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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.

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 characteristic polynomial factors as $(r-2)(r+1)$, so the complementary solution is $A{e}^{2t}+B{e}^{-t}$. Each exponential forcing resonates with a homogeneous mode; the method of undetermined coefficients therefore uses $t{e}^{2t}$ and $t{e}^{-t}$. Substitution gives the particular solution $t{e}^{2t}-t{e}^{-t}-3t$. ThusThe initial conditions give $A+B=0$ and $2A-B-3=0$. HenceThe factors of $t$ in the particular solution are essential: an unmodified exponential trial lies in the kernel of the differential operator.

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.

Define the nested integralThe order-$s^2$ equation in the inner expansion is ${\mathcal{L}}_0{Y}_2=2z{Y}_1$. The preceding result decomposes its forcing into $2z^{2}E+2bzE-2bz+2bzg$. The particular solutions combine toFor example, ${\mathcal{L}}_{0}H=E$ follows directly by differentiating the nested integral, so the equation can also be verified without relying on the supplied particular-solution formulas. At zero $Y_{2,p}=b^2$; add $-b^2+DE$ to impose the left boundary condition. As $z\to \infty$ this givesThe outer expansion in the overlap, derived in the previous section, requires the remaining constant to be $-\frac{1}{2}\log s$. Thus $D=C+b^2-\frac{1}{2}\log s$. The logarithmic matching constant is an error-function logarithmic switchback.

The complete answer through order $\epsilon$ in the inner expansion isHere $E=\mathrm{erf}z$, $g=e^{-z^2}$ and $b=2/\sqrt{\pi }$. This expression is zero at $z=0$, and its large-$z$ expansion agrees with $1+x+x^2/2-(\epsilon /2)\log x$. Together with $y_{\mathrm{out}}=e^x[1-(\epsilon /2)\log x]$, it covers both requested regions. The overlap uses $1\ll z\ll {\epsilon }^{-1/2}$, with the usual termwise ordering restrictions when matching higher terms.

If a single continuous approximation is desired, an additive composite expansion isThe apparent logarithmic divergence at $x=0$ cancels: its residual is $-(\epsilon /2)(e^x-1)\log x\to 0$. The composite therefore extends continuously to the left endpoint and satisfies the right boundary condition to the requested order.