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Notes-2nd order ODE pt2.pdf

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Mathematics
Course
MATH 251
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Unknown
Semester
Spring

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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: ay″ + by′ + cy = g(t). Where a, b, and c are constants, a ≠ 0; and g(t) ≠ 0. It has a corresponding homogeneous equation ay″ + by′ + cy = 0. © 2008, 2012 Zachary S Tseng B▯2 ▯ 1 Solution of the nonhomogeneous linear equations It can be verify easily that the difference y =1Y Y 2 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 = y + Y c where Y is any specific function that satisfies the nonhomogeneous equation, and y c C y1 1C y 2 2 is a general solution of the corresponding homogeneous equation y″ + p(t)y′ + q(t)y = 0 . (That is, 1 and y2are a pair of fundamental solutions of the corresponding homogeneous equation; C and C are arbitrary constants.) 1 2 The term yc= C y1 1C y 2 2 is called the complementary solution (or the homogeneous solution) of the nonhomogeneous equation. The termY is called the particular solution (or the nonhomogeneous solution) of the same equation. © 2008, 2012 Zachary S Tseng B▯2 ▯ 2 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 C 1= C 2 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: r1t r2t y cC e 1 +C e2 λ t λ t y c C e 1os ▯t + C e sin ▯t2 rt rt y c C e 1 C te 2 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 ▯ 3 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 ofthe 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 ▯ 4 2t Example: y″ − 2y′ − 3y = e The corresponding homogeneous equation y″ − 2y′ − 3y = 0 has characteristic equation r − 2r − 3 = (r + 1)(r − 3) = 0. So the complementary solution is y = C e + C e 3t. c 1 2 The nonhomogeneous equation has g(t) = e . 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 Ae for some unknown coefficient A. Our job is to find this as yet undetermined coefficient. Let Y = Ae 2t, thenY′ = 2Ae 2t, and Y″ = 4Ae 2. Substitute them back into the original differential equation: (4Ae ) − 2(2Ae ) − 3(Ae ) = e 2t 2t −3Ae = et 2t A = −1/3 Y(t) = −1 e 2t Hence, . 3 1 Therefore, y = y +c = C e 1 −t +C e2− et 2t. 3 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 ▯ 5 2 Example: y″ − 2y′ − 3y = 3t + 4t − 5 The corresponding homogeneous equation is still y″ − 2y′ − 3y = 0. Therefore, the complementary solution remains y = C e + C e 3t. c 1 2 2 Now g(t) = 3t + 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 eventuallyreaches 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 quadratic polynomial: Y = At + Bt + C . It follows Y′ = 2At + B , and Y″ = 2A . Substitute them into the equation: (2A) − 2(2At + B) − 3(At + Bt + C) = 3t + 4t − 5 2 2 − 3At + (− 4A − 3B)t + (2A − 2B − 3C) = 3t + 4t − 5 The corresponding terms on both sides should have the same coefficients, therefore, equating the coefficients of like terms. 2 t : 3 = − 3A A = −1 t : 4 = −4A − 3B → B = 0 1 : −5 = 2A − 2B − 3C C = 1 Therefore, Y = − t + 1 , and y = y c Y = C e +1C e − t 2 13t 2 . Thing to remember: When a polynomial appears in g(t), use a generic polynomial of the same degree for Y. That is Y = A t + An tn−1+ … + A t n n−1 1 + A 0. Note that if g(t) is a (nonzero) constant, it is considered a polynomial of degree 0, and Y would therefore also be a generic polynomial of degree 0. That is, Y is an arbitrary nonzero constant: Y = A .0Recall that the degree of a polynomial is the highest power that appears. Therefore, the rule can be stated a little differently to say that “look for the highest power of t in g(t), then list it and all the lower powers (down to the constant term) in Y. © 2008, 2012 Zachary S Tseng B▯2 ▯ 6 Example: y″ − 2y′ − 3y = 5cos(2t) Again, the same corresponding homogeneous equation as the previous −t 3t examples means that y c C 1 + C e 2s before. The nonhomogeneous term is g(t) = 5cos(2t). Cosine and sine functions do change form, slightly, when differentiated, but the pattern is simple, predictable, and repetitive: their respective forms just change to each other’s. Consequently, we should choose the form Y = Acos(2t) + Bsin(2t) . (Why do we choose to employ both cosine and sine?) Substitute Y, Y′ = −2Asin(2t) + 2Bcos(2t), and Y″ = −4Acos(2t) − 4Bsin(2t) into the equation: (−4Acos(2t) − 4Bsin(2t)) − 2(−2Asin(2t) + 2Bcos(2t)) − 3(Acos(2t) + Bsin(2t)) = 5cos(2t) (−4A − 4B − 3A)cos(2t) + (−4B + 4 A − 3B)sin(2t) = 5cos(2t) (−7A − 4B)cos(2t) + (4 A − 7B)sin(2t) = 5cos(2t) + 0sin(2t) Compare the coefficients: cos(2t): 5 = −7A − 4B → A = −7/13 sin(2t): 0 = 4 A − 7B → B = −4/13 −7 4 Therefore, Y = cos(2t)− sin(2 , and 13 13 −t 3t 7 4 y = C1e +C e 2 cos(2t)− sin(2t) 13 13 Thing to remember: When either cosine or sine appears in g(t), both cosine and sine (of the same frequency) must appear in Y. © 2008, 2012 Zachary S Tseng B▯2 ▯ 7 When g(t) is a sum of several terms When g(t) is a sum of several functions: g(t) = g 1(t) + g2(t) + … + g (tn, we can break the equation into n parts and solve them separately. Given y″ + p(t) y′ + q(t)y = g (t) + g (t) + … + g (t) 1 2 n we change it into y″ + p(t) y′ + q(t)y = g (t) 1 y″ + p(t) y′ + q(t)y = g 2t) : : y″ + p(t)y′ + q(t)y = g (n). Solve them individually to find respective particular solutions Y 1, Y 2, … , Y n. Then add up them to get Y = Y + Y + … + Y . 