#### Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 54

Answer: $(63 t+27)^{\frac{1}{3}}$

Hint: Separate the terms of x and y and then integrate them.

Given: The volume of a spherical balloon being inflated changes at a constant rate. If initially it radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds

Solution: Let V be the volume of spherical balloon

Now it is being inflated changes at a constant rate

$\because \frac{d v}{d t}=k$where k is any constant                                                         ……………….(*)

New volume of spherical balloon$=\frac{4}{3} \pi r^{3}$, r is radius

\begin{aligned} &\frac{d v}{d t}=\frac{4}{3} \pi 3 r^{2} \frac{d r}{d t}=4 \pi r^{2} \frac{d r}{d t}\\\\ &\Rightarrow k=4 \pi r^{2} \frac{d r}{d t}\\\\ &\Rightarrow d t=\frac{4 \pi}{k} r^{2} d r \end{aligned}

Integrating both sides

\begin{aligned} &\int d t=\frac{4 \pi}{k} \int r^{2} d r\\\\ &\Rightarrow t=\frac{4 \pi}{k}\left[\frac{r^{3}}{3}\right]+c\\\\ &\Rightarrow \frac{r^{8}}{3}=\frac{k}{4 \pi} t+c \end{aligned}                ...........(1)

Now given conditions are:

When$t = 0; r = 3$ and when $t = 3, r = 6$

We have to find r at $t = t$

Put in (1) we get

\begin{aligned} &\text { At } t=0, r=3 \Rightarrow \frac{3^{8}}{3}=\frac{k}{4 \pi} 0+c \Rightarrow 9=c \\ \end{aligned}                ?By(1)

$\frac{r^{8}}{3}=\frac{k}{4 \pi} t+9$                ..................(2)

Now at  $\mathrm{t}=3, \mathrm{r}=6 \Rightarrow \frac{6^{3}}{3}=\frac{k}{4 \pi} 3+9$

$\Rightarrow \frac{6 \times 6 \times 6}{3}-9=\frac{3 k}{4 \pi}$

\begin{aligned} &\Rightarrow 72-9=\frac{3 k}{4 \pi} \Rightarrow \frac{63 \times 4 \pi}{3}=k \\ &\Rightarrow k=84 \pi \end{aligned}

Put in (2)

\begin{aligned} &\frac{r^{3}}{3}=\frac{84 \pi}{4 \pi} t+9 \Rightarrow \frac{r^{3}}{3}=21 t+9 \\\\ &\Rightarrow r^{3}=63 t+27 \\\\ &\Rightarrow r=(63 t+27)^{\frac{1}{3}} \end{aligned}