I'm taking a course in stochastic differential equations, and in order to solve $dX_{t} = X_{t}\,dt + \,dB_{t}$, the book gives a hint: to multiply both sides of this equation by $e^{-t}$. (But, as explained below, "multiplication" doesn't really mean usual multiplication because this is an integral equation.)
But the original equation $dX_{t} = X_{t}\,dt + \,dB_{t}$ really means:
$$X_{t} - X_{0} = \int \limits_{0}^{t} X_{s} \,ds + \int \limits_{0}^{t} \,dB_{s}.$$
If $X_{t}$ satisfies the above equation, then according to the hint the following equation also holds.
$$e^{-t}X_{t} - e^{-t}X_{0} = \int \limits_{0}^{t} e^{-s}X_{s}\,ds + \int \limits_{0}^{t} e^{-s} \,dB_{s}$$
But why?
What the book it hinting at is the knowledge that if $$ X_t=X_0+\int_0^t X_s\,ds +B_t, $$ then $X$ is a continuous semimartingale and $$ \int_0^t e^{-s}\,dX_s=\int_0^t e^{-s}X_s\,ds+\int_0^t e^{-s}\,dB_s. $$ The differential form of an SDE is more than just a way to save ink (or electrons); it can (as here) serve as a guide to deducing one integral equation from another. Finally, the "product rule" for the stochastic integral gives us $$ \int_0^t e^{-s}\,dX_s-\int_0^t e^{-s}X_s\,ds = e^{-t}X_t-X_0, $$ from which we deduce that $e^{-t}X_t=X_0+\int_0^t e^{-s}\,dB_s$.