Question

\[\frac{dr}{dt} = - rt, r\left( 0 \right) = r_0\]

Answer 1

We have, 
\[\frac{dr}{dt} = - rt, r\left( 0 \right) = r_0\]
\[\Rightarrow \frac{1}{r}dr = - t dt\]
Integrating both sides, we get
\[\int\frac{1}{r}dr = - \int t dt\]
\[ \Rightarrow \log \left| r \right| = \frac{- t^2}{2} + C . . . . (1)\]
\[Given: t = 0, r = r_0 . \]
Substituting the values of x and y in (1), we get
\[\log \left| r_0 \right| = 0 + C\]
\[ \Rightarrow C = \log \left| r_0 \right|\]
Substituting the value of C in (1), we get
\[\log \left| r \right| = \frac{- t^2}{2} + \log \left| r_0 \right| \]
\[ \Rightarrow \log \left| r \right| - \log \left| r_0 \right| = \frac{- t^2}{2}\]
\[ \Rightarrow \log \left| \frac{r}{r_0} \right| = \frac{- t^2}{2}\]
\[ \Rightarrow r = r_0 e^\frac{- t^2}{2} \]
\[\text{ Hence, }r = r_0 e^\frac{- t^2}{2}\text{ is the required solution }.\]