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Differential Equations Ncert Solutions Maths class 12th Maths Class 12th Differential Equations

57. 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 worth after 10 years

(e^0.5 = 1.645).

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Vishal Baghel | Contributor-Level 10

4 months ago

Let P and t the principal and time respectively.

Then, increase in principal $\frac{d P}{d t} = P \times 5 %$

$\Rightarrow \frac{d P}{P} = \frac{5}{100} d t$

Integrating both sides,

$\begin{array}{l} \Rightarrow \int \frac{d P}{P} = \int \frac{1}{20} d t \\ \Rightarrow l o g P = \frac{t}{20} + c \\ \Rightarrow P = e^{\frac{t}{20} + c} \end{array}$

At, t=0, P=1000

So, $\begin{array}{l} 1000 = e^{\frac{0}{20} + c} \\ \Rightarrow e^{c} = 1000 \end{array}$

And at t=10,

$\begin{array}{l} P = e^{\frac{10}{100} + c} = e^{0.5} . e^{c} \\ \Rightarrow P = 1.648 \times 1000 = 1648 \end{array}$

P = ?1648

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57. 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 worth after 10 years

(e^0.5 = 1.645).

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57. 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 worth after 10 years (e0.5 = 1.645).

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57. 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 worth after 10 years

(e^0.5 = 1.645).