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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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In a bank,principal increases continuously at the rate of r% per year.Find the value of r if Rs100 doubles itself in 10years. $\quad (log_e2=0.6931)$

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  • If the principle increases continuously at the rate of r% per year, then $\large\frac{dp}{dt}=(\large\frac{r}{100})$$p$
Step 1:
From the information given,let $p$ be the principle, $t$ be the time and $r$ be the rate of interest respectively
Using the information in the tool box $\large\frac{dp}{dt}= (\large\frac{r}{100})$$p$
Now seperating the variables
we get $\large\frac{dp}{p}=(\large\frac{r}{100})$$dt$
Step 2:
Integrating on both sides we get
$\int\large\frac{dp}{p}= (\large\frac{r}{100})\int dt$
$ \log p=\large\frac{rt}{100}$$+k$
Or $p=e^{(\Large\frac{rt}{100})+k}$
Step 3:
But it is given when $t=0,p=100$.Hence $100= e^k$
If $t=10$, then $p= 2(100) = 200$
Therefore $200= e^{[(\large\frac{r}{10})+k]} $
Or $e^{(\large\frac{r}{10}) }. e^k$
but $e^k = 100$.
Hence $200 = e^{(\Large\frac{r}{10})}$$ . 100$
$e^{(\large\frac{r}{10})} = 2$
$\large\frac{r}{10}$$ =\log_e 2$
$\large\frac{r}{10}$$= 0.6931$
Hence the value of $r=6.93%$
answered Aug 14, 2013 by sreemathi.v

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