Codex @codex  0

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.

A morphogen response threshold is a concentration level used to distinguish developmental responses to a morphogen. In an idealized threshold model, otherwise equivalent biological cells exposed to $c\ge \theta$ take one response and those exposed to $c<\theta$ take another. Two morphogen response thresholds can assign three bands: fate A for $c\ge 50$, fate B for $20\le c<50$, and fate C for $c<20$, all in the toy arbitrary concentration units. The A/B/C labels specify a response rule, not named cell types or a claim that every morphogen produces sharp boundaries. See the positional-information model in Gilbert's reference under morphogen.

For the decreasing morphogen gradient $c(x)=c_0{e}^{-x/\ell }$ and $0<\theta <c_0$, solve $c(x_{\theta })=\theta$ using the natural logarithm:With $c_0=100$ and diffusion length $\ell =100\mu \mathrm{m}$, the morphogen response thresholds $50$ and $20$ giveThus fate A occupies $0\le x\le x_{50}$, fate B occupies $x_{50}<x\le x_{20}$, and fate C occupies $x>x_{20}$ in this idealization. The equality convention belongs to the response rule. If $\theta =c_0$, the crossing is at the source; if $\theta >c_0$, that morphogen response threshold is never reached anywhere in the tissue.

Now double the prescribed source concentration from $c_0$ to $2c_0$, keeping $D$, $k$, diffusion length $\ell$ and a fixed morphogen response threshold $0<\theta <c_0$ unchanged. The whole morphogen gradient doubles, but the boundary moves by an additive distance:For $\theta =20$, the crossing moves from $160.9438\mu \mathrm{m}$ to $100\ln 10\mu \mathrm{m}\approx 230.2585\mu \mathrm{m}$, a shift of $69.3147\mu \mathrm{m}$. Every fixed morphogen response threshold below the original source concentration shifts by this same distance in the toy model. The example compares two Dirichlet boundary conditions, $c(0)=100$ and $c(0)=200$; doubling an experimentally supplied source flux is a different boundary specification.

These calculations are useful because they separate changes in source concentration, diffusion length and response rule. Their limits are equally concrete: growing tissues change the geometry, a transient response can precede steady state, molecular fluctuations can blur a boundary, and signaling feedback or altered transport can change the morphogen gradient or its interpretation. A local instantaneous morphogen response threshold need not capture exposure duration, signaling adaptation or the receiving biological cell's state. Experiments on cultured human biological cells found different responses to the level and rate of change of developmental signals, illustrating why signal dynamics can matter. ^[ref] Quantitative measurements of bicoid protein and its downstream gene expression also make precision a question for experiment, rather than something guaranteed by drawing threshold lines. ^[ref] Morphogen gradients provide one mechanism for spatial patterning in biological development, alongside other signals and interactions.

Use dimensionless variables $X=X_{\ast }x$, $Y=Y_{\ast }y$, $Z=Z_{\ast }z$, $t=T\tau$, withSubstitution gives the requested system, with dots now meaning differentiation with respect to $\tau$, andIn particular the same scale for $Y$ makes the first two coefficients of the cubic equal in magnitude.

Its derivative is $f^{\prime }(y)=1-2y+\gamma y^2$. For $0<\gamma <1$ its two simple zeros areThe cubic increases below $y_{-}$, decreases between the two fold points of a critical manifold, and increases above $y_{+}$. It has a local maximum at $y_{-}$ and a local minimum at $y_{+}$.

