Video: Identifying a Differential Equation That Represents Linear Decay

If 𝑃(𝑑) is the number of unemployed people at time 𝑑 in a country, which of the following differential equations describes linear decay in the number of unemployed people? [A] d𝑃/d𝑑 = βˆ’150𝑃 [B] d𝑃/d𝑑 = βˆ’300𝑑 [C] d𝑃/d𝑑 = βˆ’300 𝑃² [D] d𝑃/d𝑑 = βˆ’150 𝑑² [E] d𝑃/d𝑑 = βˆ’300

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Video Transcript

If 𝑃 of 𝑑 is the number of unemployed people at time 𝑑 in a country, which of the following differential equations describes linear decay in the number of unemployed people? (a) d𝑃 by d𝑑 is negative 150 times 𝑃. (b) d𝑃 by d𝑑 is negative 300 times 𝑑. (c) d𝑃 by d𝑑 is negative 300 times 𝑃 squared. (d) d𝑃 by d𝑑 is negative 150 times 𝑑 squared. Or (e) d𝑃 by d𝑑 is negative 300.

To determine which of the differential equation describes linear decay in the number of unemployed people, we need to pull relevant information from the question. We’re told that the number of unemployed people at time 𝑑 in a country is described by the function 𝑃 of 𝑑. And this means that the number of unemployed people changes with time. We’re also told that the change in the number of unemployed people takes the form of linear decay.

We know that if a function is linear, it follows a straight line and that, in mathematical terms, decay refers to decreasing. If we sketch this behavior then, our function is a straight line with a negative slope. And as 𝑑 increases, the number of unemployed people falls or decreases. And symbolically, our function has the form of the equation of a straight line. That is, 𝑃 is equal to π‘š times 𝑑, which is our variable, plus 𝑏.

π‘š is the slope of our line. And 𝑏 is the 𝑦-intercept. We know that our slope π‘š is negative since we’re talking about linear decay. And remember, we’re trying to find which differential equation describes this linear decay. All of our differential equations have the first derivative on the left-hand side. So these are first-order differential equations. And if we differentiate our function 𝑃 with respect to time 𝑑, we have d𝑃 by d𝑑 equal to π‘š, which we know is a negative constant.

Remember that linear decay describes a constant decrease in values. And that’s our slope π‘š. Since π‘š is negative, let’s call it a negative 𝑛, where 𝑛 is a real number greater than zero. So d𝑃 by d𝑑 is negative 𝑛. So now, let’s compare this with our differential equations. In equation (a), d𝑃 by d𝑑 is negative 150 times 𝑃. Comparing this to our differential equation, there is a negative constant, negative 150. However, this is multiplying 𝑃 in equation (a). Our equation has no 𝑃 on the right-hand side. So we can discount equation (a) since this doesn’t match our differential equation.

Now, let’s look at equation (b). This has d𝑃 by d𝑑 equal to negative 300 times 𝑑. Again, we have a negative constant, negative 300. And this time, our constant multiplies 𝑑. And again, this doesn’t match our differential equation because our differential equation has no 𝑑 on the right-hand side. So we can discount equation (b).

And in fact, we have a similar situation for equation (c) and equation (d). Equation (c) has a negative constant times 𝑃 squared on the right-hand side. And equation (d) has a negative constant times 𝑑 squared on the right-hand side. Our differential equation has neither 𝑃 squared nor 𝑑 squared on the right-hand side. So we can discount both equation (c) and equation (d).

This leaves us with equation (e), where d𝑃 by d𝑑 is equal to negative 300. And if we let 𝑛 equal to 300, then this does match with our equation. On the left-hand side, we have d𝑃 by d𝑑. And on the right-hand side, we have a negative constant. The differential equation describing linear decay in the number of unemployed people is therefore equation (e) d𝑃 by d𝑑 is negative 300.

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