How should I find ## x_{2}(t) ## for this nonlinear integro-differential equation?

[Math100]

Homework Statement

Consider the nonlinear integro-differential equation $\frac{dx}{dt}=-\lambda x(t)+\epsilon x(t)\int_{0}^{\infty}f(t-s)x(s)ds, \lvert \epsilon \rvert<<1, x(0)=A$, where
$\lambda$ is a positive constant and $f(z)$ is a sufficiently well-behaved function.
Use perturbation methods to show that the solution of this equation can be expressed as $x(t)=x_{0}(t)+\epsilon x_{1}(t)+\epsilon^{2}x_{2}(t)+O(\epsilon^{3})$, where $x_{0}(t)=Ae^{-\lambda t}, x_{1}(t)=x_{0}(t)\int_{0}^{t}\int_{0}^{\infty}f(u-s)x_{0}(s)dsdu, x_{2}(t)=e^{-\lambda t}\int_{0}^{t}du e^{\lambda u}\int_{0}^{\infty}ds f(u-s)(x_{0}(u)x_{1}(s)+x_{0}(s)x_{1}(u))$. If $f(z)=e^{-\mu\lvert z \rvert}, \mu>0$, show that $x(t)=Ae^{-\lambda t}(1+\frac{A\epsilon}{\mu^{2}-\lambda^{2}}(\frac{2\mu}{\lambda}(1-e^{-\lambda t})-\frac{\mu+\lambda}{\mu}(1-e^{-\mu t})))+O(\epsilon^{2})$,
for $\mu\neq\lambda$, and $x(t)=Ae^{-\lambda t}(1+\frac{A\epsilon}{2\lambda^{2}}(3(1-e^{-\lambda t})-2\lambda te^{-\lambda t}))+O(\epsilon^{2})$, for $\mu=\lambda$.

Relevant Equations

None.

Proof:

