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For the linear partial differential equation $2y{u}_x-x{u}_y=0$, the method of characteristics gives $\dot{x}=2y$, $\dot{y}=-x$, $\dot{u}=0$. Thus $\varphi =x^2+2y^2$ is constant on each characteristic curve, whose nonzero level sets are ellipses. On a domain containing entire such ellipses, the solutions have the form $u=F(\varphi )$, subject to the desired regularity at the origin.

For the inhomogeneous equation $2y{u}_x-x{u}_y=xy(2y^2-x^2)$, a particular solution is $u_p=x^2{y}^2/2$, so the general solution is $u=x^2{y}^2/2+F(x^2+2y^2)$. Prescribing $u(x,1)=x^2$ fixes $F(s)=(s-2)/2$ only for $s\ge 2$. The inner ellipses $s<2$ never intersect that data line, so the boundary values leave their solution values undetermined. Smooth extensions must match the prescribed $F$ at $s=2$; choosing the same linear formula everywhere gives one polynomial extension.

Let $E$ be the forward shift on sequences and $P$ the polynomial defining a constant-coefficient linear recurrence relation. For a nonzero base $r$, $P(E)r^n=P(r)r^n$, so $r^n/P(r)$ is a particular solution when $P(r)\ne 0$. At a simple root of the characteristic equation of a linear recurrence, the correct trial acquires a factor $n$:The identity follows by substituting $(n+j)r^{n+j}$ into each shift term. A root of multiplicity $m$ analogously requires a degree-$m$ polynomial factor; the lower-degree homogeneous terms are supplied by the repeated characteristic root of a linear recurrence.

For $y^{\prime }+ay=g$ with real $a>0$ and $2\pi$-periodic forcing, comparing Fourier coefficients gives $(a+in){\hat{y}}_n={\hat{g}}_n$. There is no vanishing denominator, including the constant mode. With real cosine and sine coefficients $a_n,b_n$ for $g$ and $A_n,B_n$ for $y$, the equivalent equations giveWhenever termwise differentiation is justified, these formulas construct a periodic particular solution. For a forcing series with coefficients $O(n^{-2})$, the differentiated response coefficients are also summable, which supplies such a justification by uniform convergence. All other solutions add the homogeneous solution $C{e}^{-ax}$. Since this exponential cannot be periodic unless $C=0$, the periodic solution is unique. This is also a simple example of how a linear equation changes the amplitude and phase of each harmonic independently.

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.

A homogeneous linear differential equation is a linear differential equation with identically zero forcing term, $L[y]=0$. On a common interval, its solutions of a differential equation form the vector space $\ker L$, the kernel of a linear map. They include the zero function and are closed under addition and scalar multiplication. For any two solutions of a differential equation $u,v$ and constant scalars $\alpha ,\beta$, linearity gives $L[\alpha u+\beta v]=\alpha 0+\beta 0=0$: this is the superposition principle. The scalars must be constant; variable multipliers introduce extra derivatives.

For example, $y^{\prime }+2y=0$ has general solution $y=C{e}^{-2t}$, where $C$ is any constant real number. A short verification uses the product rule: $(e^{2t}y{)}^{\prime }=e^{2t}(y^{\prime }+2y)=0$, so $e^{2t}y$ is constant. For $y_1=e^{-2t}$ and $y_2=2e^{-2t}$, the sum is $3e^{-2t}$ and also solves the same homogeneous linear differential equation. At $t=1$, their values are approximately $0.13534$, $0.27067$ and $0.40601$. Every curve is a multiple of one exponential function; the solution space of a homogeneous linear differential equation has dimension one. Both $t$ and $y$ in this example and the following diagram are dimensionless mathematical variables.

