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Differential Equations : Mathematical Models

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Example Questions

Example Question #1 : Mathematical Models

Suppose a dog is carrying a virus returns to a isolated doggy day care of 40 dogs. Determine the differential equation for the number of dogs D(t) $\displaystyle D(t)$ who have contracted the virus if the rate at which it spreads is proportional to the number of interactions between the dogs with the virus and the dogs that have not yet come in contact with the virus.

Possible Answers:

$\frac{dD}{dt}=kP$ $\displaystyle \frac{dD}{dt}=kP$

$\frac{dD}{dt}=kD$ $\displaystyle \frac{dD}{dt}=kD$

$\frac{dD}{dt}=k40-D^2$ $\displaystyle \frac{dD}{dt}=k40-D^2$

$\frac{dD}{dt}=k(40-D)$ $\displaystyle \frac{dD}{dt}=k(40-D)$

$\frac{dD}{dt}=kD(40-D)$ $\displaystyle \frac{dD}{dt}=kD(40-D)$

Correct answer:

$\frac{dD}{dt}=kD(40-D)$ $\displaystyle \frac{dD}{dt}=kD(40-D)$

Explanation:

This question is asking a population dynamic type of scenario.

The total population $\displaystyle P$ in terms of time and where $\displaystyle k$ is the constant rate of proportionality, is described by the following differential equation.

$\frac{dP}{dt}=kP$ $\displaystyle \frac{dP}{dt}=kP$

For this particular function it's known that the population $\displaystyle P$ is in the form $\displaystyle D$ to represent the dogs. Since the virus spreads based on the interactions between the dogs who have the virus and those who have not yet contracted it, the population will be the product of the number of dogs with the virus and those without the virus.

Therefore the differential equation becomes,

$\frac{dD}{dt}=kD(40-D)$ $\displaystyle \frac{dD}{dt}=kD(40-D)$

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Example Question #2 : Mathematical Models

Possible Answers:

$\frac{dD}{dt}=kP$ $\displaystyle \frac{dD}{dt}=kP$

$\frac{dD}{dt}=kD(40-D)$ $\displaystyle \frac{dD}{dt}=kD(40-D)$

$\frac{dD}{dt}=k40-D^2$ $\displaystyle \frac{dD}{dt}=k40-D^2$

$\frac{dD}{dt}=kD$ $\displaystyle \frac{dD}{dt}=kD$

$\frac{dD}{dt}=k(40-D)$ $\displaystyle \frac{dD}{dt}=k(40-D)$

Correct answer:

$\frac{dD}{dt}=kD(40-D)$ $\displaystyle \frac{dD}{dt}=kD(40-D)$

Explanation:

This question is asking a population dynamic type of scenario.

The total population $\displaystyle P$ in terms of time and where $\displaystyle k$ is the constant rate of proportionality, is described by the following differential equation.

$\frac{dP}{dt}=kP$ $\displaystyle \frac{dP}{dt}=kP$

Therefore the differential equation becomes,

$\frac{dD}{dt}=kD(40-D)$ $\displaystyle \frac{dD}{dt}=kD(40-D)$

Report an Error

Example Question #2 : Mathematical Models

Possible Answers:

$\frac{dD}{dt}=k40-D^2$ $\displaystyle \frac{dD}{dt}=k40-D^2$

$\frac{dD}{dt}=k(40-D)$ $\displaystyle \frac{dD}{dt}=k(40-D)$

$\frac{dD}{dt}=kD$ $\displaystyle \frac{dD}{dt}=kD$

$\frac{dD}{dt}=kD(40-D)$ $\displaystyle \frac{dD}{dt}=kD(40-D)$

$\frac{dD}{dt}=kP$ $\displaystyle \frac{dD}{dt}=kP$

Correct answer:

$\frac{dD}{dt}=kD(40-D)$ $\displaystyle \frac{dD}{dt}=kD(40-D)$

Explanation:

This question is asking a population dynamic type of scenario.

The total population $\displaystyle P$ in terms of time and where $\displaystyle k$ is the constant rate of proportionality, is described by the following differential equation.

$\frac{dP}{dt}=kP$ $\displaystyle \frac{dP}{dt}=kP$

Therefore the differential equation becomes,

$\frac{dD}{dt}=kD(40-D)$ $\displaystyle \frac{dD}{dt}=kD(40-D)$

Report an Error

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Differential Equations : Mathematical Models

Study concepts, example questions & explanations for Differential Equations

All Differential Equations Resources

Example Questions

Example Question #1 : Mathematical Models

Example Question #2 : Mathematical Models

Example Question #2 : Mathematical Models

All Differential Equations Resources

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