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Unformatted text preview: 2 Equilibrium Solutions and Nearby Behaviors 25 The resulting differential equation has a characteristic equation of r 2 +3 r − 4 = 0 . The roots of this equation are r = 1 , − 4 . Therefore, x ( t ) = c 1 e t + c 2 e − 4 t . But, we still need y ( t ) . From the first equation of the system we have y ( t ) = 1 6 ( x ′ + x ) = 1 6 (2 c 1 e t − 3 c 2 e − 4 t ) . Thus, the solution to our system is x ( t ) = c 1 e t + c 2 e − 4 t , y ( t ) = 1 3 c 1 e t − 1 2 c 2 e − 4 t . (2.11) Sometimes one needs initial conditions. For these systems we would specify conditions like x (0) = x and y (0) = y . These would allow the determination of the arbitrary constants as before. Example 2.2. Solve x ′ = − x + 6 y y ′ = x − 2 y. (2.12) given x (0) = 2 , y (0) = 0 . We already have the general solution of this system in (2.11). Inserting the initial conditions, we have 2 = c 1 + c 2 , 0 = 1 3 c 1 − 1 2 c 2 . (2.13) Solving for c 1 and c 2 gives c 1 = 6 / 5 and c 2 = 4 / 5 . Therefore, the solution of the initial value problem is x ( t ) = 2 5 ( 3 e t + 2 e − 4 t ) , y ( t ) = 2 5 ( e t − e − 4 t ) . (2.14) 2.2 Equilibrium Solutions and Nearby Behaviors In studying systems of differential equations, it is often useful to study the behavior of solutions without obtaining an algebraic form for the solution. This is done by exploring equilibrium solutions and solutions nearby equilibrium solutions. Such techniques will be seen to be useful later in studying nonlinear systems. We begin this section by studying equilibrium solutions of system (2.4). For equilibrium solutions the system does not change in time. Therefore, equilib rium solutions satisfy the equations x ′ = 0 and y ′ = 0 . Of course, this can only happen for constant solutions. Let x and y be the (constant) equilibrium solutions. Then, x and y must satisfy the system 26 2 Systems of Differential Equations 0 = ax + by + e, 0 = cx + dy + f. (2.15) This is a linear system of nonhomogeneous algebraic equations. One only has a unique solution when the determinant of the system is not zero, i.e., ad − bc negationslash = 0 . Using Cramer’s (determinant) Rule for solving such systems, we have x = − vextendsingle vextendsingle vextendsingle vextendsingle e b f d vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle a b c d vextendsingle vextendsingle vextendsingle vextendsingle , y = − vextendsingle vextendsingle vextendsingle vextendsingle a e c f vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle a b c d vextendsingle vextendsingle vextendsingle vextendsingle . (2.16) If the system is homogeneous, e = f = 0 , then we have that the origin is the equilibrium solution; i.e., ( x ,y ) = (0 , 0). Often we will have this case since one can always make a change of coordinates from ( x,y ) to ( u,v ) by u = x − x and v = y − y . Then, u = v = 0 ....
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 Spring '13
 MRR
 Math, Equations, Constant of integration, Equilibrium point, α

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