# Differential Equations Question (1 Viewer)

#### le_420_prince

##### New Member
This question is giving me a headache starting from part a, I really dont even understand what theyre asking for...

Thankyou!

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#### notme123

##### Member
This question is giving me a headache starting from part a, I really dont even understand what theyre asking for...

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

#### Qeru

##### Well-Known Member
Could you show what Q12 is so we actually know what the variables and constants mean?

#### CM_Tutor

##### Well-Known Member
What is Py supposed to mean? Is it multiplication or a subscript? If it's a subscript then what's that notation?
$\bg_white P$ is a constant and $\bg_white y$ is a function of $\bg_white t$ (time).

#### CM_Tutor

##### Well-Known Member
We have a function $\bg_white N(t)$ where $\bg_white N(0)=N_0$, a constant, which satisfies the differential equation (DE)

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

where $\bg_white k$ and $\bg_white P$ are constants.

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

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

which can be substituted into (*) to give:

\bg_white \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 $\bg_white r=kP$, a constant, and the DE becomes:

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

with the initial value of $\bg_white y$ being

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

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

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

From the Chain Rule, we know that

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

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

\bg_white \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, $\bg_white x$ and $\bg_white y$. At $\bg_white x=0$, it has the initial value

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

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

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

from which it follows that

\bg_white \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:

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

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

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#### Qeru

##### Well-Known Member
\bg_white \begin{align*}P\frac{dy}{dt}&=kPy(P-Py) \\ &=kP^2y(1-y)\end{align*}
Is it $\bg_white P \times y$ or $\bg_white P_y$, why can you factor out a P like that?

#### CM_Tutor

##### Well-Known Member
Is it $\bg_white P \times y$ or $\bg_white P_y$, why can you factor out a P like that?
It is $\bg_white P\times y$. The point is to take a DE that calls for partial fractions and get a solution without using that approach.

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

#### Qeru

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

If it was $\bg_white P_y$ then you are correct that $\bg_white P$ would not be a common factor that could be separated. Also, we would need a definition of $\bg_white P_y$.
Yep thanks. I think thats where the confusion was for OP since the textbook makes it look like $\bg_white P_y$.

#### le_420_prince

##### New Member
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

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

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

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

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

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

Checks

When $\bg_white t=0$:

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

as expected.

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

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

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

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

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

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

And thus $\bg_white N$ is increasing at all times.

The answer to part (f) is thus:

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

and can also be expressed as:

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

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