How to obtain the following approximation?

49 Views Asked by At

In Mukhanov's "Physical foundations of cosmology" on page 102 the author considers an equation $$ \tag 1 \dot{X}_{n}(t) = -\lambda_{n\to p}(t)(1+e^{-Q/T(t)})(X_{n} - X_{n,\text{eq}}), \quad X_{n}(t\to 0) = X_{n,\text{eq}}(t), $$ where $\dot{f} \equiv \frac{df}{dt}$, $0<t<\infty$, $\lambda_{n\to p}$ is some real-valued positive function (behaving as $\lambda_{n\to p} \sim t^{-\alpha}$, where $\alpha > 0$), $Q>0$ is some constant, $T(t)>0$ is some other real valued positive function (behaving as $T \sim 1/\sqrt{t}$), $$ X_{n,\text{eq}}(t)= \frac{1}{1+e^{Q/T(t)}}, \quad X_{n,\text{eq}}(0) = \frac{1}{2} $$ The formal solution of $(1)$ is $$ \tag 2 X_{n}(t) = X_{n,\text{eq}}(t) - \int \limits_{0}^{t}d\tilde{t} \dot{X}_{n,\text{eq}}(\tilde{t})\exp\left[-\int \limits_{\tilde{t}}^{t}d\bar{t}\lambda_{n\to p}(\bar{t})(1+e^{-Q/T(\bar{t})})\right] $$ The author says that for small enough $t$ the second term tends to zero and can be considered as a perturbation. Namely, he says that it is possible to write $(2)$ as the asymptotic series in increasing powers of $\dot{X}_{n,\text{eq}}$ using the integration by parts: $$ \tag 3 X_{n}(t) = X_{n,\text{eq}}(t)\left[ 1-\frac{1}{\lambda_{n\to p}(t)(1+e^{-Q/T(t)})}\frac{\dot{X}_{n,\text{eq}}(t)}{X_{n,\text{eq}}(t)}+\dots\right] $$ I absolutely do not understand how to obtain $(3)$. Namely, after integrating by parts I obtain $$ X_{n}(t) = X_{n,\text{eq}}(0)e^{-\int \limits_{0}^{t}d\tilde{t}\lambda_{n\to p}(\tilde{t})(1+e^{-Q/T(\tilde{t})})}+ $$ $$ +\int \limits_{0}^{t}d\tilde{t}X_{n,\text{eq}}(\tilde{t})\lambda_{n\to p}(\tilde{t})(1+e^{-Q/T(\tilde{t})})e^{-\int \limits_{\tilde{t}}^{t}d\bar{t}\lambda_{n\to p}(\bar{t})(1+e^{-Q/T(\bar{t})})}, $$

and it is not clear how to obtain the result.

Could you please help me?