Consider the nonlinear integro-differential equation
$\frac{dx}{dt}=-\lambda x(t)+\epsilon x(t)\int_{0}^{\infty}f(t-s)x(s)ds, \lvert \epsilon \rvert<<1, x(0)=A$,
where $\lambda$ is a positive constant and $f(z)$ is a sufficiently well-behaved function.
Let $\epsilon=0$.
Then the unperturbed equation is $\frac{dx}{dt}=-\lambda x(t)$ with the initial condition $x(0)=A$.
This gives $\frac{dx}{dt}=-\lambda x(t)\implies \frac{dx}{x}=-\lambda dt\implies\int\frac{dx}{x} =-\lambda\int dt\implies ln\lvert x \rvert=-\lambda t+C\implies x=Ce^{-\lambda t}$ where $C$ is the constant of integration.
Since $x(0)=A$, it follows that $A=C$, so we have $x=Ae^{-\lambda t}$.
Hence, $x_{0}(t)=Ae^{-\lambda t}$.
Applying the perturbation theory produces $x(t)=x_{0}(t)+\epsilon x_{1}(t)+\epsilon^{2}x_{2}(t)+O(\epsilon^{3})$
for $\lvert \epsilon \rvert<<1$.
By direct substitution of this expansion $x(t)=x_{0}(t)+\epsilon x_{1}(t)+\epsilon^{2}x_{2}(t)+O(\epsilon^{3})$ into the original equation, we get $\frac{d}{dt}(x_{0}(t)+\epsilon x_{1}(t)+\epsilon^{2}x_{2}(t)+O(\epsilon^{3})) =-\lambda(x_{0}(t)+\epsilon x_{1}(t)+\epsilon^{2}x_{2}(t)+O(\epsilon^{3}))+\epsilon(x_{0}(t)+\epsilon x_{1}(t)+\epsilon^{2}x_{2}(t)+O(\epsilon^{3}))\int_{0}^{\infty}f(t-s)(x_{0}(s)+\epsilon x_{1}(s)+\epsilon^{2}x_{2}(s)+O(\epsilon^{3}))$ for $\lvert \epsilon \rvert<<1, x(0)=A$, where $\lambda$ is a positive constant and $f(z)$ is a sufficiently well-behaved function.
Observe that $\frac{d}{dt}(x_{0}(t)+\epsilon x_{1}(t)+\epsilon^{2}x_{2}(t)+O(\epsilon^{3})) =-\lambda x_{0}(t)-\epsilon\lambda x_{1}(t)-\epsilon^{2}\lambda x_{2}(t)-\lambda O(\epsilon^{3}) +(\epsilon x_{0}(t)+\epsilon^{2}x_{1}(t)+\epsilon^{3}x_{2}(t)+\epsilon O(\epsilon^{3}))\int_{0}^{\infty}f(t-s)(x_{0}(s) +\epsilon x_{1}(s)+\epsilon^{2}x_{2}(s)+O(\epsilon^{3}))=-A\lambda e^{-\lambda t}-\epsilon\lambda x_{1}(t)-\epsilon^{2}\lambda x_{2}(t)+O(\epsilon^{3})+(A\epsilon e^{-\lambda t}+\epsilon^{2}x_{1}(t)+\epsilon^{3}x_{2}(t)+O(\epsilon^{3}))\int_{0}^{\infty}f(t-s)(Ae^{-\lambda s} +\epsilon x_{1}(s)+\epsilon^{2}x_{2}(s)+O(\epsilon^{3}))ds$.
Now we will find $x_{1}(t)$ and $x_{2}(t)$ by grouping the terms with the same powers
of $\epsilon$ together.
Note that $-\epsilon\lambda x_{1}(t)+A\epsilon e^{-\lambda t}\int_{0}^{\infty}f(t-s)\cdot Ae^{-\lambda s}ds =\epsilon(-\lambda x_{1}(t)+Ae^{-\lambda t}\int_{0}^{\infty}f(t-s)\cdot Ae^{-\lambda s}ds)$,
so we have $\frac{dx_{1}}{dt}=-\lambda x_{1}(t)+x_{0}(t)\int_{0}^{\infty}f(t-s)x_{0}(s)ds$.
This implies that $\frac{dx_{1}}{dt}=-\lambda x_{1}(t)+x_{0}(t)\int_{0}^{\infty}f(t-s)x_{0}(s)ds \implies \frac{dx_{1}}{dt}=x_{0}(t)\int_{0}^{\infty}f(t-s)x_{0}(s)ds \implies x_{1}(t)=\int_{0}^{t}(x_{0}(u)\int_{0}^{\infty}f(u-s)x_{0}(s)ds)du \implies x_{1}(t)=x_{0}(t)\int_{0}^{t}du\int_{0}^{\infty}f(u-s)x_{0}(s)ds$.
Similarly, we have $-\epsilon^{2}\lambda x_{2}(t)+A\epsilon e^{-\lambda t}\int_{0}^{\infty} f(t-s)\cdot\epsilon x_{1}(s)ds+\epsilon^{2}x_{1}(t)\int_{0}^{\infty}f(t-s)\cdot Ae^{-\lambda s}ds =\epsilon^{2}(-\lambda x_{2}(t)+Ae^{-\lambda t}\int_{0}^{\infty}f(t-s)\cdot x_{1}(s)ds +x_{1}(t)\int_{0}^{\infty}f(t-s)\cdot Ae^{-\lambda s}ds=\epsilon^{2}(-\lambda x_{2}(t) +x_{0}(t)\int_{0}^{\infty}f(t-s)x_{1}(s)ds+x_{1}(t)\int_{0}^{\infty}f(t-s)x_{0}(s)ds$,
which implies that $\frac{dx_{2}}{dt}=-\lambda x_{2}(t)+x_{0}(t)\int_{0}^{\infty} f(t-s)x_{1}(s)ds+x_{1}(t)\int_{0}^{\infty}f(t-s)x_{0}(s)ds$.
Thus, $\frac{dx_{2}}{dt}=-\lambda x_{2}(t)+x_{0}(t)\int_{0}^{\infty} f(t-s)x_{1}(s)ds+x_{1}(t)\int_{0}^{\infty}f(t-s)x_{0}(s)ds\implies \frac{dx_{2}}{dt}=x_{0}(t)\int_{0}^{\infty}f(t-s)x_{1}(s)ds+ x_{1}(t)\int_{0}^{\infty}f(t-s)x_{0}(s)ds\implies \frac{dx_{2}}{dt}=x_{0}(u)\int_{0}^{\infty}f(u-s)x_{1}(s)ds +x_{1}(u)\int_{0}^{\infty}f(u-s)x_{0}(s)ds\implies x_{2}(t)=\int_{0}^{t}(x_{0}(u)\int_{0}^{\infty}f(u-s)x_{1}(s)ds +x_{1}(u)\int_{0}^{\infty}f(u-s)x_{0}(s)ds)du$.

