$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\const}{\operatorname{const}}$ $\newcommand{\sgn}{\operatorname{sgn}}$ $\newcommand{\rank}{\operatorname{rank}}$
The derivative expansion method is probably the most common of the various multiple scale methods. One introduces several time (or length) scales and treats them as independent variables: If $t$ is the original variable and $\varepsilon$ is the small parameter, we introduce the auxiliary time scales \begin{equation} \tau_1=\varepsilon t,\ \tau_2=\varepsilon ^2t,\ldots,\ \tau_N=\varepsilon^N t \label{eq-8.2.1} \end{equation} and consider $u(t,\varepsilon)=w(t,\tau_1,\ldots,\tau_N)$. Then \begin{equation} \frac{d \ }{dt}u= \Bigl(\frac{\partial\ }{\partial t}+ \varepsilon \frac{\partial\ }{\partial \tau_1} + \varepsilon^2 \frac{\partial\ }{\partial \tau_2}+\ldots + \varepsilon^N \frac{\partial\ }{\partial \tau_N}\Bigr)w. \label{eq-8.2.2} \end{equation} Let us apply this with $N=2$ to problem (8.1.1)--(8.1.2) \begin{align} &u''+ 2\varepsilon u'+ u=0,\qquad t>0 \label{eq-8.2.3}\\ &u(0)=1,\qquad u'(0)=0. \label{eq-8.2.4} \end{align} Then \begin{align} u' &=\frac{\partial w}{\partial t}+ \varepsilon \frac{\partial w}{\partial \tau_1} + \varepsilon^2 \frac{\partial w}{\partial \tau_2}, \label{eq-8.2.5}\\ u''&=\Bigr(\frac{\partial^2 w}{\partial t^2}+ \varepsilon \frac{\partial w}{\partial \tau_1} + \varepsilon^2 \frac{\partial w}{\partial \tau_2}\Bigr)^2w \label{eq-8.2.6}\\ &=\frac{\partial^2 w }{\partial t^2}+ 2\varepsilon \frac{\partial^2 w }{\partial t\partial \tau_1} + \varepsilon^2 \Bigl(2\frac{\partial^2w }{\partial t\partial \tau_2} + \frac{\partial^2 w }{\partial \tau_1^2}\Bigr) +O(\varepsilon^3). \notag \end{align} and we arrive to \begin{align} &\frac{\partial^2 w}{\partial t^2} + 2\varepsilon \Bigl(\frac{\partial^2 w }{\partial t\partial \tau_1}+ \frac{\partial w}{\partial t}\Bigr)+ \varepsilon^2 \Bigl(2\frac{\partial^w }{\partial t\partial \tau_2} + \frac{\partial^2 w }{\partial \tau_1^2}+ 2\frac{\partial w}{\partial \tau_1} \Bigr) + w =O(\varepsilon^3), \label{eq-8.2.7}\\ &w(0,0,0,\varepsilon)=1, \label{eq-8.2.8}\\ &\frac{\partial w}{\partial t}(0,0,0,\varepsilon)+ \varepsilon \frac{\partial w}{\partial \tau_1}(0,0,0,\varepsilon) + \varepsilon^2 \frac{\partial w}{\partial \tau_2}(0,0,0,\varepsilon)=0. \label{eq-8.2.9} \end{align} We look for a solution to this partial differential equation of the form \begin{equation} w(t,\tau_1, \tau_2,\varepsilon) = w_0(t,\tau_1, \tau_2)+ \varepsilon w_1(t,\tau_1)+ \varepsilon^2 w_2(t) +O(\varepsilon^3). \label{eq-8.2.10} \end{equation} Equalizing to $0$ coefficients at powers of $\varepsilon$ we find \begin{align} &\varepsilon^0:\ \left\{\begin{aligned} &\frac{\partial ^2w_0}{\partial t^2}+w_0=0,\\ &w_0(0,0,0)=1,\\ &\frac{\partial w_0}{\partial t}(0,0,0)=0; \end{aligned}\right. \label{eq-8.2.11}\\[4pt] &\varepsilon^1:\ \left\{\begin{aligned} &\frac{\partial ^2w_1}{\partial t^2}+w_1= -2\frac{\partial ^2w_0}{\partial t\partial\tau_1} - 2\frac{\partial w_0}{\partial t},\\ &w_1(0,0)=0,\\ &\frac{\partial w_1}{\partial t}(0,0)= - \frac{\partial w_0}{\partial \tau_1}(0,0,0); \end{aligned}\right. \label{eq-8.2.12}\\[4pt] &\varepsilon^2:\ \left\{\begin{aligned} &\frac{\partial ^2w_2}{\partial t^2}+w_2= -2\frac{\partial ^2w_0}{\partial t\partial\tau_2} - \frac{\partial^2 w_0}{\partial \tau_1^2}- 2\frac{\partial w_0}{\partial \tau_1 } - 2\frac{\partial ^2w_1}{\partial t\partial\tau_1}- 2\frac{\partial w_1}{\partial t},\\ &w_2(0)=0,\\ &\frac{\partial w_2}{\partial t}(0)= - \frac{\partial w_0}{\partial \tau_2}(0,0,0)- \frac{\partial w_1}{\partial \tau_1}(0,0). \end{aligned}\right. \label{eq-8.2.13} \end{align} The solution to problem (\ref{eq-8.2.11}) is \begin{align} &w_0(t,\tau_1, \tau_2) = A(\tau_1, \tau_2)\cos(t)+ B(\tau_1, \tau_2)\sin(t),\label{eq-8.2.14}\\ &A(0,0)=1,\quad B(0,0)=0.\label{eq-8.2.15} \end{align} Then equation in (\ref{eq-8.2.12}) is \begin{equation} \frac{\partial ^2w_1}{\partial t^2}+w_1= 2\Bigl(\frac{\partial A}{\partial\tau_1}+A\Bigr)\sin(t)- 2\Bigl(\frac{\partial B}{\partial\tau_1}+B\Bigr)\cos(t) \label{eq-8.2.16} \end{equation} and to avoid secularities we equalize coefficients here to $0$: \begin{equation} \frac{\partial A}{\partial\tau_1}+A=0,\qquad \frac{\partial B}{\partial\tau_1}+B=0 \label{eq-8.2.17} \end{equation} and with initial conditions (\ref{eq-8.2.15}) we conclude \begin{align} &A= \alpha(\tau_2) e^{-\tau_1}, && \alpha(0)=1, \label{eq-8.2.18}\\ &B= \beta(\tau_2) e^{-\tau_1}, &&\beta(0)=0. \label{eq-8.2.19} \end{align}
Now secular terms in (\ref{eq-8.2.16}) vanish and we solve (\ref{eq-8.2.16}) (and use initial conditions) and also we have (\ref{eq-8.2.21}) \begin{align} &w_1(t,\tau_1)=C(\tau_1)\cos(t)+D(\tau_1)\sin(t),&& C(0)=0, D(0)=1 \label{eq-8.2.20}\\ &w_0(t,\tau_1,\tau_2)= e^{-\tau_1}\bigl(\alpha(\tau_2)\cos(t)+\beta(\tau_2)\sin(t)\bigr), &&\alpha(0)=1, \beta(0)=0. \label{eq-8.2.21} \end{align} Equation in (\ref{eq-8.2.13}) now becomes \begin{multline*} \frac{\partial ^2w_2}{\partial t^2}+w_2=\\ \bigl[(2\alpha'+\beta)e^{-\tau_1}+2(C'+C)\bigr]\sin(t)+ \bigl[(-2\beta'+\alpha)e^{-\tau_1}-2(D'+D)\bigr]\cos(t) \end{multline*} Again we should remove secular terms: \begin{align*} &\underbracket{(2\alpha'+\beta)}+\underbracket{2e^{\tau_1}(C'+C)},\\ &\underbracket{(-2\beta'+\alpha)}-\underbracket{2e^{\tau_1}(D'+D)}=0. \end{align*} Since $\alpha,\beta$ depend only on $\tau_2$ and $C,D$ only on $\tau_1$ we have separation of variables like in PDE and then each of the marked term must be a constant. For a sake of simplicity we take them $0$: \begin{align*} &2\alpha'+\beta=0,\\ &-2\beta'+\alpha=0,\\ &C'+C=0,\\ &D'+D=0 \end{align*} and using initial conditions $\alpha(0)=1$, $\beta(0)=0$ we see that $\alpha(\tau_2)=\cos(\tau_2/2)$, $\beta(\tau_2)=\sin(\tau_2/2)$ and using initial conditions $C(0)=0$, $D(0)=1$ we see that $C(\tau_1)=0$, $D(\tau_1)=0$. Thus we arrive to \begin{align*} w_0(t,\tau_1,\tau_2)=&e^{-\tau_1}\Bigl[ \cos\bigl(\frac{\tau_2}{2}\bigr)\cos(t)+ \sin\bigl(\frac{\tau_2}{2}\bigr)\sin(t)\Bigr]= e^{-\tau_1}\cos\bigl(t-\frac{\tau_2}{2}\bigr),\\ w_1(t,\tau_1)=&e^{-\tau_1}\sin(t) \end{align*} For $w_2$ we have $w''+w=0$, $w_2(0)=0$, $w_2'(0)=0$ in virtue of (\ref{eq-8.2.13}); then $w_2=0$. Finally \begin{equation} u(t,\varepsilon)=e^{-\varepsilon t}\Bigl[ \cos\bigl(1-\frac{\varepsilon^2}{2}\bigr)t + \varepsilon \sin(t)\Bigr] \label{eq-8.2.22} \end{equation} which provides an approximation as $\varepsilon^{3}t\ll 1$.