Now keep the same differential operator $L[y]=y^{\prime }+2y$ but prescribe the forcing term $1$. The inhomogeneous linear differential equation $y^{\prime }+2y=1$ has particular solution $y_p=1/2$ and general solution $y=1/2+C{e}^{-2t}$. The initial conditions $y(0)=0,1,2$ select, respectively,At $t=1$, their values are approximately $0.43233$, $0.56767$ and $0.70300$. The midpoint $\frac{1}{2}(u_0+u_2)=u_1$ solves the same inhomogeneous linear differential equation, because $L[\frac{1}{2}(u_0+u_2)]=\frac{1}{2}(1+1)=1$. The sum $s=u_0+u_2=1+e^{-2t}$ instead satisfies $s^{\prime }+2s=2$, so it fails the equation with forcing term $1$. More generally, a linear combination of solutions of a differential equation for the same fixed nonzero forcing term retains that forcing term precisely when its scalar weights add to one. This is the affine space structure, rather than unrestricted superposition within that fixed family.

On the left, all three curves solve the same homogeneous linear differential equation. On the right, $u_0,u_1,u_2$ share the particular solution $1/2$, and the shaded difference $u_2-1/2$ is the homogeneous solution $\frac{3}{2}{e}^{-2t}$. The dashed sum has twice the forcing term, even though its starting value coincides with that of $u_2$. Adding the two initial conditions also adds their inputs; taking their midpoint averages both. This distinction makes the superposition principle useful for assembling responses to different inputs without mistaking that assembly for a solution of a differential equation with the original input. See MIT's superposition notes, including the particular solution one half.

An inhomogeneous linear differential equation has a nonzero forcing term. Its general solution is one particular solution plus the general solution of the associated homogeneous linear differential equation.

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 scalar linear differential equation of order $n\ge 1$ has the formHere the coefficients $a_j$ and the forcing term $g$ are prescribed functions of the independent variable $t$. A classical solution of a differential equation is an $n$-times differentiable function satisfying the identity at every point of its interval. The unknown function $y$ and its derivatives enter to the first power, without products between them or coefficients depending on $y$. Equivalently, the differential operator $L$ is a linear map: for constant scalars $\alpha ,\beta$,The forcing term may be zero, giving a homogeneous linear differential equation, or nonzero, giving an inhomogeneous linear differential equation. Linearity of $L$ does not mean that every function graph of a solution of a differential equation is a straight line.

To use the usual order-$n$ initial value problem theory on an interval $I$, assume that $a_n(t)\ne 0$ throughout $I$ and that all $a_j$ and $g$ are continuous functions there. Dividing by $a_n$ normalizes the highest derivative; specifying $y(t_0),y^{\prime }(t_0),\dots ,y^{(n-1)}(t_0)$ at $t_0\in I$ then selects one solution of a differential equation throughout $I$. A zero of $a_n$ needs separate analysis: one cannot divide through it and retain these conclusions. The first-order normalization appears in OpenStax's definition and standard form of linear equations; the corresponding system theorem is in MIT's existence and uniqueness notes for linear systems.

The coefficients need not be linear functions of $t$. For example, $t^2{y}^{\prime }+y=0$ is a linear differential equation on $t>0$, although $t^2$ is quadratic. Its solution of a differential equation with initial condition $y(1)=1$ is $y(t)=e^{1/t-1}$: its derivative is $y^{\prime }=-y/t^2$, so substitution gives $t^2{y}^{\prime }+y=0$, and $y(2)=e^{-1/2}\approx 0.60653$. The zero of the highest coefficient at $t=0$ explains the domain restriction. In contrast, $y^{\prime }+y^2=0$, $y{y}^{\prime }+y=0$ and $y^{\prime }+\sin y=0$ are nonlinear ordinary differential equations because of their dependence on the unknown function. Variable coefficients and nonlinear ordinary differential equations are different issues; see MIT's definitions and examples of first-order linear equations.

The practical benefit of linearity is that it separates the effect of the initial condition from the effect of the forcing term. Once one particular solution $y_p$ satisfies $L[y_p]=g$, every solution of a differential equation isHere $\ker L$ is the kernel of a linear map, consisting of solutions of a differential equation to the associated homogeneous linear differential equation. Indeed, $L[y-y_p]=g-g=0$, and conversely $L[y_p+h]=g+0=g$. Thus the family for a fixed forcing term is an affine space obtained by translating $\ker L$. If $g$ is not identically zero, the zero function is absent, so this family is not a vector space under the usual operations on functions. This short argument explains why a general solution is a particular solution plus a homogeneous solution.

