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For the homogeneous linear ordinary differential equation ${\sum }_{j=0}^n{a}_j{y}^{(j)}=0$, substitute $y=e^{rx}$. Dividing by the nonzero exponential gives the polynomial equation $P(r)={\sum }_j{a}_j{r}^j=0$. A simple root gives an exponential solution; a root of multiplicity $m$ gives $x^k{e}^{rx}$ for $0\le k<m$. Real equations admit real sine and cosine combinations of conjugate roots. These characteristic roots determine growth, decay and oscillation of the linearized modes.

For the linear ordinary differential equation $c^{\prime }=Ac-b$, a fixed left null space vector gives $(w^{T}c{)}^{\prime }=-w^{T}b$. Thus $w^{T}c$ is a conserved quantity when $w^{T}b=0$. In a box model written for concentrations per unit volume, physical tracer inventory uses the box volumes as weights. Zero column sums make $w=1$ possible, but that property must be checked rather than assumed.

A linear ordinary differential equation has constant coefficients when the multipliers $a_j$ of the unknown function and its derivatives do not depend on the independent variable. The forcing $f$ may still vary. For the homogeneous equation $f=0$, its characteristic equation of a constant-coefficient differential equation determines exponential solutions.

A connection on a vector bundle pulls back along a smooth curve $\gamma$ to differentiate its time-dependent sections. For the tangent bundle, a section $V(t)$ of the pullback tangent bundle hasThe Leibniz rule under a change of frame proves independence of the coordinates. The derivative includes the time dependence of the coefficients, so it is meaningful even at a zero tangent velocity. A vector field along a map need not extend to one ambient field if the curve self-intersects. Parallel fields satisfy $D_{t}V=0$, a linear ordinary differential equation whose unique solutions define parallel transport.

The real exponential function $\exp (x)=e^x$ converts addition into multiplication:Its values are positive, and it defines a bijection from the real numbers to the positive real numbers. The natural logarithm is its inverse function. These properties connect arithmetic with calculus: adding an amount to the input multiplies the output by a fixed factor, and the local rate of change equals the current value. The exponential series gives one construction; an equivalent characterization is the unique solution of a differential equation $y^{\prime }=y$, $y(0)=1$.

For a concrete example, $y(t)=10e^{0.2t}$ starts at $10$ and satisfies $y^{\prime }=0.2y$. The absolute derivative increases with $y$, while the relative rate $y^{\prime }/y$ stays $0.2$. At $t=5$, $y=10e\approx 27.1828$ and $y^{\prime }\approx 5.43656$. The exponential function therefore models accumulation proportional to what is already present, rather than addition of a fixed amount at every step. A positive constant rate gives exponential growth; a negative one gives exponential decay.

Every positive real base $b\ne 1$ fits the same construction through $b^x=e^{x\ln b}$. Thus its derivative is $(\ln b)b^x$: base $e$ is special because $\ln e=1$. The exponential function is also unchanged in form by differentiation and integration, helping explain its repeated appearance in linear ordinary differential equations. Its real definition, inverse relationship, and derivative rules are developed in OpenStax, integrals, exponential functions, and logarithms.

For a constant $a$ and a polynomial $P$, the product rule gives $D(e^{ax}v)=e^{ax}(D+a)v$. Iteration and linearity prove the displayed identity. In particular $(D-a{)}^m(e^{ax}v)=e^{ax}{v}^{(m)}$, which reduces a repeated-root constant-coefficient linear ordinary differential equation to repeated integration.

Why must a conducting piston move slowly? A simple model couples the first law of thermodynamics to Newton's law of cooling. Assume a fixed amount of ideal gas with uniform temperature $T(t)$ and constant whole-sample heat capacity at constant volume $C_V$, a heat bath maintained at $T_b$, negligible friction, and mechanically slow motion. Let $G>0$ be the effective thermal conductance, the reciprocal of the thermal resistance, in $\mathrm{W}{\mathrm{K}}^{-1}$. ThenPositive heat transfer must offset expansion work to keep $T$ near $T_b$. A finite $G$ and nonzero piston speed therefore require a temperature difference. This uniform-temperature model neglects spatial temperature gradients, sensor response, cylinder heat capacity and any change of $G$ with piston position; it is an illustrative calculation, not a fit to the isothermal gas syringe experiment.

For a prescribed exponential volume change $V=V_i{e}^{\lambda t}$, define $\tau =C_V/G$, $s=t/\tau$ and $ϵ=nR\lambda /G$. The dimensionless temperature $\theta =T/T_b$ obeys a linear ordinary differential equation with initial value $\theta (0)=1$:The plotted expansion uses stipulated $C_V=(5/2)nR$ and stops at $V_f=2V_i$, at $s_f=\ln 2/[(5/2)ϵ]$. During expansion, the larger $ϵ$ produces the larger temperature deficit; once the piston is held fixed, the deficit undergoes exponential decay as $e^{-(s-s_f)}$. Reducing $|ϵ|$ makes the path more nearly an isothermal process, while increasing its duration. Reversing the piston with $\lambda <0$ produces a temperature excess and outward heat transfer; the small-$|ϵ|$ regime is the intended near-isothermal approximation. Mechanical slowness alone does not ensure sufficient heat transfer.

A fundamental system for a homogeneous linear ordinary differential equation of order $n$ on an interval is a basis of its vector space of homogeneous solutions. Every solution has a unique expression ${\sum }_{j=1}^n{c}_j{y}_j$ with constant coefficients. Under the usual continuity assumptions and a nonvanishing leading coefficient, a nonzero Wronskian at one point proves that $n$ solutions form such a system: their initial-value vectors are independent and span all possible initial data, and the Picard-Lindelöf theorem supplies uniqueness. For a first-order vector system, the corresponding columns form a fundamental matrix.

The general solution of a differential equation is a family containing every solution, usually parameterized by arbitrary constants or initial data. For a forced linear ordinary differential equation, it is the sum of one particular solution and the general homogeneous solution.

On a branch where a differentiable function $y(x)$ has a nonzero derivative, the inverse function theorem allows the ordinary differential equation to be rewritten for $x(y)$. A nonlinear equation in one direction can become a linear ordinary differential equation in the other. For example, $y^{\prime }=y/(x+y+c)$ becomesThe integrating factor $1/y$ gives $(x/y{)}^{\prime }=1/y+c/y^2$, hence $x+c=y(\ln |y|+C)$. This family must retain the original denominator restriction $x+y+c\ne 0$. The separate zero solution $y=0$ on intervals avoiding $x=-c$ is lost by the inversion and must be restored.

Linear drag opposes the velocity relative to the surrounding medium with magnitude proportional to its speed, where $\alpha >0$. Its power is ${\mathbf{F}}_d\cdot \mathbf{v}=-\alpha |\mathbf{v}{|}^2$, so it dissipates kinetic energy. Under a constant external force, the resulting linear ordinary differential equation relaxes exponentially on the time scale $m/\alpha$.

Use a parameter $t$. The characteristic equations areFor a graph of a function $z=f(x,y)$ satisfying the equation, solve the first two equations after substituting $z=f(x,y)$. The chain rule and the partial differential equation give $df(x(t),y(t))/dt=f+bx-ay$, exactly the third equation. Uniqueness for the resulting linear ordinary differential equation system shows that its characteristic through the initial point agrees with this curve on the graph of a function. This proves invariance as long as the graph of a function remains defined along the curve. A claim about the entire curve additionally requires a surface continued along that flow; a restricted graph of a function patch alone cannot imply global continuation.

