$\text{ Let } X_{t}=1+t+B_{t}, \text { and } T=\inf \left\{t: X_{t}=0\right\} . \text { Define } G(t)=\int_{0}^{t \wedge T} \frac{d s}{X_{s}}. $ $\text { Let }\ \tau_{t}=G^{-1}(t) \text { be the inverse of } G, \text { and define } Y_{t}=X\left(\tau_{t}\right) . \text { Show that } Y_{t} \text { solves }\\ \text { the SDE } d Y_{t}=Y_{t} d t+\sqrt{Y}_{t} d B_{t}, Y_{0}=1, \text { for } t<\bar{T}=\inf \left\{t: Y_{t}=0\right\} \text { . }$
My attempt:
$Y_{t}=1+\int_{0}^{1} Y_{s} d s+\int_{0}^{t} \sqrt{y}_{s} d B_{s}$
Let $M_{t}=\int_{0}^{t} \sqrt{Y_{s}} d B_{S}$
$Y_{t}$ is defined as $X(\tau_{t})$ which is:
$X(\tau_{t})$ = $1 + \int_{0}^{\tau_{t}} X_{s} ds + \int_{0}^{\tau_{t}} \sqrt{X_{s}} dB_{s} $
= $1 + \int_{0}^{\tau_{t}} X_{s} ds + M(\tau_{t})$
I am not sure about this part:
By definition: $\tau_{t} = \int_{0}^{\tau_{t}} \frac{d s}{x_{s}}=t$
M has quadratic variation where $\int_{s}^{t} X_{s} d s$
Therefore by the DDS theorem: $M(\tau_{t}) = B_{t}$
$ \text{hence }X (\tau_{t}) = Y_{t} = 1+t+B_{t} $
