Find equation from differential equation?

Question

The balloon being inflated radius changes at a constant rate. If initially its radius is $3$ units and after $3$ seconds it is $6$ units. Find the radius of the balloon after $t$ seconds. The answer is in the terms of $t$. Please suggest how to proceed ?

Answer 1

Notice, let $r$ be the radius at any time $t$ then its volume is given as $$V=\frac{4\pi}{3}r^3$$ $$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$ But the balloon is inflated at constant rate say $k$ then rate of change of volume $\frac{dV}{dt}=k$ hence we get $$4\pi r^2\frac{dr}{dt}=k$$ $$r^2\frac{dr}{dt}=\frac{k}{4\pi}$$ $$\int r^2dr=\frac{k}{4\pi}\int dt$$ $$\frac{r^3}{3}=\frac{k}{4\pi}t+C$$ Now, we have the following conditions

1. initially at time $t=0$, radius $r=3$ then $$\frac{(3)^3}{3}=\frac{k}{4\pi}(0)+C\implies C=9$$
2. at time $t=3\ sec$, radius $r=6$ then $$\frac{(6)^3}{3}=\frac{k}{4\pi}(3)+9\implies k=84\pi$$ Hence, we get $$\frac{r^3}{3}=\frac{84\pi}{4\pi}t+9$$ $$\frac{r^3}{3}=21t+9$$$$\implies \color{red}{r^3=63t+27}$$