From here, how should I show/prove that $x_{2}(t)=e^{-\lambda t}\int_{0}^{t}du e^{\lambda u}\int_{0}^{\infty} ds f(u-s)(x_{0}(u)x_{1}(s)+x_{0}(s)x_{1}(u))$? Also, for the second part of the question/problem, I've constructed
a different proof as below:

Let $f(z)=e^{-\mu\lvert z \rvert}$ for $\mu>0$, where $x(t)=x_{0}(t)+\epsilon x_{1}(t)+O(\epsilon^{2})$ such that $x_{0}(t)=Ae^{-\lambda t}$
and $x_{1}(t)=x_{0}(t)\int_{0}^{t}du\int_{0}^{\infty}ds f(u-s)x_{0}(s)$.
By using direct substitution of both $x_{0}(t), x_{1}(t)$,
we have $x(t)=Ae^{-\lambda t}+\epsilon(Ae^{-\lambda t}\int_{0}^{t}du\int_{0}^{\infty}ds f(u-s)\cdot Ae^{-\lambda s})+O(\epsilon^{2})\implies x(t)=Ae^{-\lambda t}+A\epsilon e^{-\lambda t}\int_{0}^{t}\int_{0}^{\infty}f(z) \cdot Ae^{-\lambda s}dsdu\implies x(t)=Ae^{-\lambda t}(1+A\epsilon\int_{0}^{t}\int_{0}^{\infty}e^{-\mu\lvert z \rvert} \cdot e^{-\lambda s}dsdu)\implies x(t)=Ae^{-\lambda t}(1+A\epsilon\int_{0}^{t}\int_{0}^{\infty}e^{-\mu\lvert u-s \rvert}\cdot e^{-\lambda s}dsdu) \implies x(t)=Ae^{-\lambda t}(1+A\epsilon\int_{0}^{t}\int_{0}^{\infty}e^{-\mu\lvert u-s \rvert-\lambda s}dsdu)$.

And from here, how should I show/prove that $x(t)=Ae^{-\lambda t}(1+\frac{A\epsilon}{\mu^{2}-\lambda^{2}} (\frac{2\mu}{\lambda}(1-e^{-\lambda t})-\frac{\mu+\lambda}{\mu}(1-e^{-\mu t})))+O(\epsilon^{2})$ for
$\mu\neq\lambda, \mu>0$? Similarly, for $\mu=\lambda$?

 

[Mark44]

Please break up long lines of LaTeX so that readers don't have to scroll way off to the right to see the rest of the line. Each of the lines that contain an implication symbol could benefit from being split into multiple lines.

 

[Mark44]

Here's an example of what i mean by breaking up long lines. In your original post the stuff below was one long line that extended to about three times the width of my screen. My only changes were to break the line at points between equal expressions and terms of sums.

Observe that $\frac{d}{dt}(x_{0}(t)+\epsilon x_{1}(t)+\epsilon^{2}x_{2}(t)+O(\epsilon^{3}))$
$=-\lambda x_{0}(t)-\epsilon\lambda x_{1}(t)-\epsilon^{2}\lambda x_{2}(t)-\lambda O(\epsilon^{3})$
$+(\epsilon x_{0}(t)+\epsilon^{2}x_{1}(t)+\epsilon^{3}x_{2}(t)+\epsilon O(\epsilon^{3}))\int_{0}^{\infty}f(t-s)(x_{0}(s)$
$+\epsilon x_{1}(s)+\epsilon^{2}x_{2}(s)+O(\epsilon^{3}))$
$=-A\lambda e^{-\lambda t}-\epsilon\lambda x_{1}(t)-\epsilon^{2}\lambda x_{2}(t)+O(\epsilon^{3})$
$+(A\epsilon e^{-\lambda t}+\epsilon^{2}x_{1}(t)+\epsilon^{3}x_{2}(t)+O(\epsilon^{3}))\int_{0}^{\infty}f(t-s)(Ae^{-\lambda s} +\epsilon x_{1}(s)+\epsilon^{2}x_{2}(s)+O(\epsilon^{3}))ds$.

