Definite Integral technique for Differential Equation

Question

Update. I'm reordering this post to make the core question more clear.

Given the following problem:

Infection model: The number of people $N(t)$ affected by a pandemic at time $t$ is modelled by the differential equation. Where $r$ and $c$ are positive constants.

$\frac{dN}{dt} = r(1-c^{-1}N)N$

Which we had to show had a solution such that:

$e^{-rt} = \frac{N(t)^{-1} - c^{-1}}{N(0)^{-1} - c^{-1}}$

Why is the approach below using a definite Integral valid? The partial fractions and actual integration mechanics is fine. I just mean the method of turning this into a definite integral.

$\begin{align}\int_{N(0)}^{N(T)}\frac{c^{-1}}{1-c^{-1}N}dN + \int_{N(0)}^{N(T)}\frac{1}{N} dN = \int_0^Trdt \end{align}$

Additional Context

Using the example question above, and one other example, both taken from past exam papers. I show they can be solved using a method that seemed more obvious.

However i believe the definite integral method shown above, only seems to make sense for the infection model example. Therefore i want to understand:

1. Under what conditions it makes sense to do this definite integral method, why it works, and what the intuition of knowing to do this would be? Particularly concerning the range of integration.
2. I don't think this would have worked for the first example below (debt model)? If that's the case why is it applicable to one scenario and not the other.

Example 1: Debt model

Given the following debt model, first order linear ODE.

• $\frac{dD}{dt} - rD = -A$

We had to show that the solution looked like:

• $D(t) = D(0)e^{rt} + \frac{A}{r}(1 - e^{rt})$

We solve for the general solution using our integrating factor then solve for the arbitrary constant $C$ which gives $C = D(0) - \frac{A}{r}$, plug this in and rearrange giving the desired result.

• Brief experimentation suggested it wasn't obvious that the definite integral approach would have work here? Would it?

Example 2: (Infection model)

This is the question where the solution used a definite integral

The number of people $N(t)$ affected by a pandemic at time $t$ is modelled by the differential equation. Where $r$ and $c$ are positive constants.

$\frac{dN}{dt} = r(1-c^{-1}N)N$

Which we had to show had a solution such that:

$e^{-rt} = \frac{N(t)^{-1} - c^{-1}}{N(0)^{-1} - c^{-1}}$

Now this can be solved the same method as the debt model in example 1 (with a decent amount of algebra), in this case integrating the separable differential equation using partial fractions. The arbitrary constant we get is $C' = \frac{-N(0)}{N(0)-c}$

Plugging back in and rearranging gives the desired result.

Answer 1

Given the following debt model, first order linear ODE.

• $\frac{dD}{dt} - rD = -A$

We had to show that the solution looked like:

• $D(t) = D(0)e^{rt} + \frac{A}{r}(1 - e^{rt})$

let us solve the differential equation in this problem without using definite integrals: $$\begin{eqnarray} \frac{dD}{dt}=rD-A\\ \int\frac{dD}{rD-A}=\int dt \tag{1}\\ \frac{\ln|rD-A|}{r}=t+k \quad& \text{where k is the integration constant}\\ \ln|rD-A|=rt+kr \\ D(t)=\frac{Ce^{rt}+A}{r} \quad & \text{where }C\text{ is a constant}\\ \end{eqnarray}$$

solving for $C$ using the initial value condition $D=D(0)$:$\quad C=rD(0)-A$.

Thus, $$D(t)=D(0)e^{rt}+\frac{A}{r}(1-e^{rt})$$

Now let me modify $(1)$ to include definite integrals: $$\begin{eqnarray} & \int_{D(0)}^{D(T)}\frac{dD}{rD-A}=\int_{t=0}^{t=T}dt \tag{2}\\ & D(T)=D(0)e^{rT}+\frac{A}{r}\left(1-e^{rT}\right) \end{eqnarray}$$

We see that method prescribed by your test solutions works for the Debt model as well. Let us see why.

The LHS of $(2)$ shows us the difference between the total debts from $t=0$ to $t=T$. On the other hand, the RHS shows us the change in time, which is just equal to $T$. In other words, $(2)$ gives us a relationship between the total time that has elapsed and the total debt at time $t=T$ (i.e., $D(T)$) given the initial debt $D(0)$. This is exactly what we have found when we used the technique without definite integrals.