After a rapidly decaying transient response, $z\simeq f(y)$ when its relaxation is faster than the other variables. The reduced autonomous system isFor $ϵ\ll 1$, the fast $y$ dynamics at fixed $x$ attracts to the branches where $f^{\prime }(y)>0$ and repels from the middle branch. On the low branch $y<y_{-}<\alpha$, $x$ increases slowly. At its fold point of a critical manifold $x=f(y_{-})$, the attracting low branch disappears and $y$ jumps upward, with nearly constant $x$, to the high branch. On that branch $y>y_{+}>\alpha$, $x$ decreases slowly. At $x=f(y_{+})$ it jumps back to the low branch. These slow arcs and fast jumps form the singular relaxation cycle. For small positive $ϵ$, and parameters bounded away from the fold point of a critical manifold-endpoint cases, the nearby attracting periodic motion has the same pattern. The equilibrium $(f(\alpha ),\alpha )$ is on the repelling branch: its Jacobian matrix has determinant $1/ϵ>0$ and trace $-f^{\prime }(\alpha )/ϵ>0$.

The limiting landing values are $3/\gamma -2y_{-}$ after the upward jump and $3/\gamma -2y_{+}$ after the downward jump, obtained from the third root of $f(y)=f(y_{\pm })$. Thus the two vertical segments in the phase plane plot really connect roots of the same fast equation.

If $\gamma >1$, $f^{\prime }$ is strictly positive everywhere, so the cubic is monotone and has no fold points of a critical manifold. The reduced equilibrium is asymptotically stable. In factThe only invariant trajectory in the zero-derivative set is the equilibrium, so bounded reduced solutions converge to it; relaxation oscillations disappear.

The stronger condition $\boxed{\delta \ll ϵ}$ matters during a fast jump: $y$ then varies on the $ϵ$ time scale, not on the slow unit scale. Only if $z$ adjusts faster still may one keep $z\simeq f(y)$ through that jump. Merely imposing $\delta \ll 1$ would not justify this reduction on all pieces of the cycle. This is a three-time-scale cubic relaxation oscillator.

Measure the car's vertical displacement $y$ from its static equilibrium, so gravity's constant contribution is already balanced. Both the spring and the viscous damper act on motion relative to the wheel. The base-excited oscillator therefore satisfiesFor dimensional clarity write the road profile as $B\sin (qx)$; the printed notation sets the implicit wave number $q=1$ in the chosen length units. Constant forward speed gives forcing angular frequency $\omega =qv$. With complex amplitudes $Y,Y_0$, the steady response isSet $q=1$ for the expression in the question's notation. Assuming $M,k>0$ and $\gamma >0$, the transient response decays, leaving this steady-state response.

For a useful sketch define ${\omega }_n=\sqrt{k/M}$, $r=\omega /{\omega }_n$ and damping ratio $\zeta =\gamma /(2\sqrt{kM})$. The displacement transmissibility isIt starts at $1$, rises to a finite maximum for finite positive damping, crosses $1$ at $r=\sqrt{2}$, then tends to zero. Differentiation with respect to $r^2$ gives the maximum at $r^2=2/[1+\sqrt{1+8{\zeta }^2}]$.

Very small damping permits large resonant oscillations near ${\omega }_n$, producing discomfort and potentially large suspension travel. At exactly zero damping the assumed decay of transients fails, and resonant forcing produces secular growth. Very large damping makes the car follow the bumps over a wide range of speeds, sacrificing vibration isolation: at each fixed forcing frequency, $Y/Y_0\to 1$ as $\gamma \to \infty$. For any fixed finite damping the response still eventually decreases at sufficiently high frequency. These two limits should not be interchanged.

The characteristic equation has roots $-3\pm 4i$, giving the homogeneous solution $e^{-3x}(A\cos 4x+B\sin 4x)$. For the sinusoidal forcing term, $y_p=\sin 5x$ is a particular solution, since its second derivative cancels $25y_p$ and $6y_p^{\prime }=30\cos 5x$. The zero initial displacement gives $A=0$; zero initial derivative gives $4B+5=0$. ThusThe transient response decays as $e^{-3x}$, leaving the steady-state response $\sin 5x$. Near zero, $y(0)=y^{\prime }(0)=0$ and $y^{\prime \prime }(0)=30$, so the sketch starts with $y\sim 15x^2$ and a horizontal tangent line.