 

[Math100]

How to show that $x(t)=Ae^{-\lambda t}(1+\frac{A\epsilon}{\mu^{2} -\lambda^{2}}(\frac{2\mu}{\lambda}(1-e^{-\lambda t}) -\frac{\mu+\lambda}{\mu}(1-e^{-\mu t})))+O(\epsilon^{2})$
for $\mu\neq\lambda$? I know that I should integrate $\int_{0}^{\infty}e^{-\mu\lvert u-s \rvert-\lambda s}ds$ first,
but how should I integrate this?

 

[fresh_42]

I will reformat this mess so that it has a higher chance of a reply. I'm not an expert in perturbation theory so it's not an answer, but only the OP restructured. @Math100 , please hit the reply button to learn from the format for future posts and check for typos. Another remark: this has been a long text and it might have been better to split it into smaller steps and more than one thread. This also helps you to structure your work and concentrate on the steps you do not understand.

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Homework Statement: Consider the nonlinear integro-differential equation $$ \frac{dx}{dt}=-\lambda x(t)+\epsilon x(t)\int_{0}^{\infty}f(t-s)x(s)ds$$
where $|\epsilon| <<1\, , \, x(0)=A \, , \, \lambda >0$ is a constant and $f(z)$ is a sufficiently well-behaved function.

Use perturbation methods to show that the solution of this equation can be expressed as $$ x(t)=x_{0}(t)+\epsilon x_{1}(t)+\epsilon^{2}x_{2}(t)+O(\epsilon^{3}) \text{ where }$$
\begin{align*}
x_{0}(t)&=Ae^{-\lambda t}\\
x_{1}(t)&=x_{0}(t)\int_{0}^{t}\int_{0}^{\infty}f(u-s)x_{0}(s)\,ds\,du\\
x_{2}(t)&=e^{-\lambda t}\int_{0}^{t}\,du\, e^{\lambda u}\int_{0}^{\infty}\,ds \,f(u-s)(x_{0}(u)x_{1}(s)+x_{0}(s)x_{1}(u))
\end{align*}
If $f(z)=e^{-\mu\lvert z \rvert}$ with $\mu>0$, show that
$$
x(t)=\begin{cases}
Ae^{-\lambda t}(1+\frac{A\epsilon}{\mu^{2}-\lambda^{2}}(\frac{2\mu}{\lambda}(1-e^{-\lambda t})-\frac{\mu+\lambda}{\mu}(1-e^{-\mu t})))+O(\epsilon^{2}) &\text{ if }\mu\neq\lambda\\[8pt]
Ae^{-\lambda t}(1+\frac{A\epsilon}{2\lambda^{2}}(3(1-e^{-\lambda t})-2\lambda te^{-\lambda t}))+O(\epsilon^{2})&\text{ if }\mu=\lambda
\end{cases}
$$
Proof: (without repeating the above constraints)

Consider the nonlinear integro-differential equation
$$\dfrac{dx}{dt}=-\lambda x(t)+\epsilon x(t)\int_{0}^{\infty}f(t-s)x(s)\,ds.$$
If $\epsilon=0$ the unperturbed equation is $\dfrac{dx}{dt}=-\lambda x(t)$ with the initial condition $x(0)=A$. This gives
\begin{align*}
\dfrac{dx}{dt}=-\lambda x(t)&\implies \dfrac{dx}{x}=-\lambda dt \implies\int\dfrac{dx}{x}=-\lambda\int dt\\
&\implies ln\lvert x \rvert=-\lambda t+C \implies x=Ce^{-\lambda t}
\end{align*}
Since $x(0)=A$, it follows that $A=C$, so we have $x=Ae^{-\lambda t},$ i.e. $x_{0}(t)=Ae^{-\lambda t}$.

