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For $y\prime \prime +py\prime +qy=0$, the Wronskian satisfies $W\prime =-pW$.

More precisely, let $y_1,y_2$ be solutions of a differential equation for the same homogeneous linear differential equationon an interval $I$ where $p,q$ are continuous functions. For any $x_0\in I$, Abel's identity givesOnly the coefficient of $y^{\prime }$ enters the evolution of the Wronskian. In an unnormalized equation $A(x)y^{\prime \prime }+B(x)y^{\prime }+C(x)y=0$, first use $p=B/A$ on an interval where $A\ne 0$ and the coefficient ratios are continuous functions.^[ref]

The proof uses cancellation. Differentiating the Wronskian by the product rule and substituting the common equation givesMultiplying this first-order linear differential equation by the integrating factor $\mu (x)=\exp ({\int }_{x_0}^{x}p(s)ds)$ gives $(\mu W{)}^{\prime }=0$. Since $\mu (x_0)=1$, the constant is $W(x_0)$, proving Abel's identity. This argument also covers $W(x_0)=0$ without dividing by $W$.

The exponential function is positive for real $p$, so a real Wronskian either vanishes everywhere or has a fixed nonzero sign throughout $I$. In the preceding example, $p=-1$ and $W(0)=3$ give $W(x)=3e^x$ immediately. If $p=0$, every Wronskian of two solutions of a differential equation for that same equation is constant, regardless of $q$. Thus Abel's identity can certify linear independence and find the denominator used in variation of parameters without explicit formulas for both solutions of a differential equation.

A Bernoulli differential equation is a first-order ordinary differential equation of the displayed form, with $P$ and $Q$ continuous functions on an interval and a fixed real exponent $n$. A nonlinear power can hide a first-order linear differential equation: on a suitable branch, the Bernoulli power substitution $z=y^{1-n}$ removes it exactly. The transformed solution of a differential equation can be completely regular even when recovering $y$ produces finite-time blowup.

For example, the reciprocal Bernoulli equation with finite-time blowup $y^{\prime }+y=e^x{y}^2$ becomes $z^{\prime }-z=-e^x$ under $z=1/y$. The initial conditions $y(0)=1/2$ and $y(0)=-1$ select a positive trajectory blowing up at $x=2$ and a negative trajectory tending to zero as $x\to \infty$, respectively. The integrating factor exposes both behaviors with the same short calculation.

The cases $n=0$ and $n=1$ are already first-order linear differential equations: respectively $y^{\prime }+Py=Q$, with the zeroth-power term interpreted as the constant $1$, and $y^{\prime }+(P-Q)y=0$. There is no need for a nonlinear change of variable; for $n=1$, $y^{1-n}=1$ would discard the unknown entirely. For other exponents, the substitution must respect the domain of exponentiation and may lose the zero solution of a differential equation.^[ref]

For arbitrary real $n\ne 0,1$, work first on an interval where $y>0$. Define exponentiation by $y^r=\exp (r\log y)$, using the real logarithm. The change of variable is differentiable and invertible there, and the chain rule givesMultiplying the Bernoulli differential equation by $(1-n)y^{-n}$ yields the first-order linear differential equationThis is an exact transformation of nonzero solutions of a differential equation on the chosen branch; it is valid far from any reference state and does not discard higher powers as a local linearization would.

Given the initial condition $y(x_0)=y_0>0$, the integrating factor and transformed solution of a differential equation areThe product rule verifies $(\mu z{)}^{\prime }=(1-n)\mu Q$. Recover $y=z^{1/(1-n)}$ only on the connected component of $\{x:z(x)>0\}$ containing $x_0$, inside the original coefficient interval. The first-order linear differential equation can have a longer existence interval than this inverse transformation allows.^[ref]

Negative values need a separately justified real branch. For integer $n$, the required powers are real and differentiable for $y<0$ too, but the inverse must retain the sign selected by the initial condition. For $n=3$, $z=y^{-2}>0$ represents either $y=+1/\sqrt{z}$ or $y=-1/\sqrt{z}$; both square roots are taken of positive $z$. For $n=2$, $z=1/y$ retains the sign directly. For a general noninteger exponent, use the positive branch unless a specific differentiable real negative branch has been established. Never evaluate a real fractional power of a negative value merely because the linear formula has become negative.

