Differential equation - looking for a minor mistake in my solution

[DivGradCurl]

Folks, I'm looking for a minor mistake in my solution. Any help is highly appreciated.

"Some diseases (such as typhoid fever) are spread largely by carriers, individuals who can transmit the disease, but who exibit no overt symptoms. Let x and y, respectively, denote the proportion of susceptibles and carriers in the population. Suppose that carriers are identified and removed from the population at a rate \beta, so

\frac{dy}{dt}=-\beta y \mbox{.} \qquad \qquad \mbox{(i)}

Suppose also that the disease spreads at a rate proportional to the product of x and y; thus

\frac{dx}{dt}=\alpha x y \mbox{.} \qquad \qquad \mbox{(ii)}

(a) Determine y at any time t by solving Eq. (i) subject to the initial condition y(0)=y_0.

(b) Use the result of part (a) to find x at any time t by solving Eq. (ii) subject to the initial condition x(0)=x_0.

(c) Find the proportion of the population that escapes the epidemic by finding the limiting amount of x as t \to \infty.

Answers:

(a) y=y_0 e^{-\beta t}

(b) x=x_0 \exp \left[ -\frac{\alpha y_0}{\beta} \left( 1 - e^{-\beta t} \right) \right]

(c) \lim _{t\to \infty} x = x_0 \exp \left( -\frac{\alpha y_0}{\beta} \right)"

My work:

(a) THIS ONE IS OK.

\frac{dy}{dt} = - \beta y\mbox{,} \qquad y(0) = y_0

\int \frac{dy}{y} = - \beta \int dt

\ln \left| y \right| = -\beta y + \mathrm{C}

\left| y \right| = \xi e^{-\beta t} \mbox{,} \qquad \xi = e^{\mathrm{C}}

y = \xi e^{-\beta t}

y(0) = y_0 \Rightarrow \xi = y_0 \Rightarrow y = y_0 e^{-\beta t}

(b) I CAN'T FIND EXACTLY WHERE I MADE A MISTAKE DOWN HERE.

\frac{dx}{dt} = \alpha x y \mbox{,} \qquad x(0) = x_0

\frac{dx}{dt} = \alpha x y_0 e^{-\beta t}

\int \frac{dx}{x} = \alpha y_0 \int e^{-\beta t} \: dt

\ln \left| x \right| = -\frac{\alpha y_0}{\beta} e^{-\beta t} + \mathrm{C}

\left| x \right| = \theta \exp \left( -\frac{\alpha y_0}{\beta} e^{-\beta t} \right) \mbox{,} \qquad \theta = e^{\mathrm{C}}

x = \theta \exp \left( -\frac{\alpha y_0}{\beta} e^{-\beta t} \right)

x(0) = x_0 \Rightarrow \theta = x_0 \exp \left( \frac{\alpha y_0}{\beta} \right) \Rightarrow x = x_0 \exp \left[ -\frac{\alpha y_0}{\beta} \left( e^{-\beta t} -1 \right) \right]

(c) I CAN'T FIND EXACTLY WHERE I MADE A MISTAKE DOWN HERE.

\lim _{t\to \infty} x = x_0 \exp \left( \frac{\alpha y_0}{\beta} \right)


Thank you

 

[AKG]

Take the derivative of the answer given for x, and what you have for x. Just looking at it, it appears your solution satisfies the differential equation, and the given answer does not. Again, I just glanced at it, but I think that the "problem" in part c just follows from the "problem" in part b; the negative sign just carries over.

 

[DivGradCurl]

That's true, but I didn't expect it. Anyway...

If

x=x_0 \exp \left[ -\frac{\alpha y_0}{\beta} \left( e^{-\beta t} - 1 \right) \right] \mbox{,}

then

\frac{dx}{dt} = \alpha x_0 y_0 \exp \left[ -\frac{\alpha y_0}{\beta} \left( e^{-\beta t} - 1 \right) - \beta t \right]

\frac{dx}{dt} = \alpha x_0 \exp \left[ -\frac{\alpha y_0}{\beta} \left( e^{-\beta t} - 1 \right) \right] y_0 e^{- \beta t}

\frac{dx}{dt} = \alpha xy

So, (b) and (c) are also correct!

Thanks