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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.

If a nonnegative continuous function on $[0,h]$ satisfies $v(t)\le ct+A{\int }_0^{t}v(s)ds$ with $c,A>0$, put $W(t)=ct+A{\int }_0^{t}v(s)ds$. Then $v\le W$, $W(0)=0$ and $W^{\prime }\le c+AW$. Multiplication by the integrating factor $e^{-At}$ and integration prove the displayed estimate. For $A=0$ its limiting interpretation is $v(t)\le ct$.

The integral of a source over a directed line, weighted by exponential absorption between each source point and a detector. For line coordinates $x=p\mathbf{n}+s\mathbf{d}$ and a detector at the negative end, it is $\int f(p,s)\exp [-{\int }_{-\infty }^s\mu (p,r)dr]ds$. The integrating factor of a transport equation derives the weight. Reversing the detector direction changes the weight for spatially varying attenuation, so the two directed measurements need not coincide. Setting $\mu =0$ recovers the Radon transform.

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 $p\ne 1$ on a positive real branch, put $v=y^{1-p}$ and $\kappa =(1-p)a$. The Bernoulli power substitution gives $v^{\prime }-\kappa v=(1-p)q{e}^{\lambda x}$. Its integrating factor is $e^{-\kappa x}$, yieldingThe resonant case needs an extra factor x because the forcing has the same exponential dependence as the homogeneous solution. Recover $y=v^{1/(1-p)}$ only on an appropriate real branch. The case $p=1$ is already a first-order linear differential equation; a zero solution or zero-crossing behavior must be checked in the original equation separately.

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 bounded function on $[0,\infty )$ has a Laplace transform satisfying the displayed bound, by integration of $|y(t)e^{-pt}|$. Consequently its Laplace transform cannot have a pole of order two or higher at $p=0$. This can eliminate the arbitrary constant introduced when multiplication by $t$ turns an ordinary differential equation into a first-order differential equation for its Laplace transform.

For example, transforming $t{y}^{\prime \prime }+(t-1)y^{\prime }-y=0$ with $y(0)=1$ gives $p(p+1)Y^{\prime }+(3p+2)Y=2$. An integrating factor $p^2(p+1)$ yieldsThe bound forces $C=0$, selecting $y=e^{-t}$. This selection can be justified without assuming unproved endpoint regularity for every solution: $e^{-t}$ and $t-1$ have nonzero Wronskian $t{e}^{-t}$ on $(0,\infty )$, so they form a fundamental system of solutions. Boundedness removes $t-1$, and the limit at zero fixes the remaining coefficient. The endpoint limit by itself does not select a unique solution.

In the Black-Scholes model, suppose a stock pays continuous dividend yield $q$, a claim pays cash flow $f(t)S_{t}dt$, and its terminal financial payoff is $c{S}_T$. If the discounted cash-flow integral is finite, its risk-neutral pricing value is the displayed expression. To derive it, substitute $V=S\varphi (t)$ into the Black-Scholes equation with claim dividends, including stock drift $(r-q)S$. This gives ${\varphi }^{\prime }-q\varphi =-f$, $\varphi (T)=c$, which an integrating factor solves. The interest rate cancels because all payments are proportional to the same stock.

For bounded linear operators, define $C(\beta )=[A,e^{-\beta B}]$. The product rule gives $C^{\prime }+BC=[B,A]e^{-\beta B}$ and $C(0)=0$. Multiplying by the integrating factor $e^{\beta B}$ yieldsThe identity is useful when the two linear operators do not commute: it preserves their order rather than incorrectly differentiating an operator exponential as if all its factors were scalars. For unbounded linear operators, the products and differentiations need a common justified operator domain.

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.

A corotational Jeffreys fluid obeys $\stackrel{\circ }{S}+S/\tau =2{\mu }_0\stackrel{\circ }{D}+2{\mu }_{1}D/\tau$, with deviatoric stress $S$, rate-of-strain tensor $D$, Jaumann derivative and relaxation time $\tau >0$. Put $A=S-2{\mu }_{0}D$. In the corotating-frame representation of a Jaumann derivative, ${\tilde{A}}^{\prime }+\tilde{A}/\tau =2({\mu }_1-{\mu }_0)\tilde{D}/\tau$. An integrating factor gives an exponentially weighted history transported between the rotating frames. The infinite-past formula selects histories with $e^{t/\tau }\tilde{A}(t)\to 0$ as $t\to -\infty$; finite initial data also contribute an exponentially decaying homogeneous term.

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.

Switch on a constant electric field ${\mathbf{E}}_0$ at $t=0$. For prescribed initial current density $\mathbf{J}(0)$, the exact Drude model solution is the displayed formula. For electrons, the corresponding drift velocity isStarting from zero drift velocity, the current density reaches $1-e^{-1}\approx 0.632$ of its limiting value after one momentum relaxation time. Switching the electric field off at $T$ instead gives $\mathbf{J}(t)=\mathbf{J}(T)e^{-(t-T)/\tau }$ for $t\ge T$. A finite step of electric field changes the acceleration immediately but leaves drift velocity and current density continuous.

More generally, multiplying the current density equation by an integrating factor givesThis is the causal response of a first-order relaxation equation: only earlier electric fields enter the integral. It also shows why instantaneous Ohm's law is an approximation when the electric field changes appreciably during a momentum relaxation time.

For pairwise distinct rates $a,b,c$, the response isSet $v=y^{\prime }-by$. Solving $v^{\prime }-av=e^{cx}$ with $v(0)=0$ gives $(e^{cx}-e^{ax})/(c-a)$; solving the remaining first-order equation by an integrating factor gives the formula. Equivalently this is the convolution of $e^{cx}$ with the zero-displacement unit-impulse response $(e^{ax}-e^{bx})/(a-b)$ on the positive half-line. In the resonant case $c=b\ne a$, the finite limit is $y=[e^{ax}-e^{bx}-(a-b)x{e}^{bx}]/(a-b{)}^2$. The apparent singularity of the distinct-rate expression is removable, and the limit exhibits the polynomial factor of a resonant exponential particular solution.

The dissipation potential $\Omega =\alpha \Vert Q{\Vert }^{n+1}/(n+1)-\mathrm{tr}(AQ)/\tau$, with the Frobenius norm, gives $\dot{A}+A/\tau =\alpha \Vert Q{\Vert }^{n-1}{Q}^T$. An integrating factor proves the displayed memory formula. For $n>0$, its nonlinear driving term is defined continuously as zero at $Q=0$. Mechanical dissipation is $\mathrm{tr}(Q\dot{A})=\alpha \Vert Q{\Vert }^{n+1}-\mathrm{tr}(AQ)/\tau$, which must be nonnegative on thermodynamically admissible histories.

The Hasselmann stochastic climate model represents slowly responding climate variables driven by fast weather fluctuations as well as external forcing. Klaus Hasselmann's 1976 paper separates fast and slow variables, retains fluctuations lost by simple averaging, and explains why stabilizing climate feedbacks are needed to prevent indefinitely growing variability. Its general framework is broader than a single-temperature equation.^[ref]

Consider an original scalar specialization for a spatially averaged temperature anomaly $T$, with an effective ocean mixed layer heat capacity per unit area $C$:Here $F$ is imposed radiative forcing, $\lambda$ is the net restoring coefficient after the chosen climate feedbacks, and $\eta$ represents random net heat flux from unresolved weather. The anomaly energy per unit area is $CT$, so its tendency equals the net input. The weather term represents exchange with omitted degrees of freedom; it does not assert that an isolated ocean creates energy. $F$ and $\eta$ have units $\mathrm{W}{\mathrm{m}}^{-2}$, and $\lambda$ has units $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$.

First set $\eta =0$, start from $T(0)=0$, and switch to constant $F_0>0$. The integrating factor $e^{\lambda t/C}$ givesFor $C=2.0\times 10^8\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$, $\lambda =2.0\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$ and $F_0=4.0\mathrm{W}{\mathrm{m}}^{-2}$, the relaxation time is $10^8\mathrm{s}$, about $3.17$ years using 365.25 days per year. At $t=\tau$, $T=2(1-e^{-1})\mathrm{K}\simeq 1.264\mathrm{K}$, with eventual warming $2\mathrm{K}$. Doubling $C$ doubles the response time but leaves this equilibrium unchanged. This separates thermal inertia from climate sensitivity.