Division by $y^n$ excludes $y=0$. When $n>0$ and the original real power is defined at zero, $y\equiv 0$ is an additional solution of a differential equation and must be checked in the original equation. When $n<0$, the equation as written is undefined at zero. On a positive branch the right-hand side is locally Lipschitz in $y$, so the Picard-Lindelöf theorem gives local uniqueness. At zero, the same hypothesis needs a fresh check: it holds for the polynomial $n=2$ example, whereas for $0<n<1$ a nonzero power coefficient prevents local Lipschitz continuity there. If $z\downarrow 0$, the inverse sends $y\to \infty$ for $n>1$, but sends $y\downarrow 0$ for $0<n<1$. The latter endpoint can lead to finite-time extinction in a sublinear Bernoulli equation, where the original equation determines how to join to zero.

A consecutive first-order reaction mechanism passes material through a reaction intermediate. Consider a stipulated chemical reaction network $\mathrm{A}\stackrel{k_1}{\to }\mathrm{I}\stackrel{k_2}{\to }\mathrm{P}$, with constant positive rate constants, fixed temperature and volume, uniform mixing, and no reverse or competing chemical reactions. All three labels carry one conserved unit of material. Set $[\mathrm{A}](0)=A_0>0$ and $[\mathrm{I}](0)=[\mathrm{P}](0)=0$. The law of mass action givesAdding the three rate equations proves $[\mathrm{A}]+[\mathrm{I}]+[\mathrm{P}]=A_0$. Each step has a first-order reaction rate law; this does not make all three concentrations simple exponential decays.

For $k_1\ne k_2$, the exact solution isTo derive the middle expression, substitute the first into the first-order linear differential equation for I and multiply by the integrating factor $e^{k_{2}t}$. Then $d(e^{k_{2}t}[\mathrm{I}])/dt=k_1{A}_0{e}^{(k_2-k_1)t}$, which can be integrated directly with the stated initial condition. The material balance supplies P without another integration.

The apparent division by zero at equal rate constants is removable. For $k_1=k_2=k$, the same integrating factor givesThis is also the limit of the unequal-rate constant expression. Initially $[\mathrm{P}]/A_0=(kt{)}^2/2+O((kt{)}^3)$, so reaction product formation begins with zero slope despite immediate reactant loss. By contrast, a direct $\mathrm{A}⟶\mathrm{P}$ first-order reaction with the same $k$ has $[\mathrm{P}]/A_0=1-e^{-kt}$ and nonzero initial slope. Their A traces agree exactly; their P traces do not.

For the chosen equal-rate constant example, I peaks at $t=1/k=5.00\mathrm{s}$, because $d[\mathrm{I}]/dt=A_{0}k{e}^{-kt}(1-kt)$. At that time $[\mathrm{A}]=[\mathrm{I}]=A_0/e\approx 0.367879\mathrm{mmol}{\mathrm{L}}^{-1}$ and $[\mathrm{P}]\approx 0.264241\mathrm{mmol}{\mathrm{L}}^{-1}$; the direct model would already have $[\mathrm{P}]\approx 0.632121\mathrm{mmol}{\mathrm{L}}^{-1}$. The peak means formation and consumption of I are momentarily equal, not that the whole system has reached chemical equilibrium.

