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174 CHAPTER3LINEARSYSTEMS EXERCISES FOR SECTION 3.1 1. Since a > 0, Paul’s making a proÀt(x > 0) has a beneÀcial effect on Paul’s proÀts in the future because the ax term makes a positive contribution to dx/dt. However, since b < 0, Bob’s making aproÀt(y > 0) hinders Paul’s ability to make proÀt because the by term contributes negatively to dx/dt. Roughlyspeaking, business is good for Paul if his store is proÀtable and Bob’s is not. In fact, since dx/dt = x y,Paul’sproÀts will increase whenever his store is more proÀtable than Bob’s. Even though dx/dt = dy/dt = x y for this choice of parameters, the interpretation of the equation is exactly the opposite from Bob’s point of view. Since d < 0, Bob’s future proÀts are hurt whenever he is proÀtable because dy < 0. But Bob’s proÀts are helped whenever Paul is proÀtable since cx > 0. Once again, since dy/dt = x y, Bob’s proÀts will increase whenever Paul’s store is moreproÀtable than his. Finally, note that both x and y change by identical amounts since dx/dt and dy/dt are always equal. 2. Since a = 2, Paul’s making a proÀt(x > 0) has a beneÀcial effect on Paul’s future proÀts because the ax term makes a positive contribution to dx/dt. However, since b =1, Bob’s making a proÀt (y > 0) hinders Paul’s ability to make proÀt because the by term contributes negatively to dx/dt. In some sense, Paul’s proÀtability has twice the impact on his proÀts as does Bob’s proÀtability. For example, Paul’s proÀts will increase whenever his proÀts are at least one-half of Bob’s proÀts since dx/dt = 2x y. Since c = d = 0, dy/dt = 0. Consequently, Bob’s proÀts are not affected by the proÀtability of either store, and hence his proÀts are constant in this model. 3. Since a = 1andb = 0, we have dx/ dt = x. Hence, if Paul is making a proÀt(x > 0), then those proÀts will increase since dx/dt is positive. However, Bob’s proÀts have no effect on Paul’s proÀts. (Note that dx/dt = x is the standard exponential growth model.) Since c = 2andd = 1, proÀts from both stores have a positive effect on Bob’s proÀts. In some sense, Paul’s proÀts have twice the impact of Bob’s proÀts on dy/dt. 4. Since a =1andb = 2, Paul’s making a proÀt has a negative effect on his future proÀts. However, if Bob makes a proÀt, then Paul’s proÀts beneÀt. Moreover, Bob’s proÀtability has twice the impact as does Paul’s. In fact, since dx/dt =x + 2y,Paul’sproÀts will increase if x + 2y > 0or,in other words, if Bob’s proÀts are at least one-half of Paul’s proÀts. Since c = 2andd =1, Bob is in the same situation as Paul. His proÀts contribute negatively to dy/dt since d =1. However, Paul’s proÀtability has twice the positive effect. NotethatthismodelissymmetricinthesensethatbothPaulandBobperceiveeachothersproÀts in the same way. This symmetry comes from the fact that a = d and b = c. x dY 21 x dY 03 5. Y = , dt = Y6.Y= , dt= Y y 11 y 0.33π ⎛ p ⎞ dY ⎛ 3 2 7 ⎞ 7. Y = ⎜ ⎟, =⎜ ⎟Y q 206 ⎝ ⎠ dt ⎝ ⎠ r 07.32 3.1 Properties of Linear Systems and The Linearity Principle 175 8. dx =3x +2πy 9. dx =βy dt dt dy =4x y dy =γx y dt dt 10. (a) y (b) y (c) x, y 2 2 40 x(t) ❅ 20 ❅❘ ❅■ x ❅y(t) x 22 t 22 12 2 2 11. (a) y (b) y (c) x, y x(t) y(t) 2 2 3 ✠ ✠ 10 20 t x x 3 22 22 2 2 12. (a) y (b) y (c) y(t)x, y 2 20 2 ✠ 10 x t x 11 10 22 ❅■ 22 ❅x(t) 2 2 176 CHAPTER3LINEARSYSTEMS 13. (a) y (b) y (c) x, y y(t) 2 2 x(t) 1 ✠ ✠ t x 1 123 x 22 22 2 2 14. (a) If a = 0, then detA = ad bc = bc. Thus both b and c are nonzero if detA = 0. (b) Equilibrium points (x , y ) are solutions of the simultaneous system of linear equations 0 0 ⎧ ⎨ ax +by =0 0 0 ⎩ cx +dy =0. 0 0 If a = 0, the Àrst equation reduces to by = 0, and since b = 0, y = 0. In this case, the 0 0 second equation reduces to cx = 0, so x = 0 as well. Therefore, (x , y ) = (0,0) is the only 0 0 0 0 equilibrium point for the system. 15. ThevectorÀeldatapoint(x ,y )is(ax +by ,cx +dy ),soinorderforapointtobeanequilibrium 0 0 0 0 0 0 point, it must be a solution to the system of simultaneous linear equations ⎧ ⎨ ax +by =0 0 0 ⎩ cx +dy =0. 0 0 If a = 0, we know that the Àrst equation is satisÀed if and only if b x = y . 0 a 0 Nowweseethat any point that lies on this line x = (b/a)y also satisÀes the second linear 0 0 equation cx + dy = 0. In fact, if we substitute a point of this form into the second component of 0 0 the vector Àeld, we have b cx +dy =c y +dy 0 0 a 0 0 bc = +d y a 0 = adbcy a 0 = detAy a 0 =0, 3.1 Properties of Linear Systems and The Linearity Principle 177 since we are assuming that detA = 0. Hence, the line x = (b/a)y consists entirely of equilib- rium points. 0 0 If a = 0andb = 0, then the determinant condition detA = ad bc = 0 implies that c = 0. Consequently, the vector Àeld at the point (x , y ) is (by ,dy ).Sinceb = 0, we see that we get 0 0 0 0 equilibrium points if and only if y = 0. In other words, the set of equilibrium points is exactly the x-axis. 0 Finally, if a = b = 0, then the vector Àeld at the point (x , y ) is (0,cx + dy ). In this case, 0 0 0 0 weseethatapoint(x ,y )isanequilibriumpoint if and only if cx +dy = 0. Since at least one of 0 0 0 0 c or d is nonzero, the set of points (x , y ) that satisfy cx + dy = 0 is precisely a line through the origin. 0 0 0 0 16. (a) Let v = dy/dt.Thendv/dt = d2y/dt2 =qy p(dy/dt) =qy pv. Thusweobtainthe system dy =v dt dv =qy pv. dt In matrix form, this system is written as ⎛ dy ⎞ ⎜ dt ⎟= 01y . ⎝ dv ⎠ q p v dt (b) The determinant of this matrix is q. Hence, if q = 0, we know that the only equilibrium point is the origin. (c) If y is constant, then v = dy/dt is identically zero. Hence, dv/dt = 0. Also, the system reduces to ⎛ dy ⎞ ⎜ dt ⎟= 01y , ⎝ dv ⎠ q p 0 dt which implies that dv/dt =qy. Combining these two observations, we obtain dv/dt =qy = 0, and if q = 0, then y = 0. 17. The Àrst-order system corresponding to this equation is dy =v dt dv =qy pv. dt (a) If q = 0, then the system becomes dy dt =v dv =pv, dt
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