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Mathematical Methods for Physics and Engineering - Matematica.NET

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15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS(iv) If f(x) is the sum or product of any of the above then try y p (x) asthesum or product of the corresponding individual trial functions.It should be noted that this method fails if any term in the assumed trialfunction is also contained within the complementary function y c (x). In such acase the trial function should be multiplied by the smallest integer power of xsuch that it will then contain no term that already appears in the complementaryfunction. The undetermined coefficients in the trial function can now be foundby substitution into (15.8).Three further methods that are useful in finding the particular integral y p (x) arethose based on Green’s functions, the variation of parameters, <strong>and</strong> a change in thedependent variable using knowledge of the complementary function. However,since these methods are also applicable to equations with variable coefficients, adiscussion of them is postponed until section 15.2.◮Find a particular integral of the equationd 2 ydx − 2 dy2 dx + y = ex .From the above discussion our first guess at a trial particular integral would be y p (x) =be x .However, since the complementary function of this equation is y c (x) =(c 1 + c 2 x)e x (asin the previous subsection), we see that e x is already contained in it, as indeed is xe x .Multiplying our first guess by the lowest integer power of x such that the result does notappear in y c (x), we there<strong>for</strong>e try y p (x) =bx 2 e x . Substituting this into the ODE, we findthat b =1/2, so the particular integral is given by y p (x) =x 2 e x /2. ◭Solution method. If the RHS of an ODE contains only functions mentioned at thestart of this subsection then the appropriate trial function should be substitutedinto it, thereby fixing the undetermined parameters. If, however, the RHS of theequation is not of this <strong>for</strong>m then one of the more general methods outlined in subsections15.2.3–15.2.5 should be used; perhaps the most straight<strong>for</strong>ward of these isthe variation-of-parameters method.15.1.3 Constructing the general solution y c (x)+y p (x)As stated earlier, the full solution to the ODE (15.8) is found by adding togetherthe complementary function <strong>and</strong> any particular integral. In order to illustratefurther the material discussed in the last two subsections, let us find the generalsolution to a new example, starting from the beginning.495

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