More generally the peak is at $t_{\ast }=\ln (k_2/k_1)/(k_2-k_1)$, which is positive for either ordering of the two positive rate constants and tends to $1/k$ as they coincide. When $k_2\gg k_1$, a brief initial adjustment can be followed by $[\mathrm{I}]\approx (k_1/k_2)[\mathrm{A}]$ and approximately equal formation and consumption reaction rates. The exact solution states when this small-intermediate approximation can be reasonable. It is inappropriate for the illustrated equal-rate constant case: at the peak, over one third of the conserved material is I. Such timing matters if the goal is to collect a useful reaction intermediate before it becomes the final reaction product.

Take a stipulated pair of effective rods with common length $L$, linear thermal expansion coefficient ${\alpha }_L$, heat capacity $C$, projected illuminated area $A$, and thermal conductance $G$ to an environment at constant temperature $T_a$. One reference represents the symmetric polished-rod assembly. Write ${\theta }_b=T_b-T_a$ and ${\theta }_r=T_r-T_a$. Let their effective radiometric absorptances be $a_b>a_r$ for a fixed incident optical spectrum, and let $E(t)$ be the common irradiance. A spatially uniform, linear heat-balance model givesHere $G$ includes a linear approximation to the relevant losses; changing the surface treatment can also change those losses in a real device. We stipulate equal $C$ and $G$ to isolate the response to absorbed illumination, with thermal coupling from the burner neglected. Small, mechanically unconstrained extensions then give a differential displacementThe time constant $\tau$ describes thermal memory. A common imposed temperature shift cancels from the extension difference under the matched-coefficient assumptions; mismatched materials, geometry or thermal coupling can spoil that cancellation.

For a step from darkness to constant irradiance $E_0$, with initially equal rod temperatures, the first-order linear differential equation givesChoose $L=0.20\mathrm{m}$, ${\alpha }_L=20\times 10^{-6}{\mathrm{K}}^{-1}$, $A=0.0020{\mathrm{m}}^2$, $a_b=0.9$, $a_r=0.1$, $C=5.0\mathrm{J}{\mathrm{K}}^{-1}$, $G=0.20\mathrm{W}{\mathrm{K}}^{-1}$ and $E_0=600\mathrm{W}{\mathrm{m}}^{-2}$. The calculated difference of absorbed powers is $0.96\mathrm{W}$, givingFor an illustrative ideal valve that closes at $\delta =g=10\mu \mathrm{m}$, the closing time isThis threshold is crossed only when $0<g<{\delta }_{\infty }$. These are chosen model parameters, not a fit to a historical valve.

After illumination stops at $t=t_1$, the displacement decays as $\delta (t)=\delta (t_1)e^{-(t-t_1)/\tau }$. With $t_1=180\mathrm{s}$, the model reopens about $16.3\mathrm{s}$ later. Thus a sun valve responds to the thermal history of illumination; it does not read a clock or directly measure the beacon's brightness.

For $x_{n+1}=a_n{x}_n+b_n$, put $U_0=1$ and $U_n={\prod }_{j=1}^n{a}_j$, with $a_j\ne 0$. Dividing by this accumulated homogeneous growth gives $z_n=x_n/U_{n-1}$ and $z_{n+1}-z_n=b_n/U_n$. A telescoping series then provesThis is the discrete counterpart of the integrating factor for a first-order linear differential equation.

The Drude model explains finite electrical conductivity by balancing charge carrier acceleration in an electric field against loss of directed momentum through collisions. Its useful surprise is that continuous microscopic motion can produce a constant macroscopic current density: the applied electric field continually restores the directed momentum lost to the material. The simplest version has one mobile species, constant number density $n$, electric charge $q$, mass $m>0$, and momentum relaxation time $\tau >0$, in a stationary, homogeneous, isotropic material. At zero magnetic field, its mean drift velocity and current density obeyFor Electrons, $q=-e$ with positive elementary charge $e$, so ${\sigma }_0=n{e}^2\tau /m$. The Electron drift velocity points against a constant electric field, whereas conventional current points along it. Random microscopic velocities can have zero mean without being zero individually; drift velocity is this population mean, rather than each particle's speed.^[ref]

The same first-order linear differential equation predicts both Drude current relaxation after switching an electric field and the Drude frequency response to a sinusoidal electric field. Together they let a measured electrical conductivity be interpreted in terms of a density, a mass, and a time scale; a single constant-field measurement determines only their combination $n{q}^2\tau /m$.

