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# set1 - University of Waterloo AMATH 351 Differential...

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University of Waterloo AMATH 351: Differential Equations 2 Fall 2005 Lecture Notes E.R. Vrscay Department of Applied Mathematics c circlecopyrt E.R. Vrscay 2005 1

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Lecture 1 Linear Second Order Differential Equations the most general form for a linear second order ordinary differential equation in the function y ( x ) is a 2 ( x ) y ′′ ( x ) + a 1 ( x ) y ( x ) + a 0 ( x ) y ( x ) = f ( x ) , x I, (1) where I R is an interval of interest that could possibly be the entire real line. Note that the coefficients of the ODE are functions of x . As we shall see, such equations are very important in applied mathematics and theoretical physics. If we can assume that a 2 ( x ) is not zero we can divide by it to case Eq. (1) in the following so-called “standard” or “normalized” form, y ′′ + P ( x ) y + Q ( x ) y = R ( x ) , (2) where we have also, for convenience, dropped the dependence of y upon the independent variable x . Here, obviously P ( x ) = a 1 ( x ) a 1 ( x ) , Q ( x ) = a 0 ( x ) a 2 ( x ) , R ( x ) = f ( x ) a 2 ( x ) . (3) As we’ll see later, we’ll be able to relax the restriction that a 2 ( x ) does not vanish at any point x X . We expect that such singular points may give problems, but we leave this discussion to later. Associated with Eq. (2) is the homogeneous DE, y ′′ + P ( x ) y + Q ( x ) y = 0 . (4) You have already encountered special cases of Eqs. (2) and (3) in your previous course on DEs – the case of constant coefficients . In this case P ( x ) = p and Q ( x ) = q , where p and q are real constants. In this special case you will recall that there are two steps in the solution of Eq. (3): 1. Find solution of the homogeneous equation (4): We assume solutions of the form y = e rx and substitute into (4) to obtain the characteristic equation r 2 + pr + q = 0 , 2
a quadratic equation with two roots, r 1 and r 2 . Depending upon the nature of these roots (e.g., real and distinct, real and equal, complex conjugate), one may extract two linearly independent solutions of the homogeneous DE (3), y 1 ( x ) and y 2 ( x ). The general solution of (3) may then be written as y h ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) , where c 1 and c 2 are arbitrary constants. 2. Find a particular solution: In some way, e.g., the method of undetermined coefficients , you find a solution y p ( x ) of Eq. (2). Then the general solution of the DE in (4) may be written as y G ( x ) = y h ( x ) + y p ( x ) (5) = c 1 y 1 ( x ) + c 2 y 2 ( x ) + y p ( x ) . Example: The ODE y ′′ + y = cos 2 x. Using methods that you learned in AMATH 250 or MATH 228, you will find (details left as an exercise) that y G ( x ) = c 1 cos x + c 2 sin x 1 3 cos 2 x, x R . Question: What does it mean to have the general solution of a DE? A rather nonmathematical, but intuitive, answer: From y G ( x ) we can generate “all” solutions of the DE in Eq. (4). This is vague, however, since we don’t know what the word “all” really means. A more precise answer is given by the following: 3

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Mathematical answer: Since the DE is a second order one, we need two pieces of information to isolate an individual solution from the set of all solutions. One way to do this is to prescribe a set of initial conditions : If we select a point x 0 I , our interval of interest (most often, x 0
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set1 - University of Waterloo AMATH 351 Differential...

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