Applying the perturbation theory produces $x(t)=x_{0}(t)+\epsilon x_{1}(t)+\epsilon^{2}x_{2}(t)+O(\epsilon^{3})$ for $\lvert \epsilon \rvert<<1$.
By direct substitution of this expansion into the original equation, we get
\begin{align*}
\dfrac{d}{dt}&\left(x_{0}(t)+\epsilon x_{1}(t)+\epsilon^{2}x_{2}(t)+O(\epsilon^{3}) \right)\\
&=-\lambda \left(x_{0}(t)+\epsilon x_{1}(t)+\epsilon^{2}x_{2}(t)+O(\epsilon^{3})\right)\\
&\phantom{=}+\epsilon\left(x_{0}(t)+\epsilon x_{1}(t)+\epsilon^{2}x_{2}(t)+O(\epsilon^{3})\right) \int_{0}^{\infty}f(t-s)(x_{0}(s)+\epsilon x_{1}(s)+\epsilon^{2}x_{2}(s)+O(\epsilon^{3}))\,ds\\
&=-\lambda x_{0}(t)-\epsilon\lambda x_{1}(t)-\epsilon^{2}\lambda x_{2}(t)-\lambda O(\epsilon^{3})\\
&\phantom{=}+\left(\epsilon x_{0}(t)+\epsilon^{2}x_{1}(t)+\epsilon^{3}x_{2}(t)+\epsilon O(\epsilon^{3}) \right) \int_{0}^{\infty}f(t-s)(x_{0}(s)+\epsilon x_{1}(s)+\epsilon^{2}x_{2}(s)+O(\epsilon^{3}))\,ds\\
&=-A\lambda e^{-\lambda t}-\epsilon\lambda x_{1}(t)-\epsilon^{2}\lambda x_{2}(t)+O(\epsilon^{3})\\
&\phantom{=}+\left(A\epsilon e^{-\lambda t}+\epsilon^{2}x_{1}(t)+\epsilon^{3}x_{2}(t)+O(\epsilon^{3})\right)\int_{0}^{\infty}f(t-s)(Ae^{-\lambda s}
+\epsilon x_{1}(s)+\epsilon^{2}x_{2}(s)+O(\epsilon^{3}))\,ds
\end{align*}
Now we will find $x_{1}(t)$ and $x_{2}(t)$ by grouping the terms with the same powers of $\epsilon .$ Note that
$$-\epsilon\lambda x_{1}(t)+A\epsilon e^{-\lambda t}\int_{0}^{\infty}f(t-s)\cdot Ae^{-\lambda s}ds
=\epsilon(-\lambda x_{1}(t)+Ae^{-\lambda t}\int_{0}^{\infty}f(t-s)\cdot Ae^{-\lambda s}ds)$$
so we have
$$ \dfrac{dx_{1}}{dt}=-\lambda x_{1}(t)+x_{0}(t)\int_{0}^{\infty}f(t-s)x_{0}(s)\,ds .$$
This implies
\begin{align*}
\dfrac{dx_{1}}{dt}&=-\lambda x_{1}(t)+x_{0}(t)\int_{0}^{\infty}f(t-s)x_{0}(s)\,ds\\[6pt]
&\implies \dfrac{dx_{1}}{dt}=x_{0}(t)\int_{0}^{\infty}f(t-s)x_{0}(s)\,ds \\
&\implies x_{1}(t)=\int_{0}^{t}(x_{0}(u)\int_{0}^{\infty}f(u-s)x_{0}(s)\,ds)\,du\\
&\implies x_{1}(t)=x_{0}(t)\int_{0}^{t}du\int_{0}^{\infty}f(u-s)x_{0}(s)\,ds .
\end{align*}
Similarly, we have
\begin{align*}
-\epsilon^{2}\lambda x_{2}(t)&+A\epsilon e^{-\lambda t}\int_{0}^{\infty}
f(t-s)\cdot\epsilon x_{1}(s)ds+\epsilon^{2}x_{1}(t)\int_{0}^{\infty}f(t-s)\cdot Ae^{-\lambda s}\,ds\\
&=\epsilon^{2}(-\lambda x_{2}(t)+Ae^{-\lambda t}\int_{0}^{\infty}f(t-s)\cdot x_{1}(s)ds
+x_{1}(t)\int_{0}^{\infty}f(t-s)\cdot Ae^{-\lambda s}\,ds\\
&=\epsilon^{2}(-\lambda x_{2}(t)+x_{0}(t)\int_{0}^{\infty}f(t-s)x_{1}(s)ds+x_{1}(t)\int_{0}^{\infty}f(t-s)x_{0}(s)\,ds
\end{align*}
which implies that $$\dfrac{dx_{2}}{dt}=-\lambda x_{2}(t)+x_{0}(t)\int_{0}^{\infty}
f(t-s)x_{1}(s)ds+x_{1}(t)\int_{0}^{\infty}f(t-s)x_{0}(s)\,ds .$$
Thus,
\begin{align*}
\dfrac{dx_{2}}{dt}&=-\lambda x_{2}(t)+x_{0}(t)\int_{0}^{\infty} f(t-s)x_{1}(s)\,ds+x_{1}(t)\int_{0}^{\infty}f(t-s)x_{0}(s)\,ds\\
&\implies \dfrac{dx_{2}}{dt}=x_{0}(t)\int_{0}^{\infty}f(t-s)x_{1}(s)\,ds+
x_{1}(t)\int_{0}^{\infty}f(t-s)x_{0}(s)\,ds\\
&\implies \dfrac{dx_{2}}{dt}=x_{0}(u)\int_{0}^{\infty}f(u-s)x_{1}(s)\,ds
+x_{1}(u)\int_{0}^{\infty}f(u-s)x_{0}(s)\,ds\\
&\implies x_{2}(t)=\int_{0}^{t}(x_{0}(u)\int_{0}^{\infty}f(u-s)x_{1}(s)\,ds
+x_{1}(u)\int_{0}^{\infty}f(u-s)x_{0}(s)\,ds)\,du
\end{align*}
From here, how should I show/prove that
$$
x_{2}(t)=e^{-\lambda t}\int_{0}^{t}du e^{\lambda u}\int_{0}^{\infty}
ds \,f(u-s)(x_{0}(u)x_{1}(s)+x_{0}(s)x_{1}(u)) \text{ ?}
$$
Also, for the second part of the question/problem, I've constructed a different proof as below:

