* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download MTH 264 SECTION 3.3 20 DELTA COLLEGE The slope field for the
Survey
Document related concepts
Quartic function wikipedia , lookup
Fundamental theorem of algebra wikipedia , lookup
Linear algebra wikipedia , lookup
Singular-value decomposition wikipedia , lookup
Cubic function wikipedia , lookup
Elementary algebra wikipedia , lookup
Quadratic equation wikipedia , lookup
History of algebra wikipedia , lookup
Jordan normal form wikipedia , lookup
System of linear equations wikipedia , lookup
Perron–Frobenius theorem wikipedia , lookup
Transcript
MTH 264 DELTA COLLEGE SECTION 3.3 20 The slope field for the system dx dt dy dt = 2x + 6y = 2x − 2y is shown on the right. (a) Determine the type of the equilibrium point at the origin. (b) Calculate all straight line solutions. (c) Plot the x(t)- and y(t)-graphs (t ≥ 0) for the initial conditions A(1, −1), B(3, 1), C(0, −1), D(−1, 2) Solution: (a) The matrix that represents this system is A = 2 2 6 −2 . The characteristic equation of this matrix is: λ2 − 16 = 0. This equation factors into (λ − 4)(λ + 4) = 0, and so it has two distinct, real roots: λ1 = 4 and λ2 = −4. Since one eigenvalue is positive and one eigenvalue is negative, the equilibrium point at the origin is a saddle. (b) For eigenvalue λ1 = 4, we can select: −2 6 x 0 = 2 −6 y 0 For eigenvalue λ2 = −4, we can select: 6 6 x 0 = 2 2 y 0 eigenvector eigenvector ~v1 = 1 −1 3 1 ~v2 = ~A (t) = The two straight line solutions of this system (through A, B) are Y 3 1 4t e ~B (t) = and Y 1 −1 e−4t . (c) To construct solutions with initial conditions of C(0, −1) and D(−1, 2), we need to find linear combinations ~A and Y ~B that equal (0, −1) and (−1, 2), respectively, that of the initial conditions of the straight line solutions Y is solve the systems: 0 3 1 −1 3 1 = k1 + k2 and = l1 + l2 −1 1 −1 2 1 −1 The reader is left to solve these systems and verify that the solutions are k1 = −1/4 and k2 = 3/4, and l1 = 1/4 and l2 = −7/4, yielding solutions: 3 1 3 7 1 3 1 1 4t −4t 4t ~ ~ e + e and YD (t) = e − e−4t YC (t) = − 4 1 4 −1 4 1 4 −1 Below we sketch the four solutions in the phase plane, the four x(t)-graphs and then the four y(t)-graphs. It is imperative that you be able to identify which graphs come from which solutions. They are not labeled here and you should be sure that you can decide which is which.