We #160;have,drdt= rt, #160;r0=r0 #8658;1rdr #160;= t #160;dtIntegrating #160;both #160;sides, #160;we #160;get #8747;1rdr #160;= #87 ...

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Byju's Answer

Standard XII

Mathematics

Log Function

drdt=-rt, r 0...

Question

$\frac{d r}{d t} = - r t, r (0) = r_{0}$

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Solution

$We have, \frac{d r}{d t} = - r t, r (0) = r_{0}$

$\Rightarrow \frac{1}{r} d r = - t d t Integrating both sides, we get \int \frac{1}{r} d r = - \int t d t \Rightarrow \log |r| = \frac{- t^{2}}{2} + C . . . . (1) Given : t = 0, r = r_{0} . Substituting the values of x and y in (1), we get \log |r_{0}| = 0 + C \Rightarrow C = \log |r_{0}| Substituting the value of C in (1), we get$
$\log |r| = \frac{- t^{2}}{2} + \log |r_{0}| \Rightarrow \log |r| - \log |r_{0}| = \frac{- t^{2}}{2} \Rightarrow \log |\frac{r}{r_{0}}| = \frac{- t^{2}}{2} \Rightarrow r = r_{0} e^{\frac{- t^{2}}{2}} Hence, r = r_{0} e^{\frac{- t^{2}}{2}} is the required solution .$

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It has a node at $r = r_{0}$ . The value of $\frac{r_{0}}{a_{0}}$ is___

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drdt=-rt, r0=r0

$\frac{d r}{d t} = - r t, r (0) = r_{0}$