Let $f(z)=e^{-\mu\lvert z \rvert}$ for $\mu>0$, where $x(t)=x_{0}(t)+\epsilon x_{1}(t)+O(\epsilon^{2})$ such that $x_{0}(t)=Ae^{-\lambda t}$
and $x_{1}(t)=x_{0}(t)\int_{0}^{t}du\int_{0}^{\infty}ds \,f(u-s)x_{0}(s) .$ By using direct substitution of both $x_{0}(t), x_{1}(t)$, we have
\begin{align*}
x(t)&=Ae^{-\lambda t}+\epsilon(Ae^{-\lambda t}\int_{0}^{t}du\,\int_{0}^{\infty}ds\,
f(u-s)\cdot Ae^{-\lambda s})+O(\epsilon^{2})\\
&\implies x(t)=Ae^{-\lambda t}+A\epsilon e^{-\lambda t}\int_{0}^{t}\int_{0}^{\infty}f(z)
\cdot Ae^{-\lambda s}\,ds\,du\\
&\implies x(t)=Ae^{-\lambda t}(1+A\epsilon\int_{0}^{t}\int_{0}^{\infty}e^{-\mu\lvert z \rvert}
\cdot e^{-\lambda s}\,ds\,du)\\
&\implies x(t)=Ae^{-\lambda t}(1+A\epsilon\int_{0}^{t}\int_{0}^{\infty}e^{-\mu\lvert u-s \rvert}\cdot e^{-\lambda s}\,ds\,du)\\
&\implies x(t)=Ae^{-\lambda t}(1+A\epsilon\int_{0}^{t}\int_{0}^{\infty}e^{-\mu\lvert u-s \rvert-\lambda s}\,ds\,du)
\end{align*}

And from here, how should I show/prove that
$$
x(t)=Ae^{-\lambda t}\left(1+\dfrac{A\epsilon}{\mu^{2}-\lambda^{2}}
\left(\dfrac{2\mu}{\lambda}\left(1-e^{-\lambda t}\right)-\dfrac{\mu+\lambda}{\mu}\left(1-e^{-\mu t}\right)\right)\right)+O(\epsilon^{2})
$$
for $\mu\neq\lambda, \mu>0 \text{ ?}$ Similarly, for $\mu=\lambda \text{ ?}$