The original 1900 paper Zur Elektronentheorie der Metalle by Paul Drude sought a common account of electrical transport and heat transfer using mobile charged particles and ideas from the kinetic theory of gases. Its introduction, especially printed pp. 568–569, explains that comparing different transport phenomena could constrain the populations of free and bound charged particles. The modern momentum relaxation time formulation above is a useful presentation of the model, not a verbatim translation of the original notation or its historical assumptions about mass.^[ref]

For an exactly soluble example, set $L_1=L_2=L$, $R_1=R_2=R$, and $0<M<L$. Start with zero electric currents, and apply a constant electric potential difference $V_0>0$ at $t=0$. The sums $s=i_1+i_2$ and $d=i_1-i_2$ obey separate first-order linear differential equations:Their time constants are ${\tau }_{+}=(L+M)/R$ and ${\tau }_{-}=(L-M)/R$. Thus, for $t\ge 0$,Since ${\tau }_{-}<{\tau }_{+}$, $i_2(t)<0$ for $t>0$, but $i_2(t)\to 0$ as $t\to \infty$. The primary approaches $V_0/R$. The secondary response opposes the rise of linked magnetic flux, in accordance with Lenz's law. Removing mutual inductance by setting $M=0$ makes the two exponentials equal and gives zero secondary electric current.

For stipulated $L=0.1\mathrm{H}$, $M=0.06\mathrm{H}$, $R=10\Omega$ and $V_0=1\mathrm{V}$, the two time constants are $16\mathrm{ms}$ and $4\mathrm{ms}$. Setting ${\dot{i}}_2=0$ gives its largest negative magnitude atThe continuing primary electric current eventually reaches $100\mathrm{mA}$ while the secondary returns to zero. These numbers belong to the stated model, not the 1831 apparatus.

An integrating factor is a nonzero multiplier that makes a differential equation directly integrable. For a first-order linear differential equation, it gathers the unknown function and its derivative into one product rule expression. Suppose $a$ and $b$ are continuous functions on an interval containing $t_0$, and consider the initial value problemDefine $A(t)={\int }_{t_0}^{t}a(s)ds$ and $\mu (t)=e^{A(t)}$. The exact solution of a differential equation throughout that interval isThe definite integral fixes $\mu (t_0)=1$, so the initial condition appears without an extra constant to determine. If the equation originally has a coefficient multiplying $y^{\prime }$, first divide by that coefficient on an interval where it is nonzero; a zero of that coefficient needs separate attention. See OpenStax's first-order linear equations chapter for this normalization and the integrating factor procedure.

For a concrete balance, solve $y^{\prime }+2y=3$ with $y(0)=0$. Here $\mu =e^{2t}$, so multiplying the entire ordinary differential equation gives $(e^{2t}y{)}^{\prime }=3e^{2t}$. Accumulating from zero to $t$ givesThe computed values are $y(0.5)\approx 0.94818$, $y(1)\approx 1.29700$ and $y(2)\approx 1.47253$. Its derivative is $3e^{-2t}$, which verifies $y^{\prime }+2y=3$ exactly. Interpreting $3$ as a steady input and $2y$ as proportional removal explains the shape: initially all input increases $y$, then removal catches up. The constant particular solution $y_p=3/2$ balances input and removal, and the homogeneous solution $y_h=-\frac{3}{2}{e}^{-2t}$ is the transient response that corrects it to the initial condition. Because this autonomous differential equation has a time-independent right-hand side, $3/2$ is also an equilibrium point.

