MATH251-Worksheet 2

# MATH251-Worksheet 2 - MATH 251 Work sheet / Things to know...

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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 ″ - 25 y = 0 (b) y ″ + 25 y = 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 .

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© 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 = e rt always satisfies the equation above. Therefore, y = e rt will be a particular solution for each root r . Consequently, an important formula to remember for this class is (surprisingly) the quadratic formula: a ac b b r 2 4 2 - ± - = 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 r 1 and r 2 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 ′ - 12 y = 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 2 y ″ + 3 y ′ - 2 y = 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 ′ - 12 y = 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 t 0 ≠ 0? Ex. 3.3.2 y ″ + y ′ - 12 y = 0, y (45000) = 0, y ′(45000) = 2

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© 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?
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## This note was uploaded on 07/16/2011 for the course MATH 251 taught by Professor Chezhongyuan during the Spring '08 term at Pennsylvania State University, University Park.

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MATH251-Worksheet 2 - MATH 251 Work sheet / Things to know...

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