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Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equations Exercise Case Study Based Question (CSBQ) Question 2 Subquestion (ii) textbook solution.

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Answer: option(c) \log \left(\frac{P}{P_{0}}\right)=\frac{r t}{100}

Hint: Solve the equation by integrating.

Given: P_{o} is the initial principal

Solution:  \frac{d P}{d t}=\frac{P r}{100}

\frac{d P}{P}=\frac{r}{100} d t

Integrating both sides,

\int\left(\frac{1}{P}\right) d P=\int\left(\frac{r}{100}\right) d t

We know, \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c

\log P=\frac{r t}{100}+c \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(i)

Now, initially Principal, P= P_{o}

\text { i.e., at } \mathrm{t}=0, \mathrm{P}=\mathrm{P}_{0}

put values in (i)

\begin{aligned} &\log \mathrm{P}_{0}=\frac{r(0)}{100}+c \\ &\mathrm{c}=\log \mathrm{P}_{0} \end{aligned}    ........(ii)

put (ii) in (i)

\begin{aligned} &\therefore \log P=\frac{r t}{100}+\log P_{0} \\ &\therefore \log P-\log P_{0}=\frac{r t}{100} \\ &\text { Use, } \log x-\log y=\log \left(\frac{x}{y}\right) \\ &\therefore \log \left(\frac{P}{P_{0}}\right)=\frac{r t}{100} \end{aligned}

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