continuity equation therefore gives $D{\rho }_0/Dt=-3H{\rho }_0$. Every element starts with the same mass density, and hence has the same mass density at subsequent times:This proves preservation of uniformity, rather than assuming it when conserving the total mass.

By the shell theorem, only the mass interior to radius $r$ contributes to its gravitational acceleration. That mass is $4\pi {\rho }_0{r}^3/3$. The same result follows directly from the Poisson equation for Newtonian gravity: spherical symmetry gives $(r^2{\Phi }_r{)}_r=4\pi G{\rho }_0{r}^2$, and regularity at the centre sets the integration constant to zero. Hence ${\Phi }_r=4\pi G{\rho }_{0}r/3$ andThe material derivative of the background velocity is $(\dot{H}+H^2)\mathbf{r}=(\ddot{R}/R)\mathbf{r}$. There is no pressure force, and the momentum equation consequently reduces toMultiplication by $\dot{R}$ and integration, using collapse from rest, givesChoose the negative square root for contraction. Put $C^2=2K/R_0$ and $R=R_0{\cos }^2(\varphi /2)$. For $0<\varphi <\pi$, the energy integral then gives $\dot{R}=-C\tan (\varphi /2)$, while $dR/d\varphi =-R_0\sin \varphi /2$. Their ratio isIntegrating from the initial turning point proves the cycloid parametrization of pressureless homologous spherical collapse:It covers the entire contracting branch, with complete collapse at $\varphi =\pi$ and $t_c=\sqrt{3\pi /(32G{\rho }_{00})}$. At $R=0$ the uniform dust configuration is singular, so the smooth calculation is restricted to $t<t_c$.

Each Fourier mode among the comoving perturbations of a homologously collapsing cloud has its phase fixed on each fluid element. In fact, for $\mathbf{k}=\mathbf{q}/R$,Its physical wavelength contracts in proportion to $R$, while its wavelength in the comoving coordinate $\mathbf{r}/R$ stays fixed. This is the physical reason to use a time-dependent wavenumber; holding the physical wavenumber fixed would not follow the contracting pattern. Boundary effects are being excluded in this bulk Fourier mode calculation.

Let ${\rho }^{\prime }={\rho }_1(t)e^{i\mathbf{k}\cdot \mathbf{r}}$ and let the perturbed velocity and Newtonian gravitational potential have amplitudes ${\mathbf{u}}_1(t)$ and ${\Phi }_1(t)$. Linearizing the continuity equation, the uniform background eliminates ${\mathbf{u}}_1\cdot \nabla {\rho }_0$. The advected phase eliminates its own time derivative, leavingFor the density contrast $\delta ={\rho }_1/{\rho }_0$, use ${\dot{\rho }}_0=-3H{\rho }_0$ to obtain $\dot{\delta }=-i\mathbf{k}\cdot {\mathbf{u}}_1$. The linear momentum equation and Poisson equation for Newtonian gravity areDifferentiating the density contrast equation must include $\dot{\mathbf{k}}=-H\mathbf{k}$:Thus the compressive Fourier mode with $k\ne 0$ satisfiesThe pressureless approximation removes the usual acoustic restoring term of Jeans instability. A transverse velocity perturbation instead obeys ${\dot{\mathbf{u}}}_{1\perp }+H{\mathbf{u}}_{1\perp }=0$ and does not drive this density contrast.

Since $d/dt=(C/R)d/d\varphi$, the density contrast equation becomesFor a direct verification, put $s=\tan (\varphi /2)$. As $d/d\varphi =(1+s^2)d/(2ds)$, the first-derivative terms cancel, reducing the equation toThe function ${\delta }_g=s(1+s^2)/2$ has second derivative $3s$ and satisfies this equation. Transforming back gives the required growing branch:To establish what the other solution does, use reduction of order, not just the growth of this example. The coefficient of ${\delta }_{\varphi }$ is $d\log R/d\varphi$, so an independent solution is proportional to ${\delta }_g\int d\varphi /(R{\delta }_g^2)$. Choose the integration endpoint at complete collapse to select the decaying branch:For example, evaluating the integral with $a=\tan (s/2)$ gives integrand $4/[a^2(1+a^2{)}^2]$ and antiderivative $-4/a-6\arctan a-2a/(1+a^2)$. Its endpoint value gives the displayed expression. Its Wronskian with ${\delta }_g$ is $-(1+\cos \varphi {)}^{-1}$, so the two solutions are independent on $0<\varphi <\pi$.

For $ϵ=\pi -\varphi \to 0$, $R\sim R_0{ϵ}^2/4$, ${\delta }_g\sim 4{ϵ}^{-3}$ and the integral in ${\delta }_d$ is ${ϵ}^5/40+O({ϵ}^7)$. Consequently the two density modes of a collapsing pressureless sphere haveEquivalently, with $\tau =t_c-t$, the local powers are ${\tau }^{-1}$ and ${\tau }^{2/3}$. There is one independent growing branch: every solution is $A{\delta }_g+B{\delta }_d$, and only the branch with $A=0$ decays at collapse. The uniqueness is up to normalization and addition of the decaying solution; it does not mean that only one literal function can grow. A fixed nonzero initial perturbation will eventually violate $|\delta |\ll 1$, so the divergent linear result is not a nonlinear solution through collapse.

The origin is an ordinary point, since the coefficients of the second-order linear differential equation are polynomials. Seek a power series $y={\sum }_{n\ge 0}{a}_n{x}^n$. Matching the coefficients of $x^0$ and $x^1$ gives $a_2=a_3=0$. Matching the coefficient of $x^{j+2}$ givesThus the four residue classes of indices modulo four evolve separately. The initial conditions for the first solution give $a_0=1$, $a_1=0$, and henceThe second set of initial conditions gives $a_0=0$, $a_1=1$, andAn empty product equals one. In either power series, the ratio of successive nonzero terms tends to zero for every fixed $x$: the coefficient ratio is of order $1/m$, while the power increases by four. The ratio test therefore proves convergence for every $x$, and termwise differentiation verifies the differential equation. These normalized solutions are linearly independent.

For their Wronskian $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$, differentiating and substituting the differential equation givesThus the Abel identity yieldsOn the interval containing zero where $y_1$ has no zero,Since $y_2(0)/y_1(0)=0$, integration proves the requested reduction of order formula:This is a local formula around the origin, as requested. The power series define both solutions globally; an ordinary integral through a zero of $y_1$ is not intended. Reduction of order across a zero of the known solution instead uses continuation of the actual solution.

On an interval where $p,q$ are continuous, define the WronskianUsing the two second-order linear differential equations to eliminate the second derivatives givesThe integrating factor yields the Abel identityThus the Wronskian either vanishes identically or is nonzero everywhere on the interval. If $u_1,u_2$ are linearly dependent, it is identically zero. Conversely, if $W(x_0)=0$, the two initial-data vectors are dependent, so some nontrivial linear combination has both value and first derivative zero at $x_0$. The uniqueness theorem for ordinary differential equations makes that combination identically zero. Hence the solutions are linearly independent exactly when their Wronskian is nonzero at one, and therefore every, point. Continuity and the common differential equation are essential to this converse.