Answer
The estimated number of E. coli cells after 20 minutes is 28.779.
Work Step by Step
It is given that the number of E. coli cells is modelled by the initial-value problem with the following parameters:
\begin{alignat}{2}
\frac{dy}{dt}=&0.95(0.79+0.024t)^{\frac{3}{2}} \\
& y(0) = 20 \\
\end{alignat}
First, let us integrate in order to solve the initial-value problem:
\begin{align}
y = \int(0.95(0.79+0.024t)^{\frac{3}{2}})dt \\
\end{align}
Let u = 0.79 + 0.024t $\Rightarrow$ du = 0.024dt $\Rightarrow$ dt = $\frac{du}{0.024}$
\begin{align}
y = \int0.95u^{\frac{3}{2}}dt=\int39.583u^{\frac{3}{2}}du = 15.833u^{\frac{5}{2}}+C
\end{align}
Recall that u = 0.79 + 0.024t
\begin{align}
y = 15.833(0.79 + 0.024t)^{\frac{5}{2}}+C
\end{align}
By using the initial value y(0) = 20:
\begin{alignat}{3}
20 &= 15.833\times(0.79)^{\frac{5}{2}}+C \\
&C = 11.217 \\
& y = 15.833(0.79 + 0.024t)^{\frac{5}{2}}+11.217
\end{alignat}
The estimated number of E. coli after 20 minutes:
\begin{alignat}{3}
y = 15.833(0.79 + 0.024\times20)^{\frac{5}{2}}+11.217=28.779
\end{alignat}