Now set $F=0$ and idealize the weather forcing as Gaussian white noise withThe Dirac delta function has inverse-time units, so $D$ has units $(\mathrm{W}{\mathrm{m}}^{-2}{)}^2\mathrm{s}$. The rigorous stochastic differential equation iswhere $W_t$ is Brownian motion. This is an Ornstein-Uhlenbeck process. In stationary initialization its variance and autocovariance function areDefine the two-sided angular-frequency power spectral density by $R_T(u)=(2\pi {)}^{-1}{\int }_{-\infty }^{\infty }{S}_T(\omega )e^{i\omega u}d\omega$. The Fourier transform of the exponential autocovariance function, or the squared response to a sinusoidal input, givesThus fast fluctuations are suppressed; the climate output is red over the frequency range above the corner $|\omega |=1/\tau$, with a low-frequency plateau. This is a colored noise spectrum without a periodic driver. The corner is $f_c=1/(2\pi \tau )$ in cycles per second, not $1/\tau$. The continuous white input is only an approximation on times longer than weather's correlation time.

For the same $C,\lambda$ and $D=10^8(\mathrm{W}{\mathrm{m}}^{-2}{)}^2\mathrm{s}$, $v_{\infty }=0.25{\mathrm{K}}^2$ and the stationary standard deviation is $0.50\mathrm{K}$. From fixed $T(0)=0$, the variance grows as $v_{\infty }(1-e^{-2t/\tau })$, even when the forced expected value is predictable. Integrating $S_T/(2\pi )$ over all angular frequencies returns $D/(C\lambda )$, checking the normalization independently. If $\lambda =0$ the unforced variance instead grows as $2Dt/C^2$; if $\lambda <0$, deviations amplify. Neither case has the stationary spectrum above.

A red spectrum alone cannot establish attribution of recent climate change: internal variability, changing forcing and measurement effects must be distinguished. This scalar model supplies a calculable baseline for variability and persistence, not a universal spectrum for all climate timescales or a full attribution analysis.

For $y^{\prime }+a(x)y=H(x)$ with continuous $a$, the integrating factor $\mu =\exp (\int a)$ gives the displayed solution. It is continuous at the step and solves the differential equation classically on each side; the derivative jumps by one. The value assigned to the Heaviside step function at the single point zero does not affect the integral. A discontinuity in $y$ would instead produce an unintended delta term in its distributional derivative.

The displayed implicit ordinary differential equation has a singular initial point at the origin. On a nonconstant regular branch, treating $x$ as a function of $y$ gives $dx/dy+x\cot y=2\cos y$. The integrating factor $\sin y$ yields the first integral of an ordinary differential equationOne can verify this identity without dividing by $\sin y$: its derivative along a solution is $\sin y-(2\sin y-x)\cos y{y}^{\prime }=0$.

The initial point forces $C=0$, so $\sin y(x-\sin y)=0$. Near the origin this gives the two differentiable solutions $y=0$ and $y=\arcsin x$, with initial slopes zero and one. A division by $\sin y$ would lose the first solution. A piecewise switch between the two at zero would fail differentiability because their one-sided slopes differ.

There is no conflict with the Picard-Lindelöf theorem: solving explicitly for $y^{\prime }$ requires division by $2\sin y-x$, which vanishes at the initial point. A regular implicit equation or a local uniqueness theorem must be checked before claiming a unique solution through a singular point.

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 a differential one-form $Pdx+Qdy$, seek a nonzero integrating factor $\mu (x)$. Expanding $(\mu P{)}_y=(\mu Q{)}_x$ gives the displayed equation where $Q\ne 0$. Such a factor exists locally when $(P_y-Q_x)/Q$ depends only on $x$, and is $\mu =\exp \int [(P_y-Q_x)/Q]dx$. Likewise, a factor depending only on $y$ obeys ${\mu }^{\prime }/\mu =(Q_x-P_y)/P$ where $P\ne 0$. The resulting product is an exact differential form on a sufficiently small regular domain; singular lines of the factor need separate treatment.