RD Sharma Solutions Class 12 Mathematics Chapter 21 CSBQ

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RD Sharma class 12 solutions Differential Equations ex CSBQ is your one-stop answer for maths. This material contains two questions that are divided into subparts summing up to 10 questions. RD Sharma Solutions It includes case studies based on fundamental concepts and can be solved by students with a bit of basic knowledge.

RD Sharma Class 12 Solutions Chapter21 CSBQ Differential Equation- Other Exercise

Differential Equations Excercise: CSBQ

Differential Equations Exercise Case Study Based Question Question 1 (i)

Answer: Option(a) $\frac{d V}{d t}=-\lambda, \lambda>0$ is a constant
Hint: Volume of sphere $=\frac{4}{3} \pi r^{3}$ & then differentiate.
Given: Volume of a spherical balloon being deflated changes at a constant rate.
Solution: Volume of a balloon at a time, $t= V$ .
Now as volume changes at constant rate, we can say
$\begin{aligned} &\frac{d}{d t}(V) \alpha 1 \\ &\frac{d}{d t}(V)=\lambda \end{aligned}$

Where $\lambda$ is some constant.

$\frac{d V}{d t}=\lambda$
But as the volume of balloon is deflated i. e., reduced so $\lambda$ will be represented by -ve value.
Thus,
$\frac{d V}{d t}=-\lambda, \lambda>0$ is a constant .

Differential Equations Exercise Case Study Based Question Question 1 (ii)

Answer: option(a) $4 \pi r^{2} \frac{d r}{d t}=-\lambda, \lambda>0$
Hint: Use formula of volume of $\text { sphere }=\frac{4}{3} \pi r^{3}$
Given: r is the radius of balloon.
Solution: $\frac{d V}{d t}=-\lambda$ as framed above
So, using formula , $\mathrm{V}=\frac{4}{3} \pi r^{3}$
Where, r = radius of spherical balloon.
$\begin{aligned} &\frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right)=-\lambda \\ &\frac{4}{3} \pi \frac{d}{d t}\left(r^{3}\right)=-\lambda \\ &\frac{4 \pi}{3} 3 r^{2} \frac{d r}{d t}=-\lambda \\ &4 \pi r^{2} \frac{d r}{d t}=-\lambda \end{aligned}$

Thus the radius at instant t is

$4 \pi r^{2} \frac{d r}{d t}=-\lambda, \lambda>0$

Differential Equations Exercise Case Study Based Question Question 1 (iii)

Answer: option(c)
Hint: Try to understand the given equation is whether of linear or non- linear or homogeneous.
Given: Differential equation in r and t
Solution: obtained differential equation $4 \pi r^{2} \frac{d r}{d t}=-\lambda$
The above formed equation is of the order 1st. So, it is a first order differential equation.
It is of non – linear type and variable can be separated.
“So it is first order non- linear differential equation in variable separable form”.
Note: Linear equation is in the form of $\frac{d y}{d x}+y=x \ldots$
While homogeneous equation is having same degree.

Differential Equations Exercise Case Study Based Question Question 1 (iv)

Answer: option(a) $\therefore r=\left(27-\frac{3 \lambda t}{4 \pi}\right)^{\frac{1}{3}}$
Hint: Solve the differential equation & find value of radius at time t.
Given: radius of balloon is 3 units, time is t and r is the radius.
Solution: $4 \pi r^{2} \frac{d r}{d t}=-\lambda$
$4 \pi r^{2} d r=-\lambda d t$
Integrating on both sides,
$\begin{aligned} &\int 4 \pi r^{2} d r=\int-\lambda d t \\ &4 \pi \int r^{2} d r=-\lambda \int d t \\ &4 \pi \frac{r^{3}}{3}=-\lambda t+c\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(i) \end{aligned}$

Now given radius is 3 and & initially i.e., time, $t= 0$

Integrating on both sides,
$\begin{aligned} 4 \pi \frac{(3)^{3}}{3} &=-\lambda(0)+c \\ 4 \pi \frac{9 \times 3}{3} &=0+c \\ 36 \pi &=c\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (ii) \end{aligned}$

Put (ii) in (i)
$\begin{aligned} 4 \pi \frac{r^{3}}{3} &=-\lambda t+36 \pi \\ 4 \pi r^{3} &=-3 \lambda t+108 \pi \\ r^{3} &=\frac{-3 \lambda t+108 \pi}{4 \pi} \\ r^{3} &=\frac{108}{4}-\frac{3 \lambda t}{4 \pi} \\ \therefore \quad r &=\left(27-\frac{3 \lambda t}{4 \pi}\right)^{\frac{1}{3}} \end{aligned}$

Differential Equations Exercise Case Study Based Question Question 1 (v)