Consider a random walk on $\{0,1,\dots ,n\}$ with absorbing states $0,n$. At each interior state it stays in place with probability $q$, and moves one position in either direction with probability $(1-q)/2$, where $0\le q<1$. Its mean absorption time from $k$ isProof. A sequence of at most $n$ consecutive leftward moves ensures absorption, with probability at least $((1-q)/2{)}^n>0$. Dividing time into blocks of length $n$ gives a geometric series bound on the survival probabilities, so the expected value is finite. Let $e_k={\mathbb{E}}_k[T]$. The law of total expectation and first-step analysis giveThus $e_{k+1}-2e_k+e_{k-1}=-2/(1-q)$. A particular solution of this linear recurrence relation is $-k^2/(1-q)$; its homogeneous solutions are $A+Bk$. The two boundary conditions give $A=0$, $B=n/(1-q)$, proving both the formula and its uniqueness. Holding slows the usual symmetric random walk by the factor $1/(1-q)$.

Choose a finite trial family of polynomial-exponential or trigonometric functions for a particular solution, substitute it into a constant-coefficient linear equation, and solve for its coefficients. If the forcing exponent is a characteristic root of multiplicity $m$, multiply the ordinary trial by $t^m$ to leave the homogeneous kernel. The method is efficient for forcing families preserved by differentiation.

The homogeneous solution is $A\cos x+B\sin x$. The left boundary condition removes $A$, and the right boundary condition gives $B\sin l=0$. Thus the kernel of a linear map associated with this boundary value problem isThe Fredholm alternative now gives the complete classification:For necessity at $l=m\pi$, multiply by the homogeneous solution $\sin x$ and use the Lagrange identity for a Sturm-Liouville operator to obtain ${\int }_0^{l}f\sin xdx=0$. For sufficiency, the particular solutionsatisfies $z(0)=0$ and $z^{\prime \prime }+z=f$. At the other endpoint,which is zero under the compatibility condition. All solutions are then $z+B\sin x$, with $B$ arbitrary. The assumed Lipschitz continuity of $f$ in particular makes $f$ continuous, so the displayed integral defines a classical solution. The length $l=0$ is excluded: it would not define a nonempty interval for this boundary value problem.

The nonzero endpoint data contribute to the Fredholm alternative condition. With $w=e^x\sin x$, the same integration by parts identity givesFor the specified forcing the left side is $A{\int }_0^{\pi }{\sin }^{2}xdx=A\pi /2$. HenceTo verify sufficiency, set $y=e^{-x}u$. Direct differentiation gives $\mathcal{L}y=e^{-x}(u^{\prime \prime }+u)$. At this value of $A$, a particular solution of $u^{\prime \prime }+u=(2/\pi )\sin x$ is $-(x/\pi )\cos x$. The complete family obeying the two boundary conditions isThe arbitrary constant records the one-dimensional kernel; existence does not imply uniqueness here.

When $n=0$, the forcing has nonzero integral ${\int }_{-1}^{1}1dx=2$, violating the Fredholm alternative condition. There is no bounded solution. Directly, integration gives $(1-x^2)y^{\prime }=x+B$; cancellation of its logarithmic divergence at $x=-1$ requires $B=1$, while cancellation at $x=1$ requires $B=-1$.

For $n\ge 1$, the constant $1/[n(n+1)]$ is a particular solution. Subtract it from any solution. The remainder satisfies the Legendre differential equation, whose general solution is a combination of $P_n$ and the logarithmically unbounded Legendre function of the second kind. Boundedness removes the latter coefficient, givingThe constant forcing is orthogonal to $P_n$ for $n\ge 1$, consistently with Orthogonality of Legendre polynomials and the Fredholm alternative.

