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Exercises 17 and 18 concern a simple model of the national economy described by the difference equation $$Y_{k+2}-a(1+b) Y_{k+1}+a b Y_{k}=1$$Here $Y_{k}$ is the total national income during year $k, a$ is a constant less than $1,$ called the marginal propensity to consume, and $b$ is a positive constant of adjustment that describes how changes in consumer spending affect the annual rate of private investment.

Find the general solution of equation $(14)$ when $a=.9$ and $b=\frac{4}{9} .$ What happens to $Y_{k}$ as $k$ increases? [Hint: First find a particular solution of the form $Y_{k}=T,$ where $T$ is a constant called the equilibrium level of national income. 1

as $k$ increases, $Y _ { k }$ approach to $10 .$

Calculus 3

Chapter 4

Vector Spaces

Section 8

Applications to Difference Equations

Vectors

Mohamed R.

November 25, 2021

n bm m

Oregon State University

Baylor University

University of Michigan - Ann Arbor

Boston College

Lectures

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in this video, we're dealing with a model, which is a difference equation displayed here and critically. This is a non homogeneous difference equation due to the equals one on the right hand side, we're going to work with this model specifically when the coefficient days 0.9 and B is 4/9. And so the difference equation then becomes why K plus two minus eight times one plus B in this case becomes 1.3. Then it's multiplied by why K plus one and a times B becomes point for sweet right plus 0.4 times y que equals one. So next our goal here is to find the general solution to the model for the specific constants A is 0.9 and B s four nights. And so to that end, we need to find a particular solution. Since the right hand side is equal to a constant one. We suppose that see, or we can say, let see be a constant and assume that this constant C solves the difference equation. If we make that assumption and insert, see in place of y to the Power K plus two or twice up K plus two and so on. We obtained the equation that C minus 1.3 times see plus 0.4 times C is equal to the constant one. And if we collect all like terms, we have 0.1. Time C is equal to one, and this finally implies that the constant C must be equal to 10. What this work shows so far is that why K se equal to the constant 10 is a particulate er solution of this difference equation displayed here. So now that we have a particular solution, we can obtain all solutions by converting to the homogeneous equation. It's why K plus two minus 1.3 times y que plus 0.4 times y que should be a k plus one here equals zero. The auxiliary equation for this is going to be r squared minus 1.3 times are plus 0.4 equals zero. If we plug this auxiliary equation into the quadratic formula, we find then that the solutions are r equals 0.5 and pointing. But what this then means is that this set of solutions which are going to be 0.5 to the power of K and 2.8 to the power of K, contains the solutions to the homogeneous equation that is displayed above. Let's pause for a minute and considered this set here. Notice that the Homo genus difference equation is off degree, or rather, order to. That means the set of all solutions should be a dimension to, and this set of vectors that's displayed here does in fact contain two vectors, or signals, which are not multiples of each other, so that their linearly independent. This means that the set forms a basis for the solution set of this equation provided here. So now that we have the complete basis for the homogeneous equation as well as a particular solution, we can now combine both sets of answers to get the general solution. So it's right here that the general solution is we'll call it Y Que. And first I'm going to go to this set of vectors, will take a scaler C one and multiply by 0.5 to the power of K at a scaler C two and multiply it with the second solution to the homogeneous equation, which is 0.8 to the power of K. Then go to our particular solution found here and tack it on. So this is our solution Set where C one and C two are in. Are so there there any scale? Er's So this is our general solution and notice due to the 0.5 here and the point A found here. This implies that why k converges to 10 when K goes off to infinity since 0.5 to the parquet and point H to the power of K converged to zero as K increases without bound. So this is our general solution. But only in the case where a is 0.9 and B is four nights.

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