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For the normalized second-order linear differential equation $y^{\prime \prime }+P(z)y^{\prime }+Q(z)y=0$, an ordinary point $z_0$ has analytic $P,Q$ near $z_0$. At a regular singular point, which is not ordinary, $(z-z_0)P(z)$ and $(z-z_0{)}^{2}Q(z)$ are analytic. Ordinary points admit two independent convergent power series solutions. Regular singular points instead call for the Frobenius method, which permits leading powers and, in some cases, logarithms.

Here zero is an ordinary point. Substitute $y={\sum }_{n\ge 0}{a}_n{z}^n$ into the Hermite differential equation and equate coefficients:The even and odd initial coefficients are independent, givingThese are the power series solutions of the Hermite differential equation, with the present parameter equal to twice the parameter in that article. The ratio of successive terms within each nonterminating parity is asymptotically $2|z{|}^2/n$, so both series have infinite radius of convergence.

For a nonzero polynomial of degree $m$, its highest coefficient in $-2z{y}^{\prime }+\lambda y$ forces $\lambda =2m$. Conversely at that value the matching parity terminates. HenceThe polynomial solutions are scalar multiples of the Hermite polynomials. Two examples are $\boxed{\lambda =0:y=1;\lambda =2:y=z}$; direct substitution verifies both. The other independent solution generally remains nonpolynomial.

Define the Wronskian by $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$. Differentiating and using the normal-form differential equation givesIntegrating this first-order linear differential equation proves Abel's identity:Thus, on any interval avoiding zero and $m$,Absolute-value signs from real logarithms can be absorbed into the constant on each such interval. For linearly independent solutions, $C\ne 0$.

The homogeneous linear differential equation can be written in divergence form:One integration gives $u^{\prime }=C{x}^{-n-1}$. Therefore the two cases areIn each case the displayed functions have linear independence throughout the interval. Imposing both homogeneous Dirichlet boundary conditions forces $A=B=0$: for $n\ne 0$, $2^n\ne 1$, and for $n=0$, $\log 2\ne 0$. This also establishes uniqueness for the inhomogeneous boundary value problem.

Away from $y=\pm 1$, divide by the coefficient of $x^{\prime \prime }$ to obtain the normal formThus $\boxed{y=1\text{ and }y=-1}$ are the only finite singular points. For a second-order linear differential equation, $y_0$ is a regular singular point if $(y-y_0)P(y)$ and $(y-y_0{)}^{2}Q(y)$ extend analytically to $y_0$. At $y_0=1$ these functions areand at $y_0=-1$ they areEach pair is analytic at its respective point. Therefore $\boxed{\text{both finite singular points are regular}}$.

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.

Choose upward velocity positive during ascent. Gravity and linear drag both point downward, giving the linear differential equationAt the highest point $v=0$, soThe vertical ascent under linear drag time is smaller than $v_0/g$ for $\alpha >0$, because $\log (1+x)<x$. Its zero-drag limit is $v_0/g$, as required by constant-gravity projectile motion.

In vacuum, Maxwell's equations in differential form areHere $\rho$ is charge density, $\mathbf{J}$ is electric current density, $\mathbf{E}$ is electric field, and $\mathbf{B}$ is magnetic flux density. In a source-free region, taking the curl of the third equation and using $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$ gives the wave equationThe same argument gives the wave equation for $\mathbf{B}$. A concrete family satisfying all four Maxwell's equations isThe function $F$ may be any twice differentiable profile. These are transverse travelling waves; reverse the magnetic field direction for propagation in the opposite direction. Because the source-free Maxwell's equations are homogeneous linear differential equations, any finite linear combination of such paired electric fields and magnetic fields is again a solution. Infinite superpositions also work whenever the required derivatives converge. A solution of the component wave equations alone must still satisfy the divergence and mutual-curl constraints.

Put $\vartheta =T-T_0$ and assume $C>0$, $\lambda >0$. The linear differential equation for a constant forcing step isMultiplying by the integrating factor $e^{\lambda t/C}$ and integrating from zero to $t$ givesThe response initially grows at rate $F_0/C$ and tends to the equilibrium warming $F_0/\lambda$. Heat capacity per unit area and radiative-feedback units make $k_T$ an inverse time.

The two external stream velocities imposeLet $r=\Delta U/U_c\ll 1$ and seek $F=O(r)$ over $\xi =O(1)$. The nonlinear term $F{F}^{\prime \prime }$ is then $O(r^2)$, so the leading ordinary differential equation isFor $G=F^{\prime \prime }$, this is the first-order linear differential equation $G^{\prime }+\xi G=0$. Its integrating factor is $e^{{\xi }^2/2}$, giving $F^{\prime \prime }=M{e}^{-{\xi }^2/2}$. An integration using the error function givesThe two far-field boundary conditions require $E=0$ and $M=r/\sqrt{2\pi }$. A second integration usesConsequentlyThe relation between $A$ and $B$ cancels the extra $\xi e^{-{\xi }^2/2}$ term when differentiating. It gives the familiar error function mean velocity profileto first order in $r$. The additive constant $D=O(r)$ does not affect this velocity profile at the retained order; its interpretation comes from the transverse velocity.

A periodic solution is a solution of a differential equation which is a periodic function: for some $T>0$, $x(t+T)=x(t)$ at every time. This demands repetition of the state, rather than merely repetition of the coefficients in the differential equation. A nonconstant periodic solution of an autonomous differential equation traces a periodic orbit; constant equilibrium points are also periodic functions, so they must be excluded when asserting absence of periodic orbits.