The first panel adds the particular solution and homogeneous solution; the second shows the same balance in rates, $3=y^{\prime }+2y$. This links the algebra to a physical accounting rule. Newton's law of cooling has this form with $a=k$ and $b=k{T}_a$, while an electrical circuit can have a varying input. In both cases the integrating factor systematically handles the input rather than requiring a guess for a particular solution.

Variable coefficients work just as directly. For $y^{\prime }+2ty=t$, $y(0)=0$, the integrating factor is $e^{t^2}$, andThus $y(0.5)\approx 0.11060$, $y(1)\approx 0.31606$ and $y(2)\approx 0.49084$. The derivative $y^{\prime }=t{e}^{-t^2}$ verifies $y^{\prime }+2ty=t$, including at zero, where both coefficient and input vanish. No division by $t$ is necessary. The general solution is $y=1/2+C{e}^{-t^2}$: $1/2$ is a constant particular solution, and $C{e}^{-t^2}$ solves the associated homogeneous linear differential equation. Although this special forcing term admits a constant trajectory, this is a nonautonomous differential equation: it depends explicitly on time. Merely setting its right-hand side to zero at one instant would not establish a constant solution of a differential equation; at $t=0$ every state has zero slope.

The direction field shows the selected trajectory leaving its initially horizontal tangent line. For $t\ge 0$, the difference between any two solutions of a differential equation is multiplied by $e^{-t^2}$, rather than $e^{-2t}$ in the constant-rate example. This dependence on elapsed history is clearer if the exact formula is rewritten asFor $t\ge t_0$, each contribution from the input at $s$ is weighted by the accumulated coefficient between $s$ and $t$. The homogeneous solution remembers the initial condition; the particular solution given by the second term accumulates the forcing term with zero initial value. This decomposition differs from choosing a constant particular solution, but gives the same total. Initial-data memory tends to zero when ${\int }_{t_0}^{t}a\to +\infty$; a negative accumulated coefficient instead amplifies it. The reciprocal relation between integrating factor and homogeneous solution is developed in MIT's notes on solutions to linear first-order equations.

Why does the trick work? The fundamental theorem of calculus and chain rule give ${\mu }^{\prime }=a\mu$. Consequently the product rule givesIntegrating this identity and dividing by the positive exponential function $\mu$ produces the stated formula. Any nonzero constant multiple of $\mu$ works equally well; normalization is a convenience. This is the beauty of the method: a sum involving an unknown function becomes one exact derivative, leaving only an integral of known functions. See Oregon State's product-rule derivation. The related integrating factor for a differential one-form extends the idea of making an expression exact beyond this linear setting.

For the first-order ordinary differential equation $y^{\prime }=f(x,y)$ defined on an open subset $D$ of the plane, an isocline of slope $k$ is the level setHere $k$ is a constant. At every point of this isocline, the direction field prescribes a tangent vector proportional to $(1,k)$ for a solution of a differential equation. It does not generally prescribe the tangent line of the isocline itself. The level set can have several components or fail to be a smooth curve, and points outside $D$ are excluded.

For example, take $y^{\prime }=x-y$ on the whole plane. Its isoclines are the parallel straight linesEach line has slope $1$, while the direction field along it has slope $k$. Thus the isocline $y=x$ carries horizontal segments even though the line itself rises. A function graph of a solution of a differential equation follows the segments, rather than following this line. Only the isocline with $k=1$, namely $y=x-1$, is also a solution of a differential equation.^[ref]

The first-order linear differential equation has the exact familyThe initial condition $y(x_0)=y_0$ selects $A=(y_0-x_0+1)e^{x_0}$ and hence a unique solution of a differential equation for every real $x$. The picture shows $y(0)=-1,0,1,2$. For $A=0$ the solution of a differential equation stays on the $k=1$ isocline; for other $A$ it moves through different isoclines. These formulas check the pictured trajectories by direct differentiation, independently of how the direction field was sampled.

