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For a first-order linear ordinary differential equation $y^{\prime }+p(x)y=q(x)$, an integrating factor is $\exp (\int p(x)dx)$. Here $p(x)=\cot x$, soThe sine is positive on the specified interval, so this integrating factor never vanishes there. Any nonzero constant multiple is equally suitable.

The dependent variable itself does not occur, so reduce order by putting $p=y^{\prime }$. Then $p^{\prime }+2p=3x$, a first-order linear differential equation. Its integrating factor is $e^{2x}$:Integrating once more and renaming the constants givesThe two homogeneous terms are the constant solution and the exponential associated with the roots $0,-2$ of the characteristic equation.

For this first-order linear differential equation, the integrating factor is $e^x$. Thusand the general solution is

Because the original differential equation divides by $xy$, its domain requires $x\ne 0$ and $y\ne 0$. Set $u=y^2$, so $u^{\prime }=2y{y}^{\prime }$. The equation becomes the first-order linear differential equationAn integrating factor is $x^2$, since $(x^{2}u{)}^{\prime }=x^2{u}^{\prime }+2xu$. ThusReturning to $y$ gives the explicit general real solutionsEach sign is fixed on a connected interval where $x\ne 0$ and the radicand is strictly positive. The strict condition excludes $y=0$, where the original differential equation is undefined. There is no additional zero solution.

Where $\sin x\ne 0$, divide by it to obtain $y^{\prime }+2\cot xy=\csc x$. The integrating factor is ${\sin }^{2}x$, and the product rule givesThe initial condition gives $C=1$. HenceThe simplified expression has a differentiable extension through $x=0$, with $y(0)=1/2$. Direct substitution verifies the original undivided differential equation there as well: its left side is $2y(0)=1$. Thus the maximal interval containing $\pi /2$ on which this solution is finite is $\boxed{-\pi <x<\pi }$; the apparent exclusion of $0$ arose only from division by $\sin x$. This is continuation through a zero of the leading ODE coefficient.

On a nonconstant branch with $\sin y\ne 0$, regard $x$ as a function of $y$. The equation becomes the linear ordinary differential equationAn integrating factor is $\sin y$, givingTaking an antiderivative yields the implicit familyOn a regular branch this can also be written $x=\sin y+C/\sin y$. The latter form loses the constant solutions $y=n\pi$, which must be restored separately.

The implicit relation is a first integral of an ordinary differential equation: along any differentiable solution of the original equation,Conversely, a differentiable graph satisfying this relation solves the original equation wherever $\cos y\ne 0$. The original equation is undefined when $\cos y=0$; points at which $2\sin y-x=0$ also need separate attention before solving explicitly for $y^{\prime }$. This is why dividing immediately by $\tan y$ can lose valid solutions and obscure nonuniqueness.

For this first-order linear differential equation, an integrating factor on an interval avoiding zeros of $\cos x$ is $\cos x$. The product rule then givesThe initial condition gives $3/\sqrt{2}=1/\sqrt{2}+C$, hence $C=\sqrt{2}$. ThusThe displayed interval is the maximal one containing the specified initial point, since the coefficient $\tan x$ is undefined at its endpoints.

Multiplying the ordinary differential equation by the nonzero integrating factor from part (i) makes its left-hand side $dF(x,y(x))/dx$. Hence the potential is constant:Since $16+x^2>0$ for every real $x$, this is equivalent to the explicit general solutionIt is defined for all real $x$. There are no lost solutions, since the integrating factor and the coefficient of $y^{\prime }$ are everywhere nonzero.

Let $P=\mu (x)(xy-16x-x^3)$ and $Q=\mu (x)(16+x^2)$. An exact differential form must satisfy $P_y=Q_x$. These partial derivatives giveTaking antiderivatives produces $\ln |\mu |=-\frac{1}{2}\ln (16+x^2)+C$. A convenient nonzero integrating factor for a differential one-form is thereforeAn explicit potential, which also proves global exactness, isIndeed $F_x=P$ and $F_y=Q$ for the chosen integrating factor.

Multiply the differential equation by $-p(x)$ and compare the coefficients of $y^{\prime \prime }$ and $y^{\prime }$ with $-(p{y}^{\prime }{)}^{\prime }=-p{y}^{\prime \prime }-p^{\prime }{y}^{\prime }$. The required integrating factor satisfies $p^{\prime }=-bp$. Thus one may takeThen $-(p{y}^{\prime }{)}^{\prime }+qy=\lambda wy$, the desired Sturm-Liouville form, with the same homogeneous Dirichlet boundary conditions. A common positive constant multiplying $p,q,w$ is immaterial. For the usual regular Sturm-Liouville problem one also assumes real sufficiently regular coefficients and $d>0$, ensuring $p,w>0$.

