#### Please solve RD Sharma Class 12 Chapter 21 Differential Equations Exercise Case Study Based Question (CSBQ) Question 1 Subquestion (v) maths textbook solution.

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)

$\\\frac{4}{3} \pi r^{3}=84 \pi t+36 \pi\\ \begin{gathered} 4 \pi r^{3}=252 \pi t+108 \pi \\ r^{3}=63 t+\frac{108 \pi}{4 \pi} \\ r=(63 t+27)^{\frac{1}{3}} \\ \therefore r=(27+63 t)^{\frac{1}{3}} \end{gathered}$