For an explicit example, consider the initial value problemIts coefficient has period $T=2\pi$. Taking a logarithmic derivative and then applying integration givesDirect differentiation verifies both the differential equation and the initial condition. After one coefficient period,Thus the repeating coefficient generally produces a repeating shape with a changed amplitude. With $b=1$, the three cases $a=-0.2,0,+0.2$ have one-period factors approximately $0.2846,1,3.5136$, respectively. Only $a=0$ gives a periodic solution with this initial condition. For $a<0$ the positive solution of a differential equation tends to zero, and for $a>0$ it tends to infinity; neither can have any positive period. The zero solution of a differential equation, obtained with a different initial condition, is periodic for every $a$.

The exponential function separates two roles:For $b=1$, even the decaying case $a=-0.2$ initially rises, because $y^{\prime }(0)=0.8>0$. The growing case $a=0.2$ has intervals of decline, because $a+\cos t$ becomes negative. This is the distinction between the instantaneous logarithmic derivative $y^{\prime }/y$ and the Floquet growth rate measured over many periods.

More generally, for a real continuous periodic function $q$ of period $T$, the scalar linear differential equation $y^{\prime }=q(t)y$ hasA nonzero solution of a differential equation is $T$-periodic exactly when $m=1$, or equivalently when the coefficient has zero integral over a period. This scalar calculation is the simplest example of a Floquet multiplier.^[ref]

A point interaction concentrates a coefficient of a linear differential equation at a single interior point. On an interval $0<x<\ell$, with $0<c<\ell$, consider $L_{\alpha }y=y^{\prime \prime }+\alpha y(c){\delta }_c$. The domain consists of continuous functions that are twice differentiable away from $c$, with the required one-sided derivatives. Multiplication of the Dirac delta by $y$ means the unambiguous point value $y(c){\delta }_c$. It does not mean an undefined product of a delta with an arbitrary discontinuous function.

The equation is interpreted using distributional derivatives. Smooth forcing gives an ordinary equation on each side, with transmission conditions for a delta coefficient at the interaction. A delta potential in the one-dimensional Schrödinger equation is a physical example after choosing the corresponding sign and constants.

For a complete-market investment-consumption problem with constant relative risk aversion utility, the nonlinear wealth-homogeneity coefficient equation may contain $f^{\prime 2}/f$. Writing $f=g^R$ cancels this gradient square against the one from $f^{\prime \prime }$. In the index-driven correlation model the result is ${\sigma }_0^2{g}^{\prime \prime }/2+B{g}^{\prime }-\delta (m)g+1=0$, a linear differential equation; the positive economic solution gives consumption $w/g$.

For a constant-coefficient linear differential equation $P(D)y=e^{rx}$, where $D$ denotes differentiation, a simple root $P(r)=0$ forces a resonant trial. Factor $P(z)=(z-r)Q(z)$; since $(D-r)(x{e}^{rx})=e^{rx}$, commuting the differential operators gives $P(D)(x{e}^{rx})=Q(r)e^{rx}=P^{\prime }(r)e^{rx}$. Thus the displayed expression is a particular solution. More generally, if the root has multiplicity $m$, the same factorization gives $P(D)(x^m{e}^{rx})=P^{(m)}(r)e^{rx}$, so $y_p=x^m{e}^{rx}/P^{(m)}(r)$. This is the extra polynomial factor required by resonance in a differential equation.

With $s=\sqrt{z}$ and $Y(s)=y(s^2)$, the displayed linear differential equation becomes $Y_{ss}+Y=0$ away from zero. Its Frobenius solutions have exponents $0$ and $1/2$: $\cos \sqrt{z}={\sum }_{n\ge 0}(-1{)}^n{z}^n/(2n)!$ and $\sin \sqrt{z}=z^{1/2}{\sum }_{n\ge 0}(-1{)}^n{z}^n/(2n+1)!$. The first is an entire function of $z$; the second needs a local square-root branch. This illustrates how a regular singular point can arise from a degenerate change of independent variable.

The symmetric part of a real square matrix is $H=(A+A^T)/2$. It controls the quadratic form because $x^{T}Ax=x^{T}Hx$; the antisymmetric part contributes zero. In a linear differential equation $x^{\prime }=Ax$, it determines the instantaneous derivative of the squared Euclidean norm.

A simplified collisional cascade tracks large objects $S$ and small fragments $F$. A fragment can disrupt a large object and multiply, while fragment–fragment collisions remove material. With $\dot{S}=-\kappa SF$ and $\dot{F}=\alpha SF-\beta F^2$, elimination of time gives a linear differential equation, $dF/dS-(\beta /\kappa )F/S=-\alpha /\kappa$. It has solution $F=C{S}^k-(\alpha /\kappa )S/(1-k)$ for $k=\beta /\kappa \ne 1$. Its maximum occurs at $F=(\alpha /\beta )S$; early exponential growth does not imply permanent exponential growth after the large-object reservoir is depleted.

A solution is unique when any two objects satisfying the stated conditions are equal. This is an at-most-one assertion; existence is a separate requirement. A common proof assumes two solutions, subtracts their equations, and proves that their difference vanishes. For a linear differential equation this turns the task into a homogeneous problem with zero boundary or initial data.