For a nonzero solution on a branch where the real power is differentiable, set $v=u^{1-\alpha }$. Multiplying the Bernoulli differential equation by $(1-\alpha )u^{-\alpha }$ gives the linear equationIt can be solved by an integrating factor. For the specified cubic equation take $\alpha =3$, $f=g=x$, so $v=u^{-2}$ satisfies $v^{\prime }-2xv=-2x$. Its integrating factor is $e^{-x^2}$, giving $(v{e}^{-x^2}{)}^{\prime }=-2x{e}^{-x^2}$ and hence $v=1+C{e}^{x^2}$. Thus all nonzero real solutions areon connected intervals where the denominator is positive, with a fixed sign on each interval. Also $u\equiv 0$ solves the original equation and must be restored because the substitution excluded it. The choices $C=0$ give the constant solutions $u=\pm 1$. Smoothness of the original right-hand side gives uniqueness, so a nonzero solution cannot cross through the zero solution at a finite regular point.

First move the order parameter term to the right-hand side:Here $p=x^2$, $r=x$, so the integrating factor obeysChoose the irrelevant positive constant $C=1$. Multiplying the equation by $1/x$ givesThus a Sturm-Liouville theory eigenvalue formulation with the requested eigenvalue isThis is the Sturm-Liouville formulation with Bessel order as eigenvalue. The spectral parameter is the squared order of the Bessel function, with $\beta$ treated as fixed. This differs from the also familiar formulation in which ${\beta }^2$ is the eigenvalue.

The factor $1/x$ is singular at the origin. Consequently the integral defining $w$ cannot literally use lower limit zero; take a positive reference point $x_{\ast }>0$, obtain $w=x_{\ast }/x$, and absorb $x_{\ast }$ into the normalization. The differential identity holds for $x>0$. Boundary conditions and admissibility at zero and at the outer endpoint are additional ingredients needed to make a particular self-adjoint spectral problem.

Expand the two operators before comparing coefficients:Equality $L=w\tilde{L}$ requires $a=\tilde{a}$, $c=\tilde{c}$, and $(aw{)}^{\prime }=w(a^{\prime }+\tilde{b})$. Thus $a{w}^{\prime }=\tilde{b}w$, giving the positive integrating factorWith $p=aw>0$, the result is the Sturm-Liouville form $L=-(pd/dx{)}^{\prime }-wc$. The arbitrary positive multiplier $C$ rescales both sides of the eigenvalue problem and does not change its eigenvalues. The coefficient ${\tilde{a}}^{\prime }$ is already present in the divergence term of $\tilde{L}$; it must not be counted a second time in the exponential.

Move the differential terms to the left in the sign convention of part (a). Then $\tilde{a}=1+x$, $\tilde{c}=0$, and $\tilde{b}=(3+2x)-{\tilde{a}}^{\prime }=2(1+x)$. The integrating factor is therefore $w=e^{2x}$, up to a positive constant. The Sturm-Liouville problem isThe trial $y=x{e}^{-x}$ is admissible because $y(0)=0$ and $y^{\prime }=e^{-x}(1-x)$ vanishes at one. Its Rayleigh quotient has numeratorand denominatorConsequently the Rayleigh-Ritz variational principle givesThe numerator is positive for every nonzero admissible function, so the lowest eigenvalue is positive. The trial is not an exact eigenfunction: substituting it would require $x^2+2x-1=\lambda x$ for every $x$. Thus the upper inequality is actually strict. This example underlies the exponential-polynomial trial space for a mixed Sturm-Liouville problem.

The general solution contains both decay of the initial xenon and continued feeding from the initial iodine. Use $t=0$ for removal of the flux, and retain $\varphi$ as the pre-shutdown value in the initial conditions. Neglect post-shutdown nuclear fission and neutron capture. ThenMultiply this first-order linear differential equation by the integrating factor $e^{{\lambda }_{X}t}$ and integrate:With the preceding steady initial conditions, ${\lambda }_I{I}_0=G={\gamma }_I{\Sigma }_f\varphi$ and $X_0=G/({\lambda }_X+\varphi {\sigma }_{aX})$, this isIf ${\lambda }_X+\varphi {\sigma }_{aX}\gg |{\lambda }_X-{\lambda }_I|$, the coefficient of the first exponential term is small compared with the coefficients of the feeding term, yieldingThis approximation neglects the finite initial xenon population and is consequently not a small relative error at $t=0$, where its value is zero while the exact $X_0$ is positive. The exact expression supplies the initial value and satisfies $\dot{X}(0)=G-{\lambda }_X{X}_0$. For equal nuclear decay constants, the continuous limit of the exact solution is $X(t)=(X_0+{\lambda }_I{I}_{0}t)e^{-{\lambda }_{I}t}$.

Use neutron fluence $\Phi$ as the independent variable and abbreviate $a={\sigma }_{a\mathrm{Th}}$, $b={\sigma }_{a\mathrm{U}}$ and $c={\sigma }_{c\mathrm{Th}}$. Substituting the parent solution into the second linear differential equation givesThe integrating factor is $e^{b\Phi }$. Integrating from the initial state yieldsThe first term is survival of the initial population; the second accumulates production at earlier exposures, weighted by subsequent survival. For $b=a$, the continuous limit is $(N_{U0}+c{N}_{Th0}\Phi )e^{-a\Phi }$. The same formula applies to a time-varying prescribed flux through $\Phi (t)$; intervals of zero flux leave both populations unchanged.

