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PSU MATH 251 - Work sheet

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© 2011 Zachary S Tseng 1 MATH 251 Work sheet / Things to know Chapter 3 1. Second order linear differential equation Standard form: What makes it homogeneous? We will, for the most part, work with equations with constant coefficients only. What is the general form of a second order linear equation with constant coefficients? Ex. 3.1.1 Can you think of any function(s) that satisfy each equation (w/ constant coefficients) below? (a) y″ − 25y = 0 (b) y″ + 25y = 0 (c) y″ − 25 y′ = 0 The example (c) above is an instance of a second order linear equation with the y-term missing. It is essentially a first order linear equation in disguise. All equations of this type can be solved by changing it into a first order equation with the substitutions u = y′ and u′ = y″, then use the integrating factor method to solve for u, and integrate the result to find y.© 2011 Zachary S Tseng 2 2. The characteristic equation Given the equation ay″ + by′ + cy = 0, a ≠ 0, what is its characteristic equation? Any root, r, of the characteristic equation has the property that y = ert always satisfies the equation above. Therefore, y = ert will be a particular solution for each root r. Consequently, an important formula to remember for this class is (surprisingly) the quadratic formula: aacbbr242−±−= Note that the characteristic equation method does not require the given differential equation to be put into its standard form first – the quadratic formula simply doesn’t care whether or not the leading coefficient is 1. Suppose r1 and r2 are two distinct real roots of the characteristic equation, what is the general solution of the differential equation? y = Ex. 3.2.1 y″ + y′ − 12y = 0 What is its characteristic equation? What are the roots of the characteristic equation? Based on the roots, what are 2 particular solutions of the equation? The general solution is y = Ex. 3.2.2 2y″ + 3y′ − 2y = 0 y =© 2011 Zachary S Tseng 3 3. Initial Value Problems What do the initial conditions of a second order differential equation look like? A second order equation’s general solution will have 2 arbitrary constants / coefficients. Therefore, an IVP will have 2 initial conditions in order to give 2 (algebraic) equations needed to solve for them. What must the conditions look like? Ex. 3.3.1 Take the previous example y″ + y′ − 12y = 0. Find its solution satisfying (a) y(0) = 0, y′(0) = 2 (b) y(0) = 2, y′(0) = −6 How to (easily) work with initial conditions where t0 ≠ 0? Ex. 3.3.2 y″ + y′ − 12y = 0, y(45000) = 0, y′(45000) = 2© 2011 Zachary S Tseng 4 The Existence and Uniqueness Theorem (for second order linear equations) It is really the same theorem as the one we saw earlier, except this one is in the context of second order linear equation. What does it say? How to find the largest interval (that is, the interval of validity) on which a particular solution is guaranteed to exist uniquely? Ex. 3.3.3 Consider the equation given below. For each set of initial conditions, find the largest interval on which the particular solution is guaranteed to exist uniquely. (t2 + 2t – 8) y″ + sin(2t) y′ + (t + 9)3 y = t−1e5t (a) y(3) = 1, y′(3) = −9 (b) y(−1) = 9π, y′(−1) = e−1© 2011 Zachary S Tseng 5 4. The general solution of second order linear equations Structure of a general solution: homogeneous second order linear equation: y = A nonhomogeneous equation, however, will have a slightly different form of solution. Wronskian − know the determinant formula W(y1, y2)(t) = *Note that the Wronskian is defined as a function of t. Fundamental solutions What are they? Why are they important? How to determine if any two given solutions are fundamental? Ex. 3.4.1 Which pair(s) of functions below can be fundamental solutions? (1) 10, t + 2 (2) e2t, e−2t (3) et, 0 (4) 2e3t, e3t + 2 (5) cos 6t, −2sin 6t (6) 5sin t, cos(t + π/2) (7) cos 2t, sin(2t − 2π) Ex. 3.4.2 Given that y1 = te−t and y2 = 2te−t − 7ln(t) are both known solutions of a certain equation y″ + p(t)y′ + q(t)y = 0. What is the general solution of this equation?© 2011 Zachary S Tseng 6 5. The Abel’s Theorem It gives us a way to determine, up to a constant multiplier, the Wronskian of any pair of solutions of a given second order homogeneous linear equation, by looking at just the equation itself only. What does it say: W(y1, y2)(t) = (Due to their similarity) do not confuse this formula with another formula we have seen before. Which formula? Ex. 3.5.1 (Final Exam, fall 2008) Let y1(t) and y2(t) be any two solutions of the second order linear equation t2y″ − 6t y′ + cos(3t)y = 0. What is the general form of their Wronskian, W(y1, y2)(t)? Ex. 3.5.2 (Exam I, spring 2007) Consider the second order linear differential equation t y″ − 2y′ + y = 0. Suppose y1(t) and y2(t) are two fundamental solutions of the equation such that y1(1) = 2, y′1(1) = 0, y2(1) = 2, and y′2(1) = 2. Compute their Wronskian W(y1, y2)(t).© 2011 Zachary S Tseng 7 6. Characteristic equation: complex roots case (Background info) Euler’s formula: Suppose r = λ ± µi are two complex conjugate roots of the characteristic equation, what is the general solution of the differential equation? y = Ex. 3.6.1 y″ − 4y′ + 40y = 0 What are the roots of the characteristic equation? The general solution is y = What is the particular solution satisfying y(0) = 1 and y′(0) = −6? What is the particular solution satisfying y(755) = 1 and y′(755) = −6?© 2011 Zachary S Tseng 8 7. Characteristic equation: repeated real root case Suppose r is a repeated real root of the characteristic equation, what is the general solution of the differential equation? y = Ex. 3.7.1 y″ + 12y′ + 36y = 0 What are the roots of the characteristic equation? The general solution is y = What is the particular solution satisfying y(0) = 3 and y′(0) = −3? What is the particular solution satisfying


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