Interpret the statement carefully

The bacterial population initially grows as $N(t)=N_0 e^t$ (the question writes $N'(t)=N_0 e^t$, but from the context of population growth, this clearly means the population itself grows exponentially before the medicine acts).

The antibiotic takes 1 hour to reach the wound. So at $t=1$ hour, the bacterial population is $N(1)=N_0 e.$

Dynamics after 1 hour

For $t>1$, once the antibiotic reaches the wound, the population obeys $\frac{dN}{dt}=-5N^2.$

We solve this differential equation using separation of variables: $\frac{dN}{N^2}=-5\,dt.$

Integrating, $\int N^{-2}\,dN=\int -5\,dt$ $-\frac{1}{N}=-5t+C.$

Rearranging, $\frac{1}{N}=5t+C'.$

Apply the condition at $t=1$

At $t=1$, $N=N_0 e.$ Hence, $\frac{1}{N_0 e}=5(1)+C'$ $C'=\frac{1}{N_0 e}-5.$

Therefore, $\frac{1}{N}=5t+\frac{1}{N_0 e}-5.$

Find $\dfrac{N_0}{N}$ as a function of $t$

Multiply both sides by $N_0$: $\frac{N_0}{N}=5N_0 t+\frac{1}{e}-5N_0.$

So, $\boxed{\frac{N_0}{N}=5N_0(t-1)+\frac{1}{e}}$ for $t\ge 1$.

Nature of the plot

This is a straight line in $t$ with:

• positive slope $5N_0$
• value at $t=1$ equal to $\frac{N_0}{N(1)}=\frac{N_0}{N_0 e}=\frac{1}{e}.$

So after 1 hour, the graph of $\dfrac{N_0}{N}$ vs. $t$ is a linearly increasing straight line starting from $\left(1,\frac{1}{e}\right)$.

Match with the option

Hence the correct plot is the option showing an increasing straight line after $t=1$.

Therefore, the correct answer is D.