For a closed unit-branching radioactive decay chain, the rate equations areThe first gives $N_a=N_{a0}{e}^{-{\lambda }_{a}t}$. Multiply the second by its integrating factor $e^{{\lambda }_{b}t}$:Integrate from zero to $t$. For ${\lambda }_a\ne {\lambda }_b$,soThe coefficients can individually be negative; the physical population is nonnegative because the production integral is nonnegative. For equal nuclear decay constants ${\lambda }_a={\lambda }_b=\lambda$, the finite limit isA sum of two constant-coefficient copies of the same exponential function cannot represent this limit; the $t$ factor is essential.

The metric source is a gravitational frequency shift. photons following null geodesics through a time-dependent spatial metric change momentum relative to the background comoving frame. The directional projection of that metric change gives the ${\mu }^2$ factor. It is not a collision term or continuing photon scattering after recombination. The normalization and sign belong to the specified definition of $h$ and $\Delta$; a distribution perturbation should not silently be identified with temperature without fixing that normalization.

Define $\mathcal{H}=\dot{R}/R$, where dots here mean conformal-time derivatives. In an Einstein-de Sitter background,A power law $h\propto {\tau }^p$ therefore satisfies $p(p-1)+2p=6$, or $(p-2)(p+3)=0$. Thuswith the required growing mode $h=A{\tau }^2$. This is a solution in the perturbation convention used in the question, rather than a statement that every gauge-invariant gravitational potential grows as ${\tau }^2$.

Use an integrating factor for the photon equation:For the growing mode $\stackrel{...}{h}=0$. For the polynomial-source photon transport solution, a particular solution is $B_h=-2i\mu \dot{h}/k+2\ddot{h}/k^2$: direct substitution gives ${\dot{B}}_h+ik\mu B_h=2{\mu }^2\dot{h}$. The initial-value solution is consequentlyThis also follows by integrating the source twice by parts. The apparent inverse powers of $k$ cancel in the complete long-wavelength limit; the integral form remains regular. At $\mu =0$, the source vanishes and the complete expression reduces to $\Delta ={\Delta }_R$.

Physically, the exponential transports the recombination pattern along photon trajectories; its direction-dependent phase projects a Fourier perturbation onto different angular multipoles. The extra terms are the accumulated gravitational contribution, here expressible as endpoint terms because the source has this special time dependence. For endpoint monopole and dipole in photon transport, at the observer the last two terms are respectively a directional dipole and an angular monopole for each mode; they do not supply a new sky pattern at multipoles $\ell \ge 2$. The propagating recombination combination contains the intrinsic perturbation and gravitational/velocity endpoint contributions. Individual terms depend on the perturbation gauge; observable temperature anisotropy requires the consistent emitter and observer contributions.

In pure matter domination the gauge-invariant scalar potentials are constant for a growing linear mode: the matter contrast grows as $R$ while the potential scales as its contrast divided by $R$. Thus growth of this $h$ is compatible with no late Integrated Sachs-Wolfe effect in Einstein-de Sitter. When dark energy suppresses matter growth, the potentials generally evolve; photons gain and lose unequal energies entering and leaving a potential well. In a convention where $\Phi ,\Psi$ are the dimensionless Newtonian-gauge potentials, the integrated temperature source is $\int (\dot{\Phi }+\dot{\Psi })d\tau$ along the light path.

Large-angle cosmic microwave background data can constrain that late contribution, but low-multipole power alone has large cosmic variance and depends on the primordial radiation spectrum, geometry and foreground treatment. Cross-correlating the temperature map with low-redshift galaxies or other matter tracers isolates a late-time connection; decaying wells typically correlate overdensities with hotter cosmic microwave background directions. Boughn and Crittenden compared WMAP temperature data with the HEAO1 hard X-ray background and examined possible foreground origins of the correlation. This tests potential evolution under a cosmological model, not dark energy independently of all gravity and curvature assumptions.^[ref]

Put $\delta =\sqrt{ϵ}$ and write the outer asymptotic expansion as $y_0+\delta y_{1/2}+{\delta }^2{y}_1+\cdots$. Equating powers in the differential equation givesThe first two equations give $y_0=A{e}^{x^2/2}$ and $y_{1/2}=B{e}^{x^2/2}$. Since $y_0^{(3)}=A(x^3+3x)e^{x^2/2}$, an integrating factor yieldsHere $A,B,C$ are matching constants; boundary conditions have not yet been imposed.

Take the upward velocity as positive during ascent. Newton's second law with linear drag givesMultiplication by the integrating factor $e^{kt}$ and integration giveThe maximum height occurs at the first zero of this velocity, henceHere $k$ has units of inverse time. The limit $k\to 0$ gives $v_0/g$, and positive drag shortens the ascent relative to motion in a vacuum. The expression applies only up to the turning point; the drag force opposes the subsequent downward motion.