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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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In a bank, principal increases continuously at the rate of $5\%$ per year. An amount of $Rs.1000$ is deposited with this bank, how much will it work after $10$ years $(e^{0.5}=1.648)$.

$\begin{array}{1 1} 1648 \\ 1500 \\ 1300 \\ 1200\end{array} $

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1 Answer

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  • $\large\frac{dp}{dt} =\frac{ r}{100}$$.p$ where $p,r,t$ are principal,rate and time respectively
Step 1:
Given: Principal increases continuously at the rate of 5% per year
Hence $\large\frac{dp}{dt}=(\large\frac{5}{100})$$p$
$\large\frac{dp}{dt} =\frac{ p}{20}$
Now seperating the variables we get
Step 2:
Integrating on both sides we get
$\int\large\frac{dp}{p}=\frac{1}{20} $$\int dt$
$\log p = \large\frac{t}{20 }$$+C$
Hence $p=e^{(\Large\frac{t}{20}+C)}$
Step 3:
When $t=0, p=1000$
Hence $1000= e^C$
When $t=10$ , $p=e^{\Large(\frac{1}{2}+C)}$
or $p=e^{(0.5)}.e^C$
$p=1.648 (1000)$ [given $e^{0.5}=1.648]$
Therefore after 10 years the amount is worth Rs1648
answered Aug 14, 2013 by sreemathi.v

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