Differential Equations Question (1 Viewer)

le_420_prince · Jan 21, 2021

This question is giving me a headache starting from part a, I really dont even understand what theyre asking for...
Would anyone be able to please help?

Thankyou!

Qeru · Jan 21, 2021

Could you show what Q12 is so we actually know what the variables and constants mean?

CM_Tutor · Jan 22, 2021

notme123 said:
What is Py supposed to mean? Is it multiplication or a subscript? If it's a subscript then what's that notation?

$P$ is a constant and $y$ is a function of $t$ (time).

CM_Tutor · Jan 22, 2021

We have a function $N(t)$ where $N(0)=N_0$ , a constant, which satisfies the differential equation (DE)

$\frac{dN}{dt}=kN(P-N) \qquad \text{. . . . . . . . (*)}$

where $k$ and $P$ are constants.

(a) Let $y(t)$ be a function related to $N(t)$ by the equation $N(t) = P \times y(t)$ . It follows that

$\frac{dN}{dt}=P\times\frac{dy}{dt}$

which can be substituted into (*) to give:

$\begin{align*} \frac{dN}{dt}&=kN(P-N) \\ P\frac{dy}{dt}&=kPy(P-Py) \\ &=kP^2y(1-y) \\ \frac{dy}{dt}&=kPy(1-y) \end{align*}$

Now, let $r=kP$ , a constant, and the DE becomes:

$\frac{dy}{dt}=ry(1-y) \qquad \text{. . . . . . . . (**)}$

with the initial value of $y$ being

$y_0 = y(0) = \frac{N(0)}{P} = \frac{N_0}{P}$ .

(b) Let $x$ be a variable related to time by $x=rt$ , from which it follows that

$\frac{dx}{dt}=r \qquad \text{. . . . . . . . (1)}$ .

From the Chain Rule, we know that

$\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$

and we can substitute (**) and (1) into this to get

$\begin{align*} \frac{dy}{dx}&=\frac{dy}{dt}\times\frac{dt}{dx} \\ &=ry(1-y)\times\frac{1}{r} \\ &=y(1-y) \qquad \text{. . . . . . . . (***)} \end{align*}$

which is a DE in two variables, $x$ and $y$ . At $x=0$ , it has the initial value

$y = y_0 = \frac{N_0}{P} \quad \text{as} \quad t=0 \implies x=rt=r\times0=0$ .

(c) Let $v$ be a variable related to $y$ by

$v = \frac{1}{y} \implies \frac{dv}{dy} = -\frac{1}{y^2} \qquad \text{. . . . . . . . (2)$

from which it follows that

$\begin{align*} \frac{dv}{dx} &= \frac{dv}{dy} \times \frac{dy}{dx} \quad \text{using the Chain Rule} \\ &= -\frac{1}{y^2} \times y(1-y) \quad \text{using (2) and (***)} \\ &= \frac{y - 1}{y} \\ &= 1 - \frac{1}{y} \\ &=1-v \end{align*}$

The problem has now been transformed into one that can be solved by integration without the need to use partial fractions:

$\frac{dv}{1-v}=dx$

I'll leave the rest for you to work on.

Qeru · Jan 22, 2021

CM_Tutor said:
$\begin{align*}P\frac{dy}{dt}&=kPy(P-Py) \\ &=kP^2y(1-y)\end{align*}$

Is it $P \times y$ or $P_y$ , why can you factor out a P like that?

CM_Tutor · Jan 22, 2021

Qeru said:
Is it $P \times y$ or $P_y$ , why can you factor out a P like that?

It is $P\times y$ . The point is to take a DE that calls for partial fractions and get a solution without using that approach.

If it was $P_y$ then you are correct that $P$ would not be a common factor that could be separated. Also, we would need a definition of $P_y$ .

Qeru · Jan 22, 2021

CM_Tutor said:
It is $P\times y$ . The point is to take a DE that calls for partial fractions and get a solution without using that approach.

If it was $P_y$ then you are correct that $P$ would not be a common factor that could be separated. Also, we would need a definition of $P_y$ .

Yep thanks. I think thats where the confusion was for OP since the textbook makes it look like $P_y$ .

le_420_prince · Jan 22, 2021

CM_Tutor, thankyou so much for the solution!! and thankyou everyone for the input also!!! a massive help!!
i think i was quite confused by the P(subscript)y which is now agreed to be a misprint and instead P*y but now that clears it all up

CM_Tutor · Jan 23, 2021

$\frac{dv}{1-v}=dx \implies -\ln|1-v|=x+C \quad \text{for some constant $C$}$

At $x=0$ , we know that $y=N_0/P$ and so

$v=\frac{P}{N_0} \implies C=-\ln\left|1-\frac{P}{N_0}\right| \implies x=\ln\left|\frac{N_0-P}{N_0(1-v)}\right|$

$\implies v = 1-\frac{N_0-P}{N_0e^x} \implies y = \frac{N_0e^x}{N_0(e^x-1)+P}$

$\implies N = \frac{PN_0e^{kPt}}{N_0(e^{kPt}-1)+P}=\frac{PN_0}{(P-N_0)e^{-kPt}+N_0}$

Checks

When $t=0$ :

$N = \frac{PN_0}{(P-N_0)e^{0}+N_0}= \frac{PN_0}{P} = N_0$

as expected.

Further, as $t\to+\infty$ :

$N \to \frac{PN_0}{(P-N_0)\times 0+N_0} \to \frac{PN_0}{N_0} \to P^{-}$

Further, as $t\to-\infty$ :

$N \to \frac{PN_0}{(P-N_0)\times e^{+\infty}+N_0} \to 0^+$

We can see that the domain is $t\in\mathbb{R}$ and that $N$ is bounded below by 0 and bounded above by $P$ . That is, $N\in(0, P)$ . It follows that, provided $k>0$ :

$\frac{dN}{dt}=kN(P-N)>0$

And thus $N$ is increasing at all times.

The answer to part (f) is thus:

$N = \frac{PN_0}{(P-N_0)e^{-kPt}+N_0} = \frac{PN_0}{Pe^{-kPt}+N_0(1-e^{-kPt})} = \frac{PN_0e^{kPt}}{P+N_0(e^{kPt}-1)}$

and can also be expressed as:

$N = \frac{P}{1+\left(\cfrac{P}{N_0}-1\right)e^{-kPt}} = \frac{P}{1-\left(1-\cfrac{P}{N_0}\right)e^{-kPt}}$

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