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For pairwise distinct rates $a,b,c$, the response isSet $v=y^{\prime }-by$. Solving $v^{\prime }-av=e^{cx}$ with $v(0)=0$ gives $(e^{cx}-e^{ax})/(c-a)$; solving the remaining first-order equation by an integrating factor gives the formula. Equivalently this is the convolution of $e^{cx}$ with the zero-displacement unit-impulse response $(e^{ax}-e^{bx})/(a-b)$ on the positive half-line. In the resonant case $c=b\ne a$, the finite limit is $y=[e^{ax}-e^{bx}-(a-b)x{e}^{bx}]/(a-b{)}^2$. The apparent singularity of the distinct-rate expression is removable, and the limit exhibits the polynomial factor of a resonant exponential particular solution.

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.

The characteristic equation is $m^2+3m+2=(m+1)(m+2)=0$, giving the complementary function $C_1{e}^{-x}+C_2{e}^{-2x}$. For the forcing term $x$, a linear particular solution $ax+b$ gives $2ax+3a+2b=x$, hence $a=1/2$, $b=-3/4$.

The forcing $-e^{-x}$ gives resonance in a differential equation, since $e^{-x}$ already appears in the complementary function. Try $Kx{e}^{-x}$. Direct differentiation givesso $K=-1$. Combining the two particular solutions givesThe factor $x$ in the resonant exponential particular solution is required because $-1$ is already a root of the characteristic equation.

Now the characteristic equation is $(m+1{)}^2=0$, with a repeated root, so the complementary function is $(C_1+C_{2}x)e^{-x}$. For the polynomial forcing, substitute $a{x}^2+bx+c$ to obtainThus $a=1$, $b=-4$, $c=6$. For the forcing $2e^{-x}$, both $e^{-x}$ and $x{e}^{-x}$ already occur in the complementary function; the resonant exponential particular solution therefore starts with $K{x}^2{e}^{-x}$. The identitygives $K=1$. The resulting general solution is

Now $c=b$, so the forcing coincides with a homogeneous exponential and requires a resonant exponential particular solution. With $v=y^{\prime }-by$ as above, $v^{\prime }-av=e^{bx}$ and $v(0)=0$ give $v=(e^{ax}-e^{bx})/(a-b)$. ThereforeIntegrating from zero yieldsIts numerator starts at order $x^2$, ensuring both initial conditions. It is also the finite coalescing-rate limit $c\to b$ of the nonresonant exponential forcing response with zero initial data.

The characteristic polynomial of the homogeneous equation is $r^2+7r+12=(r+3)(r+4)$. Thus the complementary function is $C_1{e}^{-3x}+C_2{e}^{-4x}$. The forcing resonates with the first term, so a resonant exponential particular solution is $y_p=Kx{e}^{-3x}$. Substitution gives $y_p^{\prime \prime }+7y_p^{\prime }+12y_p=K{e}^{-3x}$, hence $K=2$.

The full solution is $y=(C_1+2x)e^{-3x}+C_2{e}^{-4x}$. At zero, the two initial conditions give $C_1+C_2=1$ and $2-3C_1-4C_2=0$. Solving yields $C_1=2$, $C_2=-1$, so

The one-sided Laplace transform is $\bar{u}(p)={\int }_0^{\infty }{e}^{-pt}u(t)dt$ for sufficiently large $\mathrm{Re}p$. Integration by parts, with the exponential suppressing the upper boundary, givesDifferentiation under the integral gives $d\bar{u}/dp=-{\int }_0^{\infty }t{e}^{-pt}u(t)dt$, proving $\overline{tu}=-d\bar{u}/dp$. Start with $\overline{e^{-t}}=1/(p+1)$ and apply this rule repeatedly:With all three initial data zero, transforming the forced equation gives $(p+1{)}^3\bar{y}=1/(p+1)$. Thus $\bar{y}=(p+1{)}^{-4}$ and the listed inverse transform givesThe cubic factor is also the resonant exponential particular solution associated with the triple characteristic root $-1$.

In a linear differential equation, resonance in a differential equation occurs when the forcing belongs to a mode annihilated by the homogeneous operator, so an ordinary trial of the same shape cannot be inverted. For $P(D)y=e^{rt}$ with $P(r)=0$, a polynomial factor in $t$ supplies a resonant exponential particular solution. The root multiplicity determines the required power.

For $y^{\prime \prime }+y=\cos t$, $y_p=(t/2)\sin t$ is a particular solution. Direct differentiation gives $y_p^{\prime \prime }+y_p=\cos t$. By contrast, the resonant equation $y^{\prime }+y=e^{-t}$ has $y_p=t{e}^{-t}$, which still decays: resonance enhances a mode relative to its natural exponential, without always causing an unbounded physical response. Under logarithmic coordinates in a Cauchy-Euler equation, polynomial factors in transformed time become logarithms of $x$.