# Thread: Urgent Differential Equation Modelling

1. ## Urgent Differential Equation Modelling

Hi everyone,
I have this problem that I really need somehelp with.

the rate of spread of a rumour is proportional to the product of the fraction y of the population L that has heard the rumour and the fraction who have not.

(1) Find the differential equation that is satisfied by y.

dy/dt = k * [y(L-y) / (L^2)]

(2) Show that the solution with initial condition y(0) = y0 is given by

y = (Ly0e^(kLt)) / ((L-y0)+y0e^(kLt))

I am lost on this.

(3) A school has 100 students. At 9am, 20 students have heard a rumour. At 11am 50 students have heard the rumour. At what time will 80 students will have heard the rumour.

I dont know what to do here either, however, im guessing these are the initial conditions. i.e.

y(0) = 20
y(2) = 50

2. Originally Posted by arguabsysi
Hi everyone,
I have this problem that I really need somehelp with.

the rate of spread of a rumour is proportional to the product of the fraction y of the population L that has heard the rumour and the fraction who have not.

(1) Find the differential equation that is satisfied by y.

dy/dt = k * [y(L-y) / (L^2)]

(2) Show that the solution with initial condition y(0) = y0 is given by

y = (Ly0e^(kLt)) / ((L-y0)+y0e^(kLt))

I am lost on this.

(3) A school has 100 students. At 9am, 20 students have heard a rumour. At 11am 50 students have heard the rumour. At what time will 80 students will have heard the rumour.

I dont know what to do here either, however, im guessing these are the initial conditions. i.e.

y(0) = 20
y(2) = 50

I tried this differential equation out, and it worked:

$\frac{dy}{dt}=ky(L-y)$

This is not separable. However, if I write it as follows:

$\frac{dy}{dt}=kLy-ky^2 \implies \color{red}\boxed{\frac{dy}{dt}-kLy=-ky^2}$.

This equation has the form of a Bernoulli's Equation (I cover this in my Differential Equations Tutorial):

$\frac{dy}{dt}+P(t)y=Q(t)y^n$ (general form of Bernoulli Equation)

If we make a substitution $v=y^{1-n}$ the Bernoulli Eqn will become a linear equation we know how to solve.

$v=y^{1-2}=y^{-1} \implies y=v^{-1}$

We also need a substitution for $\frac{dy}{dt}$.

We can get this by applying a definition of the chain rule: $\frac{dy}{dt}=\frac{dy}{dv}\cdot \frac{dv}{dt} \implies \frac{dy}{dt}=-\frac{1}{v^2}\frac{dv}{dt}$.

Substituting this into the DE, we get:

$-\frac{1}{v^2}\frac{dv}{dt}-kL\frac{1}{v}=-k\frac{1}{v^2}$

Multiply both sides by $-v^2$ to get a linear DE:

$\frac{dv}{dt}+kLv=k$

Apply Integrating factor:

$\rho(t)=e^{\int P(t)\,dt}=e^{\int kL \,dt}=e^{kLt}$

It should be known by now: when we multiply through by the int. factor, we get on the left side of the equation the derivative of $\rho(t)\cdot v$

Thus, our DE becomes:

$\left[e^{kLt}v\right]^{/}=ke^{kLt}$

Solving the DE, we get:

$e^{kLt}v=\frac{1}{L}e^{kLt}+C \implies v=\frac{1}{L}+Ce^{-kLt}$

We don't want v. We want y. So sub back in $v=y^{-1}$

$\therefore \frac{1}{y}=\frac{1}{L}+Ce^{-kLt}\implies y=\frac{1}{\frac{1}{L}+Ce^{-kLt}}=\frac{e^{kLt}}{\frac{e^{kLt}}{L}+C}$

When $y(0)=y_0$, we have:

$y_0=\frac{1}{\frac{1}{L}+C}\implies y_0\left(\frac{1}{L}+C\right)=1\implies \frac{1}{L}+C=\frac{1}{y_0} \implies C=\frac{1}{L}-\frac{1}{y_0}=\frac{L-y_0}{Ly_0}$

$\therefore y=\frac{e^{kLt}}{\frac{e^{kLt}}{L}+\frac{L-y_0}{Ly_0}}\implies y=\frac{e^{kLt}}{\frac{y_0e^{kLt}+(L-y_0)}{Ly_0}} \implies \color{red}\boxed{y=\frac{Ly_0e^{kLt}}{(L-y_0)+y_0e^{kLt}}}$

I believe in this case $y(0)=y_0=20$ and $L=100$

Thus,

$y=\frac{2000e^{100kt}}{(80)+20e^{100kt}}$

Since y(2)=50, we see that

$50=\frac{2000e^{200k}}{(80)+20e^{200k}}$

My calculator tells me that $k=\frac{\ln(2)}{100}$

Thus, $y=\frac{2000e^{\ln(2)t}}{(80)+20e^{\ln(2)t}}$.

Now find t when y(t)=80

$80=\frac{2000e^{\ln(2)t}}{(80)+20e^{\ln(2)t}} \implies \color{red}\boxed{t=4}$.

So at 1 PM, 80 students will have heard the rumor.

(I hope my math was right! It does make sense!)

3. Originally Posted by Chris L T521
This is not separable. However, if I write it as follows:

$\frac{dy}{dt}=kLy-ky^2 \implies \color{red}\boxed{\frac{dy}{dt}-kLy=-ky^2}$.

I hate to say this but it is separable...

$
\frac{dy}{dt}=kLy-ky^2 \iff \frac{dy}{dt}=k(L-y)y
$

$\frac{dy}{(L-y)y}=kdx$

You can use partial fractions on the left hand side or my favorite trick

$\frac{1}{(L-y)y}=\frac{1}{L}\cdot\frac{L}{(L-y)y}=\frac{1}{L}\cdot\frac{(L-y)+y}{(L-y)y}=\frac{1}{L}\left( \frac{(L-y)}{(L-y)y}+\frac{y}{(L-y)y}\right)=$

$\frac{1}{L}\frac{1}{y}+\frac{1}{L}\frac{1}{(L-y)}$

Rock on!!!

4. Originally Posted by TheEmptySet
I hate to say this but it is separable...

$
\frac{dy}{dt}=kLy-ky^2 \iff \frac{dy}{dt}=k(L-y)y
$

$\frac{dy}{(L-y)y}=kdx$

You can use partial fractions on the left hand side or my favorite trick

$\frac{1}{(L-y)y}=\frac{1}{L}\cdot\frac{L}{(L-y)y}=\frac{1}{L}\cdot\frac{(L-y)+y}{(L-y)y}=\frac{1}{L}\left( \frac{(L-y)}{(L-y)y}+\frac{y}{(L-y)y}\right)=$

$\frac{1}{L}\frac{1}{y}+\frac{1}{L}\frac{1}{(L-y)}$

Rock on!!!
I keep forgetting that...

Thanks for catching that!

Both ways are acceptable, but I think your way is easier.