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The characteristic equation of the constant-coefficient linear ordinary differential equation iswhose roots of a polynomial are $r=-1\pm i/2$. Taking real and imaginary parts of the corresponding exponential functions gives the real general solution

Since $x>-1$, divide by $x+1$ to obtain the first-order linear ordinary differential equationAn integrating factor is $\mu (x)=e^{-3\log (x+1)}=(x+1{)}^{-3}$. Multiplication by this factor givesAfter integration, $(x+1{)}^{-3}y=\frac{1}{2}(x+1{)}^2+C$, so

Seek a particular solution $y_p=ax+b$. Substitution into the linear ordinary differential equation gives $6ax+2b=6x+2$, so $a=b=1$. Add the complementary solution from part (a):The two prescribed values yield $A+B=5$ and $A/2+B/4=3/2$. Thus $2A+B=6$, so $A=1$ and $B=4$. ThereforeDirect substitution gives $y(1)=7$ and $y(2)=9/2$, and the homogeneous solution contributes zero to the left-hand differential operator.

On an interval avoiding $x=0$, divide by $x$ to obtain the first-order linear ordinary differential equationAn integrating factor is $\mu (x)=\exp \left(\int -2/xdx\right)=x^{-2}$. Multiplying through and using the product rule givesBy integration by parts, $\int x\cos xdx=x\sin x+\cos x+C$. ConsequentlyThe formula differentiates back to the original equation. Although division excluded $x=0$, this formula has a smooth extension there with $y(0)=0$, satisfying the original equation at zero as well.

Let ${\hat{u}}_0$ be the Fourier transform of the initial temperature. Solving the scalar linear ordinary differential equation givesRecover $u$ using the Fourier inversion theorem, with inverse factor $(2\pi {)}^{-2}$ for this two-dimensional convention. Equivalently, take the convolution of the initial data with the two-dimensional heat kernel. On a periodic square, the same exponential applies to each discrete Fourier mode, with $k_x=2\pi m/L_x$ and $k_y=2\pi n/L_y$.

For a first-order meromorphic linear ordinary differential equation $Y_t=A(t)Y$, with a coefficient pole of order $p>1$ in a fixed local coordinate and gauge, its Poincaré rank is $p-1$. At infinity one must transform the differential system, not just the matrix-valued function: $Y_{\lambda }=A_0{\lambda }^{m}Y+\cdots$ becomes $Y_t=-A_0{t}^{-m-2}Y+\cdots$ under $t=1/\lambda$, giving rank $m+1$. A singular gauge can lower an apparent rank; distinct leading eigenvalues prevent the elementary degeneracy in the unramified setting.

At a regular point of a second-order linear ordinary differential equation, a nontrivial solution cannot have both $x$ and $x^{\prime }$ zero, by initial-value uniqueness. Its zeros are therefore simple. Under the linearization of a Riccati equation, such a zero $t_{\ast }$ produces $y=x^{\prime }/(cx)\sim 1/[c(t-t_{\ast })]$. Zeros separate the intervals on which the Riccati equation solution is finite.

On an interval where the substitution is real and $x\ne 0$, a factor formed by exponentiation can remove inverse powers from the coefficients of a linear ordinary differential equation. Forthe substitution $y=x^{n}z$ givesIt becomes a constant-coefficient differential equation exactly when $2n+b=0$, $an+d=0$ and $n(n-1)+bn+e=0$. Derive this by the product rule; checking only the coefficient of $z^{\prime }$ is insufficient.

For constant acceleration vector $\mathbf{g}$ and $\gamma >0$, a projectile initially at ${\mathbf{r}}_0$ with velocity ${\mathbf{v}}_0$ hasThe integrating factor $e^{\gamma t}$ solves the first-order linear ordinary differential equation for the velocity; one more integration proves the position formula. The terminal velocity is $\mathbf{g}/\gamma$, and $1/\gamma$ is the relaxation time. For $\mathbf{g}=(0,-g)$ and ${\mathbf{r}}_0=0$, eliminating time when $V_{0x}\ne 0$ givesFor finite $t\ge 0$, the physical domain is $0\le \gamma x/V_{0x}<1$. If $V_{0x}=0$, the path is vertical and cannot generally be represented as a single-valued graph $z(x)$. Expanding at fixed $x$ as $\gamma \to 0$ recovers $z=(V_{0z}/V_{0x})x-g{x}^2/(2V_{0x}^2)$.

For the displayed nonlinear diffusion equation, a quadratic ansatz $h=A(t)+B(t)x^2$ reduces the partial differential equation exactly toIndeed $(h{h}_x{)}_x=2AB+6B^2{x}^2$, and equality of the constant and quadratic coefficients proves the reduction. For nonzero curvature, integration gives $B=-1/[6(t-t_0)]$. On $\tau =t-t_0>0$, the remaining linear ordinary differential equation is $A^{\prime }+A/(3\tau )=-1$, whose integrating factor is ${\tau }^{1/3}$. Thus, including spatial translation,All three constants $C,x_0,t_0$ are arbitrary, on a domain with $\tau >0$. When the quadratic coefficient vanishes, the more general affine ansatz gives $h=px+(p^2-1)t+q$.

