Propagación de un virus y ejemplo

How can differential equations model the spread of a virus?

Differential equations are powerful tools in the analysis of complex phenomena such as the spread of a virus. Their ability to model these processes allows us to predict the evolution of dynamic situations and helps to implement preventive measures. This content details the method used to solve real problems, such as the spread of a virus in an animal population.

What approach do we take to define the problem?

To model the spread of a virus, several key variables are identified:

• Total population (N): Number of animals at the start of the study.
• Infected population (P): Number of infected animals at a given time.
• Time (T): Period in days from the beginning of the observation.

In the problem considered, after four days, ten animals were infected. This serves as a basis for structuring our mathematical model.

How did we create the mathematical model?

We identified that the speed of spread of the virus, equivalent to the rate of change of the infected population ( P ), is proportional to both:

• The current number of infected animals ( P ).
• The elapsed time ( T ).

The initial differential equation is:

[ \frac{dP}{dT} = kP ]

where ( k ) is a proportionality constant representing the propagation velocity.

How do we solve the differential equation?

To solve the equation, we apply the method of separation of variables:

1. We rewrite the equation as:

   [ \frac{dP}{P} = k , dT ]

2. We integrate both sides:

   • The integral of (\frac{1}{P}) is ( \ln(P) ).
   • The integral of (kT) with respect to (T) is ( \frac{kT^2}{2} ).

   Therefore, we obtain:

   [ \ln(P) = \frac{kT^2}{2} + C ]

3. We apply the exponential to solve for ( P ):

   [ P = e^{frac{kT^2}{2} + C} = e^{frac{kT^2}{2}} \cdot e^C ]

   Because ( e^C ) is constant, we simplify to ( P = Ce^{\{frac{kT^2}{2}} ).

How do we determine the constant ( C )?

Knowing that initially there was one infected animal (( P(0) = 1 )), the value of the constant ( C ) is determined:

[ 1 = Ce^0 ]

Which implies that ( C = 1 ).

How do we find the constant ( k )?

Knowing that after four days there are ten infected animals, this is substituted into the model:

[ 10 = e^{16k/2} = e^{8k} ]

We apply natural logarithm:

[ \ln(10) = 8k ]

That gives ( k = \frac{\ln(10)}{8} \approx 0.287 ).

How long does it take for the entire population to become infected?

For a total population of 1000 animals, we calculate the time ( T ) in which the entire population will be infected by substituting into the model:

[ 1000 = e^{0.287 \cdot T^2/2} ]

Applying natural logarithm and solving for ( T ):

[ T = \sqrt{0.287}{0.287}{0.287} \approx 6.92 ]

This indicates that at approximately 7 days the entire population would be infected.

The applications of this mathematical model are broad and allow us to foresee critical scenarios in public health, offering crucial tools to implement preventive measures and control the spread of contagious diseases. With knowledge, we can make better decisions and effective strategies, so keep exploring more examples in your projects!