The denominator of the travelling response vanishes atFor the positive speeds in the paper, a finite travelling response exists whenever $c\ne c^{\ast }$. At $c=c^{\ast }$, the dynamic condition would reduce to $p_1/\rho =0$, which is impossible. The finite-depth dispersion relation of a surface gravity wave with wavenumber $k$ is ${\omega }^2=gk\tanh (kh)$. At $k=1$, its phase speed $\omega /k$ equals $c^{\ast }$: the moving pressure pattern drives a natural surface gravity wave at its own speed, producing resonance.

The restriction concerns a bounded travelling response. To distinguish this from existence of a solution of an initial value problem, set $T=\tanh h$, $P=p_1/\rho$, and ${\omega }_{\ast }^2=gT$. Ifthe linearized free-surface boundary conditions reduce to the forced harmonic oscillatorFor example, if the forcing is switched on at $t=0$ from a flat surface with zero initial velocity, then, for $c\ne {\omega }_{\ast }$,This satisfies $H(0)=\dot{H}(0)=0$ and contains both free oscillations and the forced component. At $c={\omega }_{\ast }$ a solution of the initial value problem still exists, but its particular solution containswhich grows linearly with time. Thus the ideal inviscid flow model predicts resonant growth rather than a finite constant amplitude. Additional free oscillations can set the initial conditions but cannot remove that secular growth.

The characteristic equation of the homogeneous ordinary differential equation is $r^2-2r+5=0$, with roots $1\pm 2i$. Therefore $y_h=e^x(C\cos 2x+D\sin 2x)$. For the forcing term, try the particular solution $y_p=e^{-x}(ax+b)$. Substitution gives $8a=1$, $8b-4a=0$, so $a=1/8$, $b=1/16$. The two initial conditions give $C=-1/16$ and $D=0$. ConsequentlyBoth the value and the first derivative vanish at zero; direct differentiation verifies the inhomogeneous equation.

The characteristic equation of the homogeneous linear differential equation is $m^2+m-2=(m-1)(m+2)=0$, giving the homogeneous solution $A{e}^t+B{e}^{-2t}$. For the first forcing, a particular solution of the form $C{e}^{-t}$ gives $-2C=1$, soThe initial conditions impose $A+B=1/2$ and $A-2B=-1/2$, whence $A=1/6$, $B=1/3$. Thus the first solution isFor the second forcing, $e^t$ is already a homogeneous solution, so the usual exponential trial must be multiplied by $t$. Substitution of the particular solution $Ct{e}^t$ gives $3C{e}^t=e^t$, hence $C=1/3$. Now $y=A{e}^t+B{e}^{-2t}+t{e}^t/3$, and the initial conditions give $A+B=0$, $A-2B=-1/3$. Therefore the resonant solution isThe additional factor $t$ reflects resonance in a differential equation.

Two functions are linearly dependent if constants $c_1,c_2$, not both zero, satisfy $c_1{y}_1+c_2{y}_2=0$ identically. Differentiating also gives $c_1{\dot{y}}_1+c_2{\dot{y}}_2=0$. Since the determinant of these two equations is the nonzero Wronskian $W$, both constants must vanish. Hence the two solutions are linearly independent.

Differentiating the Wronskian and substituting the homogeneous linear differential equation givesIntegrating proves Abel's identity:For variation of parameters, impose ${\dot{a}}_1{y}_1+{\dot{a}}_2{y}_2=0$ on the proposed particular solution. Then $\dot{y}=a_1{\dot{y}}_1+a_2{\dot{y}}_2$, and substitution into the inhomogeneous differential equation leaves ${\dot{a}}_1{\dot{y}}_1+{\dot{a}}_2{\dot{y}}_2=f$. Solving this two-by-two linear system givesArbitrary constants of integration merely add a homogeneous solution.

For the constant positive coefficient $q={\omega }^2$, take $\omega >0$. Direct differentiation verifies that $y_1=\cos \omega t$ and $y_2=\sin \omega t$ solve the homogeneous linear differential equation; their Wronskian is $W=\omega$. The requested integrands areTaking $t_0=0$ yieldsThus $|a_1(t)|\sim t/(2\omega )$ grows with power one, whereas $a_2$ is bounded. Their combination simplifies to $y_p=-t\cos (\omega t)/(2\omega )+\sin (\omega t)/(2{\omega }^2)$; the final sine term is itself a homogeneous solution. The secular term is the response to resonance in a differential equation.

