Solution without using integrating factor: Using the method of D operator Consider a differntial equation of the form $$\frac{dx}{dt}+\gamma x=F(t)$$ which can also be written as $$(D+\gamma)x(t)=F(t)$$ where $D\equiv d/dt$ and $\gamma$ is a real constant. Symbolically, the solution (the particular integral, to be specific) can be written as $$x(t)=\frac{1}{D+\gamma}F(t).$$ If $F(t)$ were  known function of specific type (like sine, cosine, exponential etc) this could be solved by the method of D-operators. But for an arbitrary function $F(t)$, can we express the solution $x(t)$ in terms of an integral by expanding $(D+\gamma)^{-1}$ into a power series or something similar.
I know how to do this using the method of integrating factors. But I want to know how far can we proceed towards the solution of the particular integral by the method of D operators when $F(t)$ is not specified.
 A: The first requirement for something like that to be possible is that the operator $D+\gamma{I}$ be invertible. For this, one would want to restrict the operator from $C^1(I)$ to some proper subset. Notice that if one defines $$U_{\gamma}[g](t)=e^{\gamma{t}}g(t),$$ then $$U_{-\gamma}DU_{\gamma}=D+\gamma{I},$$ so the equation can be rewritten as $$U_{-\gamma}DU_{\gamma}[x]=F.$$ This means $$DU_{\gamma}[x]=U_{\gamma}[F],$$ which implies $$e^{\gamma{t}}x(t)-e^{\gamma{a}}x(a)=\int_a^tU_{\gamma}[F]\,\mathrm{d}s,$$ and this means $$x(t)=e^{\gamma{(a-t)}}x(a)+U_{-\gamma}\int_a^tU_{\gamma}[F]\,\mathrm{d}s.$$ This is precisely a rehashing of the method by integrating factor, but the reason I am discussing it is because it is important to start here to know under what conditions one can use the power series approach you suggest. Using integration by parts, one has that $$U_{-\gamma}\int_a^tU_{\gamma}[F]\,\mathrm{d}s=\frac{F(t)}{\gamma}-\frac{e^{\gamma{(a-t)}}F(a)}{\gamma}-\frac1{\gamma}U_{-\gamma}\int_a^tU_{\gamma}[F']\,\mathrm{d}s.$$ This assumes that $F\in{C^1(I)},$ whereas the method by integration factor only assumes $F\in{C^0(I)}.$ If $F\in{C^2(I)},$ then $$U_{-\gamma}\int_a^tU_{\gamma}[F]\,\mathrm{d}s=\frac{F(t)}{\gamma}-\frac{e^{\gamma{(a-t)}}F(a)}{\gamma}-\frac{F'(t)}{\gamma^2}+\frac{e^{\gamma(a-t)}F'(a)}{\gamma^2}+\frac1{\gamma^2}U_{-\gamma}\int_a^tU_{\gamma}[F'']\,\mathrm{d}s.$$
One can use induction to demonstrate that if $F\in{C^k(I)},$ then $$U_{-\gamma}\int_a^tU_{\gamma}[F]\,\mathrm{d}s=\sum_{m=0}^{k-1}\frac1{\gamma^{m+1}}(-D)^m[F](t)-e^{\gamma(a-t)}\sum_{m=0}^{k-1}\frac1{\gamma^{m+1}}(-D)^m[F](a)$$ $$+\frac{(-1)^k}{\gamma^k}U_{-\gamma}\int_a^tU_{\gamma}D^k[F]\,\mathrm{d}s.$$ As such, one has that $$x(t)=e^{\gamma(a-t)}\left[x(a)-\sum_{m=0}^{k-1}\frac1{\gamma^{m+1}}(-D)^m[F](a)\right]+\sum_{m=0}^{k-1}\frac1{\gamma^{m+1}}(-D)^m[F](t)$$ $$+\frac{(-1)^k}{\gamma^k}U_{-\gamma}\int_a^tU_{\gamma}D^k[F]\,\mathrm{d}s.$$ For the power series expansion of $$\left(I+\frac{D}{\gamma}\right)^{-1}$$ to be a valid approach, what you require is that $$e^{\gamma(a-t)}\left[x(a)-\sum_{m=0}^{k-1}\frac1{\gamma^{m+1}}(-D)^m[F](a)\right]+\frac{(-1)^k}{\gamma^k}U_{-\gamma}\int_a^tU_{\gamma}D^k[F]\,\mathrm{d}s\to0$$ as $k\to\infty,$ which has, as a necessary condition, that $F\in{C^{\omega}(I)},$ and it also means that there are certain convergence criteria $F$ ought to satisfy.