1 2 n Comment: The above is a consequence of the general version of the * Superposition Principle : General Principle of Superposition: If y is a solution of the equation 1 y″ + p(t)y′ + q(t)y = g (t1, and y 2s a solution of the equation y″ + p(t)y′ + q(t)y = g (t2. Then, for any pair of constants C and C1, the f2nction y = C y + C y is1a 1 2 2 solution of the equation y″ + p(t)y′ + q(t)y = C g 1t)1+ C g (t2.2 *Note that when 1 (t)2= g (t) = 0, the above becomes the homogeneous linear equation version of the Superposition Principle seen in an earlier section. © 2008, 2012 Zachary S Tseng B▯2 ▯ 8 2t 2 Example: y″ − 2y′ − 3y = e + 3t + 4t − 5 + 5cos(2t) Solve each of the sub▯parts: 2t −1 2t y″ − 2y′ − 3y = e → Y 1t) = e 3 y″ − 2y′ − 3y = 3t+ 4t − 5 → Y 2t) = − t + 1 −7 4 y″ − 2y′ − 3y = 5cos(2t) → Y 3t) = cos(2t)− sin(2t) 13 13 Then add up the partial results: −1 2t 2 7 4 Y(t) = 3 e −t +1− 13 cos(2t)−13 sin(2t). The general solution is −t 3t 1 2t 2 7 4 y(t) =C 1 +C e −2e −t +1− cos(2t)− sin(2t) . 3 13 13 © 2008, 2012 Zachary S Tseng B▯2 ▯ 9 If initial conditions are present, only apply the initial values after the general solution is found to find the particular solution. While it might be tempting to solve for the coefficients C an1 C as s2on as they appear (they would appear with the complementary solution y , at the very beginning), we c nevertheless could not have found them without knowing Y. Since the initial values consist of contribution from both parts y c and Y. Therefore, we must wait until we have found the general solution in its entirety before applying the initial values to find C 1and C .2 Example: y″ − 2y′ − 3y = 3t + 4t − 5, y(0) = 9, y′(0) = −4 First find the general solution: y = C e1+ C e − t2+ 1t 2 . Then use the initial conditions to find that C = 7 1nd C = 1. 2 −t 3t 2 The particular solution is: y = 7e + e − t + 1 . © 2008, 2012 Zachary S Tseng B▯2 ▯ 10 A (possible) glitch? There is a complication that occurs under a certain circumstance… 3t Example: y″ − 2y′ − 3y = 5e −t 3t 3t The old news is that y = c e + 1 e . Sin2e g(t) = 5e , we should be able to use the form Y = Ae , just like in the first example, right? 3t 3t But if we substitute Y, Y ′ = 3Ae , and Y″ = 9Ae into the differential equation and simplify, we would get the equation 3t 0 = 5e . That means there is no solution for A. Our method (that has worked well thus far) seems to have failed. The same outcome (an inability to find A) also happens when g(t) is a multiple of e . But, for any other exponent our choice of the form for Y works. What i3tso spec−tl about these two particular exponential functions, e and e , that causes our method to misfire? (Hint: What is the complementary solution of the nonhomgeneous equation?) The answer is that those two functions are exactlythe terms in y . Being a c part of the complementary solution (the solution of the corresponding homogeneous equation) means that any constant multiple of either functions will ALWAYS results in zero on the right▯hand sideof the equation. Therefore, it is impossible to match the given g(t). The cure: The remedy is surprisingly simple: multiply our usual choice by t. In the above example, we should instead use the form Y = Ate . 3t In general, whenever your initial choice of the form of Y has any term in common with the complementary solution, then you must alter it by multiplying your initial choice of Y by t, as many times as necessary but no more than necessary. © 2008, 2012 Zachary S Tseng B▯2 ▯ 11 3t Example: y″ − 6y′ + 9y = e 3t 3t 3t The complementary solution is y = C c + C 1e . g(t)2= e , therefore, the initial choice would be Y = Ae . But wait, that is the 3t same as the first term of y c, so multiply Y by t to get Y = Ate . However, the new Y is now in common with the second term of y . c Multiply it by t again to get Y = A t e . That is the final, correct choice of the general form of Y to use. (Exercise: Verify that neither Y = Ae , nor Y = Ate would yield an answer to this problem.) Once we have established that Y = At e 2 3t, then Y′ = 2Ate + 3t 2 3t 3t 3t 2 3t 3At e , and Y″ = 2Ae + 12Ate + 9At e . Substitute them back into the original equation:
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