© 2008, 2012 Zachary S Tseng B-2 - 1 Second Order Linear Nonhomogeneous Differential Equations; Method of Undetermined Coefficients We will now turn our attention to nonhomogeneous second order linear equations, equations with the standard form y″ + p(t) y′ + q(t) y = g(t), g(t) ≠ 0. (*) Each such nonhomogeneous equation has a corresponding homogeneous equation: y″ + p(t) y′ + q(t) y = 0. (**) Note that the two equations have the same left-hand side, (**) is just the homogeneous version of (*), with g(t) = 0. We will focus our attention to the simpler topic of nonhomogeneous second order linear equations with constant coefficients: a y″ + b y′ + c y = g(t). Where a, b, and c are constants, a ≠ 0; and g(t) ≠ 0. It has a corresponding homogeneous equation a y″ + b y′ + c y = 0.© 2008, 2012 Zachary S Tseng B-2 - 2 Solution of the nonhomogeneous linear equations It can be verify easily that the difference y = Y1 − Y2, of any two solutions of the nonhomogeneous equation (*), is always a solution of its corresponding homogeneous equation (**). Therefore, every solution of (*) can be obtained from a single solution of (*), by adding to it all possible solutions of its corresponding homogeneous equation (**). As a result: Theroem: The general solution of the second order nonhomogeneous linear equation y″ + p(t) y′ + q(t) y = g(t) can be expressed in the form y = yc + Y where Y is any specific function that satisfies the nonhomogeneous equation, and yc = C1 y1 + C2 y2 is a general solution of the corresponding homogeneous equation y″ + p(t) y′ + q(t) y = 0. (That is, y1 and y2 are a pair of fundamental solutions of the corresponding homogeneous equation; C1 and C2 are arbitrary constants.) The term yc = C1 y1 + C2 y2 is called the complementary solution (or the homogeneous solution) of the nonhomogeneous equation. The term Y is called the particular solution (or the nonhomogeneous solution) of the same equation.© 2008, 2012 Zachary S Tseng B-2 - 3 Comment: It should be noted that the “complementary solution” is never actually a solution of the given nonhomogeneous equation! It is merely taken from the corresponding homogeneous equation as a component that, when coupled with a particular solution, gives us the general solution of a nonhomogeneous linear equation. On the other hand, the particular solution is necessarily always a solution of the said nonhomogeneous equation. Indeed, in a slightly different context, it must be a “particular” solution of a certain initial value problem that contains the given equation and whatever initial conditions that would result in C1 = C2 = 0. In the case of nonhomgeneous equations with constant coefficients, the complementary solution can be easily found from the roots of the characteristic polynomial. They are always one of the three forms: trtrceCeCy2121+= yc = C1 e λ t cos µ t + C2 e λ t sin µ t yc = C1 e rt + C2 t e rt Therefore, the only task remaining is to find the particular solution Y, which is any one function that satisfies the given nonhomogeneous equation. That might sound like an easy task. But it is quite nontrivial. There are two general approaches to find Y : the Methods of Undetermined Coefficients, and Variation of Parameters. We will only study the former in this class.© 2008, 2012 Zachary S Tseng B-2 - 4 Method of Undetermined Coefficients The Method of Undetermined Coefficients (sometimes referred to as the method of Judicious Guessing) is a systematic way (almost, but not quite, like using “educated guesses”) to determine the general form/type of the particular solution Y(t) based on the nonhomogeneous term g(t) in the given equation. The basic idea is that many of the most familiar and commonly encountered functions have derivatives that vary little (in the form/type of function) from their parent functions: exponential, polynomials, sine and cosine. (Contrast them against log functions, whose derivatives, while simple and predictable, are rational functions; or tangent, whose higher derivatives quickly become a messy combinations of the powers of secant and tangent.) Consequently, when those functions appear in g(t), we can predict the type of function that the solution Y would be. Write down the (best guess) form of Y, leaving the coefficient(s) undetermined. Then compute Y ′ and Y ″, put them into the equation, and solve for the unknown coefficient(s). We shall see how this idea is put into practice in the following three simple examples.© 2008, 2012 Zachary S Tseng B-2 - 5 Example: y″ − 2y′ − 3y = e2t The corresponding homogeneous equation y″ − 2y′ − 3y = 0 has characteristic equation r2 − 2r − 3 = (r + 1)(r − 3) = 0. So the complementary solution is yc = C1 e−t + C2 e3t. The nonhomogeneous equation has g(t) = e2t. It is an exponential function, which does not change form after differentiation: an exponential function’s derivative will remain an exponential function with the same exponent (although its coefficient might change due to the effect of the Chain Rule). Therefore, we can very reasonably expect that Y(t) is in the form A e2t for some unknown coefficient A. Our job is to find this as yet undetermined coefficient. Let Y = A e2t, then Y ′ = 2A e2t, and Y ″ = 4A e2t. Substitute them back into the original differential equation: (4A e2t) − 2(2A e2t) − 3(A e2t) = e2t − 3A e2t = e2t A = −1 / 3 Hence, tetY231)(−=. Therefore, tttceeCeCYyy232131−+=+=−. Thing to remember: When an exponential function appears in g(t), use an exponential function of the same exponent for Y.© 2008, 2012 Zachary S Tseng B-2 - 6 Example: y″ − 2y′ − 3y = 3t2 + 4t − 5 The corresponding homogeneous equation is still y″ − 2y′ − 3y = 0. Therefore, the complementary solution remains yc = C1 e−t + C2 e3t. Now g(t) = 3t2 + 4t − 5. It is a degree 2 (i.e., quadratic) polynomial. Since polynomials, like exponential functions, do not change form after differentiation: the derivative of a polynomial is just another polynomial of one degree less (until it eventually reaches zero). We expect that Y(t) will, therefore, be a polynomial of the same degree as that of g(t). (Why will their degrees be the same?) So, we will let Y be a generic
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