Differential Equations Exam 2 Summary

Exam 2 Summary Notes The exam will cover material from Section 3.1 to 3.8. There are two sets of formulas that will be provided- One is the system of equations from which we get the Variation of Parameters. The second is the cosine sum formula used in the two trig review handouts. Structure and Theory (Mostly 3.2) The goal of the theory was to establish the structure of solutions to the second order DE: y 00 + p(t)y 0 + q(t)y = g(t) We saw that two functions form a fundamental set of solutions to the homogeneous DE if the Wronskian is not zero (at the initial value of time). 1. Vocabulary: Linear operator, general solution, fundamental set of solutions, linear combination of a set of functions. 2. Theorems: • The Existence and Uniqueness Theorem for y 00 + p(t)y 0 + q(t)y = g(t). • Principle of Superposition. • Abel’s Theorem. If y1 , y2 are solutions to y 00 + p(t)y 0 + q(t)y = 0, then the Wronskian is either always zero or never zero on the interval for which the solutions are valid. That is because the Wronskian may be computed as: − W (y1 , y2 )(t) = Ce R p(t) dt • The Fundamental Set of Solutions: y 00 + p(t)y 0 + q(t)y = 0 We can guarantee that we can always find a fundamental set of solutions. We did that by appealing to the Existence and Uniqueness Theorem for the following two initial value problems: – y1 solves y 00 + p(t)y 0 + q(t)y = 0 with y(t0 ) = 1, y 0 (t0 ) = 0 – y2 solves y 00 + p(t)y 0 + q(t)y = 0 with y(t0 ) = 0, y 0 (t0 ) = 1 3. The Structure of Solutions to y 00 + p(t)y 0 + q(t)y = g(t), y(t0 ) = y0 , y 0 (t0 ) = v0 Given a fundamental set of solutions to the homogeneous equation, y1 , y2 , then there is a solution to the initial value problem, written as: y(t) = C1 y1 (t) + C2 y2 (t) + yp (t) where yp (t) solves the non-homogeneous equation. In fact, if we have: y 00 + p(t)y 0 + q(t)y = g1 (t) + g2 (t) + . . . + gn (t), we can solve by splitting the problem up into smaller problems: 1 • y1 , y2 form a fundamental set of solutions to the homogeneous equation. • yp1 solves y 00 + p(t)y 0 + q(t)y = g1 (t) • yp2 solves y 00 + p(t)y 0 + q(t)y = g2 (t) and so on.. • ypn solves y 00 + p(t)y 0 + q(t)y = gn (t) and the full solution is: y(t) = C1 y1 + C2 y2 + yp1 + yp2 + . . . + ypn Finding the Homogeneous Solution We had two distinct equations to solveay 00 + by 0 + cy = 0 y 00 + p(t)y 0 + q(t)y = 0 or First we look at the case with constant coefficients, then we look at the more general case. Constant Coefficients To solve ay 00 + by 0 + cy = 0 we use the ansatz y = ert . Then we form the associated characteristic equation: √ −b ± b2 − 4ac 2 ar + br + c = 0 ⇒ r= 2a so that the solutions depend on the discriminant, b2 − 4ac in the following way: • b2 − 4ac > 0 ⇒ two distinct real roots r1 , r2 . The general solution is: yh (t) = c1 er1 t + c2 er2 t If a, b, c > 0 (as in the Spring-Mass model) we can further say that r1 , r2 are negative. We would say that this system is OVERDAMPED. • b2 − 4ac = 0 ⇒ one real root r = −b/2a. Then the general solution is: yh (t) = e−(b/2a)t (C1 + C2 t) If a, b, c > 0 (as in the Spring-Mass model), the exponential term has a negative exponent. In this case (one real root), the system is CRITICALLY DAMPED. • b2 − 4ac < 0 ⇒ two complex conjugate solutions, r = α ± iβ. Then the solution is: yh (t) = eαt (C1 cos(βt) + C2 sin(βt)) If a, b, c > 0, then α = −(b/2a) < 0. In the case of complex roots, the system is said to the UNDERDAMPED. If α = 0 (this occurs when there is no damping), we get pure periodic motion, with period 2π/β or circular frequency β. 2 Solving the more general case We had two methods for solving the more general equation: y 00 + p(t)y 0 + q(t)y = 0 but each method relied on already having one solution, y1 (t). Given that situation, we can solve for y2 (so that y1 , y2 form a fundamental set), by one of two methods: • By use of the Wronskian: There are two ways to compute this, – W (y1 , y2 ) = Ce− – W (y1 , y2 ) = y1 y20 R p(t) dt − (This is from Abel’s Theorem) y2 y10 Therefore, these are equal, and y2 is the unknown: y1 y20 − y2 y10 = Ce− R p(t) dt • Reduction of order, where y2 = v(t)y1 (t). Finding the particular solution. Our two methods were: Method of Undetermined Coefficients and Variation of Parameters. • Method of Undetermined Coefficients This method is motivated by the observation that, a linear operator of the form L(y) = ay 00 + by 0 + cy, acting on certain classes of functions, returns the same class. In summary, the table from the text: if gi (t) is: Pn (t) Pn (t)eαt αt Pn (t)e sin(µt) or cos(µt) The ansatz ypi is: ts (a0 + a1 t + . . . an tn ) ts eαt (a0 + a1 t + . . . + an tn ) ts eαt ((a0 + a1 t + . . . + an tn ) sin(µt) + (b0 + b1 t + . . . + bn tn ) cos(µt)) The ts term comes from an analysis of the homogeneous part of the solution. That is, multiply by t or t2 so that no term of the ansatz is included as a term of the homogeneous solution. • Variation of Parameters: Given y 00 + p(t)y 0 + q(t)y = g(t), with y1 , y2 solutions to the homogeneous equation, we write the ansatz for the particular solution as: yp = u1 y1 + u2 y2 From our analysis, we saw that u1 , u2 were required to solve: u01 y1 + u02 y2 = 0 u01 y10 + u02 y20 = g(t) (these equations will be provided on any exam or quiz) From which we get the formulas for u01 and u02 : −y2 g y1 g u01 = u02 = W (y1 , y2 ) W (y1 , y2 ) 3 Analysis of the Oscillator Model Given mu00 + γu0 + ku = F (t) we should be able to determine the constants from a given setup for a spring-mass system. 1. Unforced (F (t) = 0) (a) No damping: Natural frequency is q k/m (b) With damping: Underdamped, Critically Damped, Overdamped 2. Forced (a) With no damping, Periodic forcing: Determine when Beating and Resonance occur. (b) With damping: Identify (or construct) the transient and steady-state part of the solution. With a small amount of damping, understand that we can get a slightly different kind of resonance, with the forcing frequency close to the frequency of the undamped, unforced system. Other Material 1. Be familiar with complex numbers, their polar form, and basic operations using complex numbers. 2. Know and use Euler’s Formula. Formula Page • For Variation of Parameters, the system of equations was: u01 y1 + u02 y2 = 0 u01 y10 + u02 y20 = g(t) • For the trig identities, we considered: A cos(ωt) + B sin(ωt) = R cos(ωt − δ) And the cosine formula: B−A A+B cos(A) + cos(B) = 2 cos cos 2 2  B−A A+B cos(A) − cos(B) = 2 sin sin 2 2    4