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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.

For the harmonic oscillator equation $y^{\prime \prime }+{\omega }^{2}y=J\delta (t-t_{\ast })$, with $\omega >0$, the displacement is continuous and the velocity increases by $J$ at the impulse. If the incoming displacement is zero and $J=-y^{\prime }(t_{\ast }-)$, the outgoing displacement and velocity both vanish. The subsequent homogeneous solution is identically zero. Equivalently, the causal forced response $(J/\omega )H(t-t_{\ast })\sin (\omega (t-t_{\ast }))$ cancels the pre-existing oscillation, where $H$ is the Heaviside step function.

A particular solution is any one solution of a forced linear differential equation. Every solution is the sum of that particular solution and an arbitrary homogeneous solution.

The integrating factor is $e^{-2x}$, so $(e^{-2x}y{)}^{\prime }=e^{(\lambda -2)x}$. Integrating gives the general solutionThe particular solution is only defined up to a homogeneous solution. Subtract $e^{2x}/(\lambda -2)$ from it, absorbing that multiple into the arbitrary constant. The same family can then be writtenHolding $\tilde{C}$ fixed, the quotient tends to $x{e}^{2x}$, the derivative of $e^{\lambda x}$ with respect to $\lambda$ at $2$. Thus the resonant family is $y=e^{2x}(\tilde{C}+x)$. Direct differentiation gives $y^{\prime }-2y=e^{2x}$, so it is the full general solution at resonance. This resonant exponential forcing in a first-order equation limit concerns a reparametrized family, not a fixed value of the original divergent constant $C$. The initial condition gives $\tilde{C}=2e^{-2}-1$, hence

Let $N=a+b$ and measure Lionel's fortune in unit stakes. This is gambler's ruin: the fortune moves by $+1$ with probability $p$ and $-1$ with probability $q$, stopping at $0$ or $N$. Let $m_i$ be the expected value of the absorption time starting at $i$. It is finite: from any interior state, a run of $N$ wins or losses has a fixed positive probability of absorption within $N$ steps, giving a geometric bound on survival over successive blocks. First-step analysis therefore givesFor $p\ne q$, the homogeneous solution of the recurrence is $A+B(q/p{)}^i$, and a particular solution is $-i/(p-q)$. Put $\rho =q/p$. Imposing the two endpoint boundary conditions gives the expected duration of biased gambler's ruinConsequently the expected number of games isThe denominator and numerator have matching signs, so this is positive. As a check its continuous limit at $p=q=1/2$ is $ab$, the expected duration of symmetric gambler's ruin.

Away from the impulse, the harmonic oscillator equation has a homogeneous solution. Before $x=\pi /2$, the initial conditions select $y=\cos x$. The jump condition for an impulse requires continuity of $y$ and a unit jump in $y^{\prime }$: a jump in $y$ would produce a derivative of the Dirac delta distribution, which is absent from the forcing, while integrating through the impulse gives $[y^{\prime }]=1$.

Just before the impulse, $y=0$ and $y^{\prime }=-1$. Just afterwards, $y=0$ and $y^{\prime }=0$, so uniqueness for the harmonic oscillator equation makes the subsequent solution identically zero. Equivalently the causal response is $H(x-\pi /2)\sin (x-\pi /2)$, with $H$ the Heaviside step function, and it cancels the original oscillation.

The solution isThe impulse cancellation of a harmonic oscillator produces a continuous graph with a corner at $(\pi /2,0)$, followed by a horizontal zero line.

Transform in time and use the zero interior initial conditions. The Laplace transform of a derivative formula turns the wave equation into $c^2{Y}_{xx}=p^{2}Y$. Its transformed endpoint conditions are $Y(0,p)=a/p$ and $Y(L,p)=0$. A homogeneous solution satisfying the right condition is a multiple of $\sinh [p(L-x)/c]$; the left condition fixes that multiple. Thus, for $\mathrm{Re}p>0$,Here $L,c>0$ and the denominator has no zeros in this half-plane. The initial-boundary corner at $(0,0)$ is deliberately discontinuous: the applied step launches propagating fronts, so one should not demand a globally smooth solution across them.

Write $E=\mathrm{erf}z$, $g=e^{-z^2}$ and $b=2/\sqrt{\pi }$. The first correction in the inner expansion satisfies ${\mathcal{L}}_0{Y}_1=2zE$. Direct differentiation verifies that $zE+bg$ is a particular solution. Adding the constant and error function homogeneous solutions gives $Y_1=zE+bg+A+BE$.

The left boundary condition forces $A=-b$. Re-expanding the outer expansion at $x=sz$ givesTherefore $Y_1\sim z$, with no constant term in its large-$z$ overlap. Since $E\to 1$ and $g\to 0$, this sets $B=b$. Thus the required half-order correction isIt vanishes at zero and has the correct linear overlap. Keeping this $O(\sqrt{\epsilon })$ term is essential even though the outer solution has no correction of that order.

An integrating factor explains the second hint and the first-order inner calculation. Formultiplication by $e^{z^2}$ gives $(e^{z^2}{Y}^{\prime }{)}^{\prime }=e^{z^2}(2a_{1}z+a_2+a_3\mathrm{erf}z)$. The term $a_{1}z$ already produces $2a_{1}z$, and integrating the remaining terms twice gives $a_2{E}_0+a_3{E}_1$, in terms of the Gaussian drift primitives. The homogeneous solutions are a constant and the error function. HenceFor the piecewise forcing in the inner problem, $F$ replaces $a_{1}z$ and provides the continuously matched particular solution for $2zH(-z)$. The coefficients $2c$ and $2d$ then account for the $2Y_0$ forcing.

The homogeneous solution isFor the constant and $\cos 2\varphi$ forcing terms, a particular solution isIndeed, applying $d^2/d{\varphi }^2+1$ multiplies the $\cos 2\varphi$ coefficient by $-3$. ThusThe $B\sin \varphi$ term only changes the definition of $b$ at this order, so one may set $B=0$.

For $y\prime -ay=e^{\lambda x}$, subtracting an appropriate homogeneous solution from the particular solution removes its apparent singularity as $\lambda \to a$. The resulting derivative in the parameter produces the resonant particular solution $x{e}^{ax}$. The arbitrary constant must be reparametrized before taking the limit.