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If π of π‘ is the number of unemployed people at time π‘ in a country, which of the following differential equations describes linear decay in the number of unemployed people?
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(a) dπ by dπ‘ is negative 150 times π.
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(b) dπ by dπ‘ is negative 300 times π‘.
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(c) dπ by dπ‘ is negative 300 times π squared.
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(d) dπ by dπ‘ is negative 150 times π‘ squared.
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Or (e) dπ by dπ‘ is negative 300.
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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.
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Weβre told that the number of unemployed people at time π‘ in a country is described by the function π of π‘.
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And this means that the number of unemployed people changes with time.
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Weβre also told that the change in the number of unemployed people takes the form of linear decay.
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We know that if a function is linear, it follows a straight line and that, in mathematical terms, decay refers to decreasing.
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If we sketch this behavior then, our function is a straight line with a negative slope.
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And as π‘ increases, the number of unemployed people falls or decreases.
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And symbolically, our function has the form of the equation of a straight line.
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That is, π is equal to π times π‘, which is our variable, plus π.
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π is the slope of our line.
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And π is the π¦-intercept.
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We know that our slope π is negative since weβre talking about linear decay.
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And remember, weβre trying to find which differential equation describes this linear decay.
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All of our differential equations have the first derivative on the left-hand side.
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So these are first-order differential equations.
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And if we differentiate our function π with respect to time π‘, we have dπ by dπ‘ equal to π, which we know is a negative constant.
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Remember that linear decay describes a constant decrease in values.
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And thatβs our slope π.
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Since π is negative, letβs call it a negative π, where π is a real number greater than zero.
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So dπ by dπ‘ is negative π.
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So now, letβs compare this with our differential equations.
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In equation (a), dπ by dπ‘ is negative 150 times π.
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Comparing this to our differential equation, there is a negative constant, negative 150.
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However, this is multiplying π in equation (a).
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Our equation has no π on the right-hand side.
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So we can discount equation (a) since this doesnβt match our differential equation.
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Now, letβs look at equation (b).
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This has dπ by dπ‘ equal to negative 300 times π‘.
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Again, we have a negative constant, negative 300.
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And this time, our constant multiplies π‘.
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And again, this doesnβt match our differential equation because our differential equation has no π‘ on the right-hand side.
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So we can discount equation (b).
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And in fact, we have a similar situation for equation (c) and equation (d).
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Equation (c) has a negative constant times π squared on the right-hand side.
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And equation (d) has a negative constant times π‘ squared on the right-hand side.
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Our differential equation has neither π squared nor π‘ squared on the right-hand side.
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So we can discount both equation (c) and equation (d).
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This leaves us with equation (e), where dπ by dπ‘ is equal to negative 300.
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And if we let π equal to 300, then this does match with our equation.
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On the left-hand side, we have dπ by dπ‘.
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And on the right-hand side, we have a negative constant.
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The differential equation describing linear decay in the number of unemployed people is therefore equation (e) dπ by dπ‘ is negative 300.