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Consider the following equation $\frac{dx}{dt}=-\lambda x +\epsilon x(t-a), \quad x(0)=1,\quad |\epsilon|\ll1$ where $a$ and $\lambda$ are positive constants and $x(t-a)$ means the function $x(t)$ evaluated at $t-a$. The task is to obtain a solution as a series in powers of $\epsilon$. The book suggests to show that $x(t)=e^{-\lambda t} w(t)$ satisfies $\frac{dw}{dt}=\delta w(t-a)\quad w(0)=1,\quad \delta=\epsilon e^{\lambda a}$ and show that the perturbation expansion is given by the following expression: $w(t)=1+t\sum_{n=1}^{\infty}\frac{\delta^n}{n!}\left(t-na\right)^{n-1}$ hence $x(t)=e^{-\lambda t}\left(1+t\sum_{n=1}^{\infty}\frac{\delta^n}{n!}\left(t-na\right)^{n-1}\right)$ Since the method given in the solution is basically verification by substitution and proof by induction, I have been looking for a more systematic approach. The presence of time delay in RHS suggests that Laplace transform method could be appropriate here. So I proceed as follows. Apply Laplace transform to both sides of the original equation observing the initial conditions: $x(t)\fallingdotseq X(s)$ $sX(s)-x(0)=-\lambda X(s) +\epsilon e^{-sa}X(s)$ Solve for $X(s)$: $X(s)=\frac{x(0)}{s+\lambda-\epsilon e^{-sa}}=\frac{1}{s+\lambda-\epsilon e^{-sa}}$ Expand in powers of $\epsilon$: $X(s)=\frac{1}{s+\lambda}\left(\frac{1}{1-\epsilon\frac{e^{-sa}}{s+\lambda}}\right)=\frac{1}{s+\lambda}\left(1+\sum_{n=1}^{\infty}\frac{\epsilon^n e^{-nas}}{(s+\lambda)^n}\right)=\frac{1}{s+\lambda}+\sum_{n=1}^{\infty}\frac{\epsilon^n e^{-nas}}{(s+\lambda)^{n+1}}$ Taking inverse transforms $\frac{1}{s+\lambda} \risingdotseq e^{-\lambda t}$ $\frac{e^{-nas}}{(s+\lambda)^{n+1}} \risingdotseq \frac{(t-na)^n e^{-\lambda(t-na)}}{n!}$ But then I lose the factor of $t$ before the sum as compared to the book solution. Where am I making the slip? Thanks in advance.

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    I realised that this is a possibly outdated symbol specific to old Russian textbooks for example [this one](http://books.google.co.uk/books/about/Ordinary_differential_equations.html?id=dzPvAAAAMAAJ). Could not find any I am aware of available in preview2012-05-29

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This might be a slightly different approach, but here is how I would tackle this.

Let $D=\frac{\mathrm{d}}{\mathrm{d}t}$, then we can rewrite your equation as $ (D+\lambda-\epsilon e^{-aD})x=0\tag{1} $ To invert the operator, consider the expansion: $ \begin{align} \frac{1}{\lambda+x-\epsilon e^{-ax}} &=\frac{1}{\lambda+x}\frac{1}{1-\epsilon\frac{e^{-ax}}{\lambda+x}}\\ &=\frac{1}{\lambda+x}+\epsilon\frac{e^{-ax}}{(\lambda+x)^2}+\epsilon^2\frac{e^{-2ax}}{(\lambda+x)^3}+\epsilon^3\frac{e^{-3ax}}{(\lambda+x)^4}+\dots\tag{2} \end{align} $ Using integrating factors we get that $ (\lambda+D)^{-1}f=e^{-\lambda t}\int e^{\lambda t}f\,\mathrm{d}t\tag{3} $ which leads us to $ (\lambda+D)^{-k-1}0=e^{-\lambda t}\sum_{j=0}^kc_j\frac{t^{k-j}}{(k-j)!}\tag{4} $ and therefore the coefficient of $\epsilon^k$ would be $ e^{-kaD}(\lambda+D)^{-k-1}0=e^{-\lambda (t-ka)}\sum_{j=0}^kc_j\frac{(t-ka)^{k-j}}{(k-j)!}\tag{5} $

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    Thank you, I will give it a close look. Surprisingly, your approach links to my [other question](http://math.stackexchange.com/questions/151360/leftd2-fracz2a2-rightu-fraczagx-operator-solution) here2012-05-29