The zero-slope isocline $f(x,y)=0$ is called a nullcline. Below the nullcline $y=x$ in this example, $y^{\prime }>0$; above it, $y^{\prime }<0$. For $A>0$, the solution of a differential equation meets it at $x=\log A$, where $y^{\prime }=0$ and $y^{\prime \prime }=1$, a strict local minimum. The nullcline is therefore a useful guide to turning points, although it is not itself a solution of a differential equation. In the equivalent planar autonomous differential equation $\dot{x}=1$, $\dot{y}=x-y$, it is the nullcline for the second component; there is no equilibrium point, because the first component never vanishes.

This makes isoclines useful even when no explicit solution of a differential equation is available: organize a direction field by common slopes, read where function graphs rise or fall, and identify possible turning points on nullclines.

For every real parameter $a$,where $F(0)=0$. For $a>0$, differentiation under the integral sign and a partial fraction decomposition giveSolving this first-order linear differential equation yields $F(a)=a[\pi \ln (1+a)+C]$. To fix the constant, dominated convergence gives ${\lim }_{a\to 0^{+}}F(a)/a={\int }_0^{\infty }\ln (x^2)/(1+x^2)dx=0$; the last absolutely convergent integral changes sign under $x=1/t$. Thus $C=0$. Oddness in $a$ supplies the result for negative parameters. The initial value $F(0)=0$ alone does not determine $C$, since every solution $a[\pi \ln (1+a)+C]$ tends to zero as $a\to 0^{+}$.

An integrating factor isMultiplication by this integrating factor turns the first-order linear differential equation into $(e^{x^3}y{)}^{\prime }=x^2{e}^{x^3}$. Integrating from zero to $x$ givesThus the initial value problem has the solutionIndeed, direct differentiation gives $y^{\prime }=-3x^2(a-1/3)e^{-x^3}$, which verifies both the differential equation and the initial condition.

Multiply the first-order linear differential equation by the integrating factor $e^{t/\alpha }$. ThenIf $y\to 0$ in the remote past, then $e^{t/\alpha }y\to 0$ there. Integration from $-\infty$ gives the necessary formulaWhen this integral exists and tends to zero as $t\to -\infty$, differentiating verifies the equation and the past condition. Two such solutions differ by $C{e}^{-t/\alpha }$, which diverges at $-\infty$ unless $C=0$, proving uniqueness. This is the causal response of a first-order relaxation equation, a convolution with its decaying impulse response.

The phrase “general function” implicitly requires compatibility with the past condition: it does not guarantee existence for arbitrary $f$. A sufficient condition is bounded forcing with $f(t)\to 0$ as $t\to -\infty$; after setting $r=t-s$, the integral is bounded by the supremum of $|f|$ before $t$. For example, $f\equiv 1$ gives $y=1+C{e}^{-t/\alpha }$, so no solution has the stated past limit. Both forcings below vanish in the past and meet the intended causal condition, the impulse being interpreted distributionally.

Here the singular perturbation has an exact conservative structure:The integrating factor for this first-order linear differential equation givesAt $x=1$ and $x=-1$ the integral terms have opposite signs while the exponential factors agree. Equal boundary data therefore force $C=0$, and normalization gives the unique exact solutionThis exposes the Gaussian interior layer of a conservative drift equation. Its central scale is $x=O(\sqrt{ϵ})$, where $x=\sqrt{ϵ}\xi$ gives $Y^{\prime \prime }+\xi Y^{\prime }+Y=0$. The central Gaussian function profile is $e^{-{\xi }^2/2}$ with amplitude $e^{1/(2ϵ)}$.

Away from the center, $|x|\gg \sqrt{ϵ}$, use exponential WKB approximation branches rather than an assumed bounded algebraic outer solution. The reduced ordinary differential equation $x{y}_0^{\prime }+y_0=0$ would give $C/x$; matching both equal positive endpoint values with that algebraic branch is impossible. The exact solution selects its zero coefficient and the exponentially large branch with a Gaussian function profile instead. Near either endpoint, a distance $O(ϵ)$ changes $y$ by order one relative to its endpoint value: with $r=(1-|x|)/ϵ$, $y=e^{r-ϵr^2/2}$. These endpoint scales describe normalization of the same global exponential solution.

