1. Epidemic (Diff Equation)

The spread of a disease in a community is modelled by the following differential equation:

dy/dx = 0.2y - 0.02x where y is the number of infected individuals in thousands, and x the time in days.

1) Show the equation that for the family of 'curves' in the plane for which dy/dx is a constant, is the form y=mx+c.

2) Solve the equation using the linear 1st order method, given that initially there are on thousand infected individuals.

2. Originally Posted by Natasha1
The spread of a disease in a community is modelled by the following differential equation:

dy/dx = 0.2y - 0.02x where y is the number of infected individuals in thousands, and x the time in days.

1) Show the equation that for the family of 'curves' in the plane for which dy/dx is a constant, is the form y=mx+c.

2) Solve the equation using the linear 1st order method, given that initially there are on thousand infected individuals.
You have,
$\displaystyle \frac{dy}{dx}-(.2)y=-(.02)x$
This is a Linear-First Order Differencial Equation
Thus,
$\displaystyle P(x)=-.2$ and $\displaystyle Q(x)=-.02x$
Thus,
$\displaystyle I(x)=\exp \left( \int -.2dx \right)=e^{-.2x}$
Thus, the solutions are,
$\displaystyle y=\frac{1}{I(x)} \left( \int I(x)Q(x) dx \right)$
Substitute,
$\displaystyle y=e^{.2x}\int (-.02)e^{-.2x}x dx$
Thus,
$\displaystyle y=-(.02)e^{.2x}\int xe^{-.2x}dx$
Integrate by parts,
$\displaystyle y=-(.02)e^{.2x}\left( -5xe^{-.2x}+5\int e^{-.2x}dx \right)$
Thus,
$\displaystyle =(-.02)e^{.2x} \left( -5xe^{-.2x}-25e^{-.2x} \right)$
Thus,
$\displaystyle y=.1x+.5+C$
Thus all solutions are,
$\displaystyle y=.1x+C$

Maybe I made a mistake I did not check my work, need to go soon.

3. Assuming I did everything correctly and that,
$\displaystyle f(x)=.1x+C$
Then initially ya have, $\displaystyle f(0)=2000$
Thus,
$\displaystyle f(0)=.1(0)+C=2000$
Thus,
$\displaystyle C=2000$

Thus, the unique equation that satisfies this OED is,
$\displaystyle y=.1x+2000$

4. Originally Posted by ThePerfectHacker
You have,
$\displaystyle \frac{dy}{dx}-(.2)y=-(.02)x$
This is a Linear-First Order Differencial Equation
Thus,
$\displaystyle P(x)=-.2$ and $\displaystyle Q(x)=-.02x$
Thus,
$\displaystyle I(x)=\exp \left( \int -.2dx \right)=e^{-.2x}$
Thus, the solutions are,
$\displaystyle y=\frac{1}{I(x)} \left( \int I(x)Q(x) dx \right)$
Substitute,
$\displaystyle y=e^{.2x}\int (-.02)e^{-.2x}x dx$
Thus,
$\displaystyle y=-(.02)e^{.2x}\int xe^{-.2x}dx$
Integrate by parts,
$\displaystyle y=-(.02)e^{.2x}\left( -5xe^{-.2x}+5\int e^{-.2x}dx \right)$
Thus,
$\displaystyle =(-.02)e^{.2x} \left( -5xe^{-.2x}-25e^{-.2x} \right)$
Thus,
$\displaystyle y=.1x+.5+C$
Thus all solutions are,
$\displaystyle y=.1x+C$

Maybe I made a mistake I did not check my work, need to go soon.
Which can't be right, as the initial condition gives $\displaystyle C=1000$, and
substituting the solution back into the equation gives:

$\displaystyle 0.1=200$.

RonL

5. Originally Posted by CaptainBlack
Which can't be right, as the initial condition gives $\displaystyle C=1000$, and
substituting the solution back into the equation gives:

$\displaystyle 0.1=200$.

RonL
Where is the mistake?
I hate making posts when I got to got in 20 minutes and they take like 30 minutes to write.

6. Originally Posted by ThePerfectHacker
Where is the mistake?
I hate making posts when I got to got in 20 minutes and they take like 30 minutes to write.
I will have to examine the solution carefully to find it, you are using a method
I don't usually employ so it will need more carefully examination that it would
normally take.

RonL

7. I don't use that method either, well at least I haven't been taught it like that so for me it's double dutch as they say in England! Which means I still don't get it

RoL help

8. Natasha, the method I used was,
$\displaystyle \frac{dy}{dx}+P(x)y=Q(x)$
Then,
$\displaystyle y=\frac{1}{I(x)}\left( \int I(x)Q(x)dx \right)$

9. Originally Posted by Natasha1
The spread of a disease in a community is modelled by the following differential equation:

dy/dx = 0.2y - 0.02x where y is the number of infected individuals in thousands, and x the time in days.

1) Show the equation that for the family of 'curves' in the plane for which dy/dx is a constant, is the form y=mx+c.

2) Solve the equation using the linear 1st order method, given that initially there are on thousand infected individuals.
We have a linear first order ODE with constant coefficients.

$\displaystyle \frac{dy}{dx}=0.2y-0.02x$

The method I normaly employ to solve these equations is to find the general
solution $\displaystyle y_1(x)$ of the homeogeneous equation:

$\displaystyle \frac{dy_1}{dx}=0.2y_1$

Then if $\displaystyle y_2(x)$ is any solution (Particular Integral) of the original equation
then the general solution of the original equation $\displaystyle y(x)=y_1(x)+y_2(x)$.

Now to solve the homogeneous equation we try a solution of
the form: $\displaystyle y_1(x)=Ae^{kx}$, so:

$\displaystyle k\ Ae^{kx}=0.2e^{kx}$

therefore $\displaystyle k=0.2$, and $\displaystyle y_1(x)=Ae^{0.2x}$.

For the particular integral a reasonable guess would be of the form
$\displaystyle y_2(x)=ax+c$. Substituting this into the equation gives:

$\displaystyle a=0.2ax+0.2c-0.02x$,

which means that $\displaystyle a=0.1$ and $\displaystyle c=0.5$

So the general solution to the DE is:

$\displaystyle y(x)=Ae^{0.2x}+0.1x+0.5$.

Now when $\displaystyle x=0$ $\displaystyle y=1000$, so:

$\displaystyle 1000=A+0.5$,

hence $\displaystyle A=999.5$, and the solution is:

$\displaystyle y(x)=999.5e^{0.2x}+0.1x+0.5$.

Given the wording of the problem I suspect that the intention is that you
use the method that PH was proposing (as it applies to general first order
ODEs)

RonL

10. I finally found my mistake
It was a common one, notice I forgot the Constant after the derivative.