But we still need y t from the first equation of the

Info icon This preview shows pages 3–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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 0 and y (0) = y 0 . 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 0 and y 0 be the (constant) equilibrium solutions. Then, x 0 and y 0 must satisfy the system
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
26 2 Systems of Differential Equations 0 = ax 0 + by 0 + e, 0 = cx 0 + dy 0 + 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 0 = vextendsingle vextendsingle vextendsingle vextendsingle e b f d vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ab cd vextendsingle vextendsingle vextendsingle vextendsingle , y 0 = vextendsingle vextendsingle vextendsingle vextendsingle ae cf vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ab cd 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 0 ,y 0 ) = (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 0 and v = y y 0 . Then, u 0 = v 0 = 0 . Next we are interested in the behavior of solutions near the equilibrium solutions. Later this behavior will be useful in analyzing more complicated nonlinear systems. We will look at some simple systems that are readily solved. Example 2.3. Stable Node (sink) Consider the system x = 2 x y = y. (2.17) This is a simple uncoupled system. Each equation is simply solved to give x ( t ) = c 1 e 2 t and y ( t ) = c 2 e t .
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern