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

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}