Let $\mathbf{k}=(a,b,c)$, $\mathbf{\ell }=(l,m,n)$ and $\mathbf{r}=(x,y,z)$. Then $\dot{\mathbf{r}}=\mathbf{r}-\mathbf{k}\times \mathbf{r}$. Since $|\mathbf{k}|=1$,Along the characteristic starting from $s\mathbf{\ell }$ these give $\mathbf{k}\cdot \mathbf{r}=e^{t}s(\mathbf{k}\cdot \mathbf{\ell })$ and $|\mathbf{r}|=e^t|s|$. Consequently the characteristic surface generated by the given line satisfiesTo see that the whole displayed cone is generated, not just contained in it, decompose $\mathbf{\ell }=\kappa \mathbf{k}+\mathbf{v}$ with $\mathbf{v}\perp \mathbf{k}$, $\kappa =\mathbf{k}\cdot \mathbf{\ell }$. The full solution isWhen $0<|\kappa |<1$, variation of $t$ gives all azimuths about the axis, while the freely chosen $s$ supplies every positive radius on both halves. This is precisely the two-sided circular cone, withThe edge cases matter: $\kappa =0$ gives the plane $ax+by+cz=0$ (semi-angle $\pi /2$), and $|\kappa |=1$ generates only the axis (semi-angle zero), not a two-dimensional surface. In the latter case the initial line is itself characteristic, so it does not determine a unique integral surface. The nondegenerate cone is smooth away from its vertex and can be used as a graph of a function only on patches whose tangent plane is not vertical. Thus the printed geometric conclusion has its ordinary characteristic-surface meaning in the nondegenerate case, with these domain and degeneracy qualifications; it cannot be asserted for every global smooth graph of a function without them.

On an interval with $x>0$, use the power-law ansatz $y=x^r$. Substitution gives $r(r-1)+2r+1=r^2+r+1=0$, whose roots are $r=-1/2\pm i\sqrt{3}/2$. Taking real and imaginary parts of $x^r$ givesThese two real solutions are linearly independent: their Wronskian is $(\sqrt{3}/2)x^{-2}\ne 0$. They therefore span the solution space of the second-order linear ordinary differential equation on that interval.

For $x<0$, replace $x^{-1/2}$ by $|x{|}^{-1/2}$ and $\log x$ by $\log |x|$, with independent real constants on the negative interval. This follows equally by the change of variable $t=\log |x|$, which turns the Cauchy-Euler equation into $d^{2}y/d{t}^2+dy/dt+y=0$. The coefficient of $y^{\prime \prime }$ vanishes at zero, so the general formula is stated separately on either side. Every nonzero mode has an unbounded oscillatory envelope as $x\to 0$; a continuous extension through zero requires the solution to vanish identically on both sides.

The characteristic equation of the linear ordinary differential equation is $r^2+2r+2=0$, with roots $-1\pm i$. A real pair of homogeneous solutions is thereforeTheir Wronskian, with the convention $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$, isThus the two functions are linearly independent throughout the interval and span its homogeneous solutions.

The only nonconstant entry of the metric tensor is $g_{\varphi \varphi }=R^2{\sin }^2\theta$, whose partial derivative is ${\partial }_{\theta }{g}_{\varphi \varphi }=2R^2\sin \theta \cos \theta$. The nonzero Christoffel symbols areAlong a fixed latitude with $0<\theta <\pi$, parallel transport givesSet $X=V^{\theta }$ and $Y=\sin \theta V^{\varphi }$. Since $\theta$ is fixed, this linear ordinary differential equation becomesIt is a rotation at constant rate $-\cos \theta$. Its general solution, also valid at the equator where the rate is zero, isThe initial data give $A=1$ and $B=0$. At $\theta =\pi /3$, a complete circuit has $\varphi \cos \theta =\pi$, soThe physical squared length uses the metric tensor, rather than the Euclidean sum of coordinate components:It is constant and equals $R^2$ for these data. More generally, metric compatibility gives $dg(V,V)/d\varphi =({\nabla }_{{\partial }_{\varphi }}g)(V,V)+2g({\nabla }_{{\partial }_{\varphi }}V,V)=0$. The final change of direction is holonomy, exemplified by parallel transport around a latitude of the unit sphere; parallel transport preserves length without requiring the direction to return unchanged.