Thus a central $O(\sqrt{ϵ})$ turning region, exponential outer regions on both sides, and $O(ϵ)$ endpoint normalization scales give the complete regional description. The peak is exponentially large, so no uniformly bounded regular outer expansion should be presumed. The exact expression already provides a uniform solution without further matching calculations.

Introduce $\omega =\dot{\theta }$. The pendulum with quadratic damping is the autonomous planar systemIts equilibrium points are $(k\pi ,0)$. Since the derivative of $\omega |\omega |$ is zero at zero, the linearization of a dynamical system at each equilibrium point isAt an odd multiple of $\pi$ its eigenvalues are $\pm 1$, so it is a saddle equilibrium; the stable and unstable tangents are $\omega =-(\theta -k\pi )$ and $\omega =\theta -k\pi$. At an even multiple the eigenvalues are $\pm i$. This linearization is a centre and does not decide the nonlinear stability. In particular it would be incorrect to infer a linear damped focus from this Jacobian.

For the full system the mechanical energy is a Lyapunov function near an even equilibrium point:A sufficiently small sublevel set is compact and contains only that equilibrium point. The only invariant motion with $\dot{H}=0$ has $\omega =0$ and $\sin \theta =0$, hence is the equilibrium point itself. The LaSalle invariance principle proves asymptotic stability. Near it the trajectories turn clockwise and spiral inward, as the near-linear angular motion persists while energy decreases. Thus even equilibria are nonhyperbolic asymptotically stable spiral points; odd equilibria remain saddles.

For explicit trajectories put $W={\omega }^2$ on a branch where $\omega \ne 0$. The chain rule gives $dW/d\theta =2\dot{\omega }$, so the two signs yield the first-order linear differential equations $W^{\prime }+2cW=-2\sin \theta$ for $\omega >0$, and $W^{\prime }-2cW=-2\sin \theta$ for $\omega <0$. Their integrating factors giveOnly portions with a nonnegative right-hand side are physical. At a turning point the constants of the two branches are matched using $\omega =0$; neither expression is one global trajectory through both signs. When $c=0$ they reduce to energy contours, with closed orbits around the even equilibrium points. For every $c>0$, the quadratic damping destroys those closed orbits and makes the centres attracting spirals. Its effect is higher order in amplitude, so decay is weaker near rest than with linear damping. The saddle equilibria remain unstable, though their separatrices change shape.

At the first half-temperature time, the solution of Newton's law of cooling givesSince $\alpha >2$, this time is positive and finite. The instantaneous temperature increase supplies the new initial condition $T(t_1+)=\alpha T_0/2+\beta$. Restart the same first-order linear differential equation at $t_1$:This is Newton cooling with an instantaneous temperature jump. The time translation in the exponential decay is essential: its prefactor is the excess immediately after the jump.

With a positive cooling rate $k$, Newton's law of cooling gives $T^{\prime }=-k(T-T_0)$. The temperature excess $U=T-T_0$ satisfies the first-order linear differential equation $U^{\prime }=-kU$. Applying the initial condition to its exponential decay solution givesThe temperature approaches the ambient value $T_0$, rather than zero. The intended cooling situation has $T_0>0$ and $k>0$.

The defining Wronskian identity gives the first-order linear differential equationWork first on an interval where $y_1\ne 0$. Dividing by $y_1$ recognizes a derivative:This is reduction of order. Choose the nonzero Wronskian from the Abel identity to obtain linear independence. The constant $C$ only adds a multiple of $y_1$. If $y_1$ has zeros, this quotient formula is local; the resulting solution can be continued across ordinary zeros using the original second-order linear differential equation.