Q.The particular integral of the differential equation $f(D)y=e^{ax}$ where $f(D)=(D-a)g(D)$, $g(a)\neq 0$ is :

The particular integral in this resonance case is $\dfrac{x\,e^{ax}}{g(a)}$.

1. For $f(D)y=e^{ax}$, the standard rule gives $\text{PI}=\dfrac{1}{f(D)}e^{ax}=\dfrac{e^{ax}}{f(a)}$, provided $f(a)\ne0$.
2. Here $f(D)=(D-a)g(D)$, so $f(a)=(a-a)\,g(a)=0$ — the direct substitution rule fails because $D=a$ is a root of $f(D)=0$ (the resonant / repeated-root case).
3. In this case the operator is split as $\dfrac{1}{(D-a)g(D)}e^{ax}=\dfrac{1}{g(D)}\left[\dfrac{1}{D-a}e^{ax}\right]$. …