Take the membrane particle flux density to be positive from cytoplasm into the cell nucleus. With membrane permeability $\alpha$,For fixed compartment volumes, conservation of mass and the absence of exterior flux giveSubtract the two concentration derivatives:This linear ordinary differential equation has solutionIts relaxation time is $\boxed{T=V_c{V}_n/[\alpha A_n(V_c+V_n)]}$. Here $\alpha$ has dimensions length/time, so $\alpha A_n/V$ has dimensions inverse time. A positive initial concentration difference decays monotonically toward zero; a negative one approaches zero from below.

The printed “otherwise” branch requires correction. Its factor $3/4$ is the probability of ending at any of the three bases different from the initial base, not of ending at one specified different base. Assigning that probability to each of three alternatives would make a row of the transition matrix sum to $5/2-(3/2)e^{-\mu t}$, which exceeds one for $t>0$. This distinction is present in the source PDF, not introduced by the transcription.

For a consistent Jukes-Cantor model, let each ordered change from one nucleotide to a specified different nucleotide have instantaneous rate $q$. Its continuous-time Markov chain has diagonal generator entries $-3q$ and off-diagonal entries $q$. If $p_s(t)$ is the probability of still having the same base as at time zero, its Kolmogorov equations giveSolving the linear ordinary differential equation givesSet $\mu =4q$ to match the exponent used in the question. Symmetry then gives the properly normalized transition probabilitiesSumming the three different-base outcomes yieldsThe assumptions are a time-homogeneous Markov process, the same change rate between every distinct ordered base pair, identical rates across independent sites, and no insertions or deletions. The stationary distribution is uniform over the four bases; starting the process in that distribution is not required for this conditional calculation.

There is also a parameter-convention issue. The actual total mutation rate out of a base is $r=3q=3\mu /4$. Thus the exponent parameter $\mu$ in (b)–(d) is $4/3$ times the event rate used in the elementary counting interpretation of (a). Keeping this conversion explicit avoids changing the meaning of “mutation rate” halfway through the calculation.

Admissions add to the expected value at rate $\alpha$, while the expected aggregate discharge rate is $\delta \mathbb{E}[X(t)]$. ConsequentlyMultiplying this linear ordinary differential equation by the integrating factor $e^{\delta t}$ givesIntegrating from zero, using $N(0)=0$, gives $e^{\delta t}N(t)=\alpha (e^{\delta t}-1)/\delta$. HenceIt starts at zero, has initial slope $\alpha$, and approaches $\alpha /\delta$ as $t\to \infty$, as the admission and discharge rates balance.

An integrating factor for $u^{\prime }+P(x)u=Q(x)$ is a nonzero function $\mu (x)=\exp (\int P(x)dx)$, chosen so that $(\mu u{)}^{\prime }=\mu Q$. Integrating this total derivative reduces the first-order linear ordinary differential equation to one quadrature.

For the given nonlinear equation, let $u=y^2$, so $u^{\prime }=2y{y}^{\prime }$. Multiplication by $2y$ gives $u^{\prime }+u=x^2+2x$. Using the integrating factor $e^x$, or recognizing the particular solution $u=x^2$, yields $u=x^2+C{e}^{-x}$. The initial value gives $C=1$ and selects the positive branch:The radicand is strictly positive for every real $x$. This solution never reaches the singular denominator $y=0$ of the original equation, so multiplication by $y$ has introduced no forbidden crossing. The negative branch would violate $y(0)=1$.

For $x\ne 0$, divide by $x^2$ to obtain the linear ordinary differential equation $y^{\prime }+3y/x=x^{-2}$. An integrating factor is $x^3$ on either interval not containing zero. Thus $(x^{3}y{)}^{\prime }=x$ andSubstituting gives $x^2{y}^{\prime }+3xy=1$ directly. No classical solution can include $x=0$, since the original equation there would assert $0=1$; constants on the two separate intervals need not agree.