After multiplication by ${\sec }^{2}x$, recognize both sides as derivatives:The nontrivial first integral is thereforeIt necessarily depends on the derivative, correcting the printed notation $h(x,y)$. A second integration giveson the specified interval. Since $y(0)=0$, $C_2=0$. A finite derivative at zero also implies $y{y}^{\prime }\to 0$ there, forcing $C_1=0$. A positive particular solution is consequentlyThe negative of this function is the other sign choice. Near zero, $-2\log (\cos x)=x^2+x^4/6+O(x^6)$, so the positive branch has $y=x+x^3/12+O(x^5)$ and $y^{\prime }(0)=1$; the negative branch has derivative $-1$.

For the positive branch, $y^{\prime }=\tan x/y>0$. Its second derivative has numerator $-2\log \cos x-{\sin }^{2}x$, multiplied by a positive factor. This numerator is zero at zero and has derivative $2{\sin }^{3}x/\cos x>0$, so the curve is convex for positive $x$. It starts tangent to $y=x$ and rises without bound as $x\uparrow \pi /2$, with $y\sim \sqrt{-2\log (\pi /2-x)}$. The third panel of the sketch above shows this branch.

Write the linear differential equation as $\dot{\mathbf{x}}=M\mathbf{x}$, withIts characteristic polynomial is $(\lambda +4{)}^2+9$. Thus a choice of eigenvalues and eigenvectors isThe distinct eigenvalues give two linearly independent solutions, and the general complex solution isReal solutions have $a_2=\overline{a_1}$. Equivalently,They spiral counterclockwise towards the origin, which is a stable spiral.

In a forced linear differential equation, resonance in a differential equation occurs when exponential forcing excites a mode with the same exponent as its eigenvalue, producing a polynomial factor such as $t$ multiplying that exponential. Merely matching an exponent is not enough: its projection onto that mode must be nonzero. Here $t{e}^{{\lambda }_{j}t}$ still decays in absolute magnitude because the real part of ${\lambda }_j$ is negative; resonance refers to this extra factor, not necessarily to unbounded growth.

For each forcing vector, resolve it in the eigenvector basis:If $u_k$ is the coefficient of the $k$th eigenvector, its scalar equation is ${\dot{u}}_k={\lambda }_k{u}_k+{\sum }_j{c}_{kj}{e}^{{\lambda }_{j}t}$. Multiplying by $e^{-{\lambda }_{k}t}$ and integrating shows that the $j=k$ contribution is $c_{kk}t{e}^{{\lambda }_{k}t}$, whereas a $j\ne k$ contribution is $c_{kj}{e}^{{\lambda }_{j}t}/({\lambda }_j-{\lambda }_k)$. Therefore the exact conditions for no resonant response areUnder these conditions one particular solution iswhich contains no factor of $t$. For a real forcing, $p_2=\overline{p_1}$ and $q_2=\overline{q_1}$, and the two boxed conditions are conjugate to each other.

The complementary solution comes from the characteristic equation $r^4-a^4=0$, whose roots are $a,-a,ia,-ia$. It is thereforeFor the resonant exponential particular solution, the forcing is already a homogeneous exponential, so the method of undetermined coefficients needs the resonant trial $y_p=Kx{e}^{-ax}$. If $P(r)=r^4-a^4$, the identity $P(D)[x{e}^{rx}]=P^{\prime }(r)e^{rx}$ when $P(r)=0$ gives $P^{\prime }(-a)=-4a^3$. Hence $K=-1/(4a^3)$ andThe decay condition first eliminates the growing exponential, $C_1=0$. A nonzero combination of the sine and cosine has a fixed nonzero amplitude and cannot tend to zero, so $C_3=C_4=0$ as well. Both remaining exponential terms decay because $a>0$. Finally $y(0)=C_2=1$, giving the unique answerThis is an instance of resonance in a differential equation: multiplication by $x$ supplies the particular solution when the forcing exponent is a simple characteristic root.