From the question it is given that,
The interest is compounded continuously at 6% per annum
So, let us assume P be the principal,
\( (\dfrac{dP}{dt}) = \dfrac{Pr}{100}\)
Then,
\( \dfrac{dP}{dt}=(\dfrac{r}{100})dt\)
Integrating on both side we get,
\( ∫ \dfrac{dP}{P} = ∫ \dfrac{r}{100} dt\)
Log P = (\( \dfrac{rt}{100}\)) + c … [equation (i)]
Let us assume Po be the initial principal at t = 0
log (Po) = 0 + C
C = log (Po)
Now, substitute the value of C in equation (i)
log (P) = \( \dfrac{rt}{100}\) + log (Po)
\(log (\dfrac{P}{P_0}) = \dfrac{rt}{100}\)
Now in case 1:
Po = 1000, t = 10 years and r = 6
\( log (\dfrac{P}{1000}) = \dfrac{(6 × 10)}{100}\)
log P – log 1000 = 0.6
log P = loge0.6 + log 1000
Taking log common in both terms,
log P = log (e0.6 + 1000)
log P = log (1.822 + 1000)
log P = log 1822
Then,
P = ₹ 1822
₹ 1000 will be ₹ 1822 after 10 years,
Now in case 2:
Let us assume t1 ne the time to double ₹ 1000,
P = 2000, Po = 1000, r = 6%
\( log (\dfrac{P}{P_0}) = \dfrac{rt}{100}\)
\( log\dfrac{2000}{1000}=\dfrac{6t_1}{100}\)
\( \dfrac{100log_2}{6}=t_1\)
\( \dfrac{(100 × 0.6931)}{6}=t_1\)
t1 = 11.55 years
Therefore, it will take 12 years approximately to double
Answered by Aaryan | 1 year agoFind the particular solution satisfying the given condition \( (\dfrac{dy}{dx}) = (x – 1)\dfrac{dy}{dx} = 2x^3y\)
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