A narrower ansatz $h=a{t}^{-1/3}(1-x^2/t^{2/3})-bt$ fixes two coefficients together and permits exactly $(a,b)=(1/6,3/4)$ or $(0,1)$. The zero value of $a$ is a valid spatially constant branch, not an algebraic exception to discard. These finite-parameter families are exact solutions; they are not the complete general solution of the nonlinear partial differential equation, and their sums need not be solutions.

For a radial function in three dimensions, $(\Delta -1)y=g(x)$ becomes $y^{\prime \prime }+2y^{\prime }/x-y=g(x)$. Setting $u=xy$ gives $u^{\prime \prime }-u=xg(x)$, reducing it to a constant-coefficient linear ordinary differential equation. If $g=e^{-3x}/x^2$, a decaying particular solution obtained by variation of parameters isThe small-argument expansion of the exponential integral givesThe undetermined decaying homogeneous coefficient $\alpha$ is fixed by a boundary condition or by a matched asymptotic expansion.

Consider an original ideal model with sample volume $V_s$, bath volume $V_b$, and initial sample solute concentration $c_0$. Assume each volume is fixed, a freely permeant, nonbinding solute, no chemical reactions, complete mixing on each side, and equal concentrations at equilibrium. Each fresh bath initially contains none of this solute. The protein is perfectly retained in this model.

The first equilibration redistributes the initial amount of substance $c_0{V}_s$ through both compartments. Conservation givesThis denominator includes the sample itself. For the chosen $V_s=1\mathrm{mL}$ and $V_b=9\mathrm{mL}$, $c_1/c_0=1/10$, rather than $1/9$. Nine-tenths of the initial solute is in the bath and one-tenth remains in the sample; nothing has vanished during equilibration.

Discarding the equilibrated bath removes its solute. A second fresh bath starts with only $c_1{V}_s$ of solute in the entire system, so the same argument applies again. Induction gives the geometric progressionHere $n$ counts completed equilibrations with fresh baths: $n=1$ is the first bath, $n=2$ follows one replacement, and $n=3$ follows two replacements. The normalized sample concentrations are respectively $0.1$, $0.01$ and $0.001$. Leaving the first bath in place for longer cannot lower the ideal equilibrium value below $0.1$.

The same conservation argument handles a fresh bath already containing a permeant solute at fixed concentration $b$:Thus the sample approaches $b$, not necessarily zero. For example, if a hypothetical new buffer component starts at zero in the sample and at $b$ in every bath, its sample concentration reaches $0.9b$, $0.99b$ and $0.999b$ in the three equilibrations. Outward removal and inward replacement follow the same mathematics.

Equilibrium fractions do not specify elapsed time. An illustrative well-mixed linear ordinary differential equation model within one bath isHere $K$ is an effective transfer coefficient with units of volume per time. Adding the two equations conserves $V_s{c}_s+V_b{c}_b$; subtracting their normalized forms givesThe difference undergoes exponential decay. A bath replacement resets $c_b$, renewing the driving difference without making $c_s$ jump instantly. The plotted points are equilibrium endpoints, not a measured time course or a timetable for an arbitrary protein.

Real sample volume can change through osmosis; proteins can be lost through leakage or adsorption to the semipermeable membrane. Consequently, a measured decrease in protein concentration alone cannot distinguish dilution from loss of protein. Measuring recovered volume as well as protein concentration permits an estimate of recovered protein mass.^[ref]

The part of a response that dies away after an initial condition or an input change. In a stable linear ordinary differential equation, the decaying homogeneous solution gives a transient response relative to a chosen particular solution. A homogeneous solution that grows or persists is not a decaying transient.

Viscous damping means a force proportional to, and opposite to, velocity: $F_{\mathrm{d}}=-b\dot{x}$ with $b\ge 0$. Its coefficient has units $\mathrm{kg}{\mathrm{s}}^{-1}$, equivalently $\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$. The name describes a force law; eddy-current damping can approximate the same law without a viscous fluid. Constant $m$, $b$ and $k$ produce a linear ordinary differential equation, allowing superposition of responses. Kinetic friction with approximately constant magnitude and quadratic drag obey different laws, so their decay need not have the exponential decay or critical threshold derived here.

Suppose two complex amplitudes satisfy ${\partial }_t|A{|}^2\le 2a|A||B|-2d|A{|}^2$ and ${\partial }_t|B{|}^2\le 2c|A||B|-2d|B{|}^2$, with positive $a,c$ and $d\ge 0$. Then $N=c|A{|}^2+a|B{|}^2\ge 2\sqrt{ac}|A||B|$ gives $\dot{N}\le 2(\sqrt{ac}-d)N$. A positive weighted norm avoids treating two coupling rates of different size as a single arithmetic sum. This is useful for nonautonomous linear ordinary differential equations and the bounded-modulation alpha-Omega growth estimate.