Answer: option(b) $r=(27+63 t)^{\frac{1}{3}}$
Hint: Here the case is given of inflate , so use formula $4 \pi r^{2} \frac{d r}{d t}=-\lambda$ and solve it.
Given: radius of balloon is 6 units.
Solution: as the case seems to be of radius increasing, after 3 seconds the differential equation gets modified as ,
$\begin{aligned} &4 \pi r^{2} \frac{d r}{d t}=-\lambda \\ &4 \pi r^{2} d r=-\lambda d t \end{aligned}$
Integrating both the sides.
$\begin{aligned} &\int 4 \pi r^{2} d r=\int \lambda d t \\ &4 \pi \int r^{2} d r=\lambda \int d t \\ &4 \pi \frac{r^{3}}{3}=\lambda t+c \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(i) \end{aligned}$
Now as stated before the radius was initially 3 units.
$\begin{aligned} 4 \pi \frac{r^{3}}{3} &=\lambda t+c \\ 4 \pi \frac{(3)^{3}}{3} &=\lambda(0)+c \\ 36 \pi &=c\; \; \; \; \; \; \; \; \; \; \; \; \; .......(ii) \end{aligned}$
Put (ii) in (i)
$\therefore 4 \pi \frac{r^{3}}{3}=\lambda t+36 \pi$
Now it is given that after 3 seconds radius is 6 times.
$\begin{gathered} 4 \pi \frac{(6)^{3}}{3}=\lambda(3)+c \\ 288 \pi=3 \lambda+c \\ 288 \pi-3 \lambda=c\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ......(iii) \end{gathered}$
Put (ii) in (iii)
$\begin{aligned} &288 \pi-3 \lambda=36 \pi \\ &288 \pi-36 \pi=3 \lambda \\ &\therefore \lambda=96 \pi-12 \pi \\ &\therefore \lambda=84 \pi \; \; \; \; \; \; \; \; \; \; \; \; ....(iv) \end{aligned}$
Now put (ii) & (iv) in (i)
$\\\frac{4}{3} \pi r^{3}=84 \pi t+36 \pi\\ \begin{gathered} 4 \pi r^{3}=252 \pi t+108 \pi \\ r^{3}=63 t+\frac{108 \pi}{4 \pi} \\ r=(63 t+27)^{\frac{1}{3}} \\ \therefore r=(27+63 t)^{\frac{1}{3}} \end{gathered}$

Differential Equations Exercise Case Study Based Question Question 2 (i)

Answer: option(b) $\frac{d P}{d t}=\frac{P R}{100}$
Hint: Solve the problem by forming differential equation.
Given: principal= p
Time= t
Rate of interest= r% per annum
Solution: We know,
$I=\frac{P R T}{100}$
Where, I = rate of interest.
Interest = change in principal w .r .t to time.
So we say,
$\begin{aligned} &\frac{d}{d t}(P)=\frac{P R}{100} \\ &\therefore \frac{d P}{d t}=\frac{P R}{100} \end{aligned}$

Differential Equations Exercise Case Study Based Question Question 2 (ii)

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

Differential Equations Exercise Case Study Based Question Question 2 (iii)

Answer: option(a) $t=20 \log _{\mathrm{e}} 2$
Hint: Just put values in obtained equation.
Given $r= 5%$ ,
Solution: If $P_{0}=100 \text { then } P=2\; P_{2}=200$
Put all the values in ,
$\begin{aligned} &\log \left(\frac{P}{P_{0}}\right)=\frac{r t}{100} \\ &\log \left(\frac{200}{100}\right)=\frac{5 \times t}{100} \\ &\log 2=\frac{t}{20} \end{aligned}$
$\therefore t=20 \log _{\mathrm{e}} 2$

Differential Equations Exercise Case Study Based Question Question 2 (iv)

Answer: option(c) $r=10 \log _{\mathrm{e}} 2 \%$
Hint: Put all the values & obtain value of ‘r’
Given: Rs 100 doubles itself in 10 years.
Solution : $P_{0}=100 \text { then } P=200=2 P_{0}$
$t = 10$ years
put values in following equation,
$\begin{aligned} &\log \left(\frac{200}{100}\right)=\frac{r \times 10}{100} \\ &\log 2=\frac{\mathrm{r}}{10} \\ &\therefore r=10 \log _{\mathrm{e}} 2 \% \end{aligned}$

Differential Equations Exercise Case Study Based Question Question 2 (v)

Answer: (a) $P=1000 e^{0.5}$
Hint: Substitute values in given equation & get the solution.
Given:
$\begin{aligned} &P_{0}=1000 \\ &r=5 \% \\ &t=10 \text { years } \end{aligned}$
Solution: we use the given data accordingly,
$\begin{aligned} &\log \left(\frac{P}{P_{0}}\right)=\frac{r t}{100} \\ &\log \left(\frac{p}{1000}\right)=\frac{5 \times 10}{100} \\ &\log P-\log 1000=\frac{50}{100} \\ &\log \left(\frac{P}{1000}\right)=0.5 \end{aligned}$
$\begin{aligned} &\frac{P}{1000}=e^{0.5} \\ \\&\therefore P=1000 e^{0.5} \end{aligned}$

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Frequently Asked Questions (FAQs)

1. What are the concepts covered in chapter 21 of the class 12 maths book?

CBSE Class 12 Chapter 21 in mathematics named as Differential Equations. The concepts covered under the chapter are general and particular solutions of a differential equation, first order, First degree differential equation and many more.

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