 

[pasmith]

Use an integrating factor: $\dot x + \lambda x = e^{-\lambda t}\frac{d}{dt}(xe^{\lambda t})$. Hence $$\begin{split} e^{-\lambda t}\frac{d}{dt}(xe^{\lambda t}) &= \epsilon \int_0^\infty f(t-s) x(t) x(s)\,ds \\ x(t) &= Ae^{-\lambda t} + \epsilon e^{-\lambda t} \int_0^t e^{\lambda u} \int_0^\infty f(u-s) x(u) x(s)\,ds\,du.\end{split}$$ It is now straightforward to substitute $x = x_0 + \epsilon x_1 + \epsilon^2 x_2 + O(\epsilon^3)$ and read off the terms at $O(\epsilon^0)$ etc.

To integrate $\int_0^\infty h(|t-s|)g(s)\,ds$, split the range of integration at $t$: $$\begin{split} \int_0^\infty h(|t-s|)g(s)\,ds &= \underbrace{\int_0^t h(|t-s|)g(s)\,ds}_{0 \leq s \leq t} + \underbrace{\int_t^\infty h(|t-s|)g(s)\,ds}_{t \leq s < \infty} \\ &= \int_0^t h(t-s)g(s)\,ds + \int_t^\infty h(s-t)g(s)\,ds.\end{split}$$

 

[Math100]

Use an integrating factor: $\dot x + \lambda x = e^{-\lambda t}\frac{d}{dt}(xe^{\lambda t})$. Hence $$\begin{split} e^{-\lambda t}\frac{d}{dt}(xe^{\lambda t}) &= \epsilon \int_0^\infty f(t-s) x(t) x(s)\,ds \\ x(t) &= Ae^{-\lambda t} + \epsilon e^{-\lambda t} \int_0^t e^{\lambda u} \int_0^\infty f(u-s) x(u) x(s)\,ds\,du.\end{split}$$ It is now straightforward to substitute $x = x_0 + \epsilon x_1 + \epsilon^2 x_2 + O(\epsilon^3)$ and read off the terms at $O(\epsilon^0)$ etc.

To integrate $\int_0^\infty h(|t-s|)g(s)\,ds$, split the range of integration at $t$: $$\begin{split} \int_0^\infty h(|t-s|)g(s)\,ds &= \underbrace{\int_0^t h(|t-s|)g(s)\,ds}_{0 \leq s \leq t} + \underbrace{\int_t^\infty h(|t-s|)g(s)\,ds}_{t \leq s < \infty} \\ &= \int_0^t h(t-s)g(s)\,ds + \int_t^\infty h(s-t)g(s)\,ds.\end{split}$$

I've got
\begin{align*}
\int_{0}^{\infty} e^{-\mu\lvert u-s \rvert-\lambda s} \, ds &= \int_{0}^{t} e^{-\mu\lvert u-s \rvert-\lambda s} \, ds + \int_{t}^{\infty} e^{-\mu\lvert u-s \rvert-\lambda s} \, ds \\
&= \int_{0}^{t} e^{-\mu(u-s)-\lambda s} \, ds + \int_{t}^{\infty} e^{-\mu(s-u)-\lambda s} \, ds \\
&= e^{-\mu u} \int_{0}^{t} e^{(\mu-\lambda)s} \, ds + e^{\mu u} \int_{t}^{\infty} e^{-(\mu+\lambda)s} \, ds \\
&= e^{-\mu u} \left[\frac{e^{(\mu-\lambda)s}}{\mu-\lambda}\right]_{0}^{t} + e^{\mu u} \left[-\frac{e^{-(\mu+\lambda)s}}{\mu+\lambda}\right]_{t}^{\infty} \\
&= e^{-\mu u} \left(\frac{e^{(\mu-\lambda)t} - 1}{\mu - \lambda}\right) + e^{\mu u} \left(\frac{e^{-(\mu+\lambda)t}}{\mu + \lambda}\right).
\end{align*}
Is this correct?