Lecture 17: Nonhomogeneous equations. 1 Undetermined coefficients method to find a particular

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1 Lecture 17: Nonhomogeneous equations 1 Undetermined coefficients method to find a particular solution The method of undetermined coefficients (sometimes referred to as the method of justicious guessing) is a systematic way (almost, but not quite, like using educated guesses) to determine the general form or type of the particular solution y p (t) based on the nonhomogeneous term f(t) in the given equation. Consider the second order nonhomogeneous linear differential equation with constant coefficients: Ay +By +Cy = f(t). (1.1) The characteristic polynomial (only for the left hand side) is P(λ) = Aλ 2 +Bλ+C. (1.2) Assume P(λ) = 0 has two roots λ 1,2. Fordifferent formsofright handside f(t), wehave different guesses forthesolutions. 1. If f(t) is a polynomial in t, then y(t) is also a polynomial in t (same degree). 2. If f(t) = Acosωt+Bsinωt, then y(t) = M cosωt+n sinωt. 3. If f(t) = e st, then y(t) = Ye st. However, if P(s) = 0, then we need to figure out the multiplicity of the root λ = s and modify the y(t) by an extra factor t. (a) If P(s) 0, then the particular solution could be y p (t) = 1 P(s) est. (1.3) (b) If P(s) = 0 with multiplicity m, then the particular solution can be chosen as y p (t) = tm P (m) (s) est. (1.4) 4. If f(t) is a sum of several functions: f(t) = f 1 (t)+f 2 (t)+ +f n (t), then we can break the equation into n parts and solve them separately. 1 Copy right reserved by Yingwei Wang

2 5. If f(t) is a product of two or more simple functions, e.g. f(t) = t 2 e 5t cos(3t), then our basic choice (before multiplying by t, if necessary) should be a product consist of the corresponding choices of the individual components of f(t). One thing to keep in mind: that there should be only as many undetermined coefficients in y p (t) as there are distinct terms (after expanding the expression and simplifying algebraically). For example, if f(t) = t 3 e 5t cos(3t) in (1.1), then the form of y p (t) must be a product of their corresponding choices: a generic degree 3 polynomial, an exponential function, and both cosine and sine. More precisely, we have (a) Correct form: y p (t) = (At 3 +Bt 2 +Ct+D)e 5t cos(3t)+(et 3 +Ft 2 +Gt+H)e 5t sin(3t). (b) Wrong form: y p (t) = (At 3 +Bt 2 +Ct+D)e 5t (cos(3t)+sin(3t)). Example 1. Find a particular solution to each of the following non-homogeneous equations: y +3y +2y = t+2, (1.5) y +3y +2y = e t, (1.6) y +3y +2y = e t, (1.7) y +3y +2y = e t +t+2. (1.8) Solution: Recall the Example 1 in Lecture note 15, characteristic polynomial is P(λ) = λ 2 +3λ+2. And the general solution to homogeneous equation y +3y +2y = 0 is y c (t) = c 1 e t +c 2 e 2t. 1. The paricular solution to (1.5) can be chosen in the form y p (t) = at+b. y p (t) = at+b, y p(t) = a, y p(t) = 0, 3a+2at+2b = t+2, 2 Copy right reserved by Yingwei Wang

3 2a = 1, 3a+2b = 2, a = 1 2, b = 1 4, y p (t) = 1 2 t Moreover, the complete solution to (1.5) is y(t) = c 1 e t +c 2 e 2t t The paricular solution to (1.6) can be chosen in the form y p (t) = ae t. y p (t) = ae t, y p (t) = aet, y p, (1+3+2)ae t = e t, a = 1 6, y p (t) = 1 6 et. Besides, y p (t) can also be obtained directly from (1.3). Moreover, the complete solution to (1.6) is y(t) = c 1 e t +c 2 e 2t et. 3. The right hand side f(t) = e t. We find that λ = 1 is one of the roots of the characteristic polynomial P(λ). Thus, paricular solution to (1.7) should be chosen in the form y p (t) = ate t. Due to (1.4), the coefficient a can be obtained by a = 1 P (λ) λ= 1 = Moreover, the complete solution to (1.7) (1.6) is 1 (2λ+3) λ= 1 = 1. y(t) = c 1 e t +c 2 e 2t +te t. 4. The paricular solution to (1.8) is the summation of the one to (1.5) and the one to (1.6), i.e., y p (t) = 1 6 et t Copy right reserved by Yingwei Wang

4 Example 2. Find a particular solution to each of the following non-homogeneous equations: y +y +y = cost, (1.9) y +y +y = tsint, (1.10) Solution: Recall the Example 4 in Lecture note 15, characteristic polynomial is P(λ) = λ 2 +λ+1. And the general solution to homogeneous equation y +y +y = 0 is ( ) ( ) 3 3 y c (t) = c 1 e t/2 cos 2 t +c 2 e t/2 sin 2 t. 1. The paricular solution to (1.9) can be chosen in the form y p (t) = Acost+Bsint. y p +y p +y p = ( A+B +A)cost+( B A+B)sint = cost, A+B +A = 1, B A+B = 0, A = 0, B = 1. It implies that y p (t) = sint and the complete solution to (1.9) is ( ) ( ) 3 3 y(t) = c 1 e t/2 cos 2 t +c 2 e t/2 sin 2 t +sint. 2. The paricular solution to (1.10) can be chosen in the following form y p (t) = (At+B)cost+(Ct+D)sint. Then the y p +y p +y p = tsint gives the following Ccost Asint+( A Ct D)sint+( At B +C)cost +(A+Ct+D)cost+( At B +C)sint+(Ct+D)sint = tsint. Mathcing the coefficients of tcost,tsint,cost,sint gives the following linear system A+C +A = 0, C A+C = 1, C B +C +A+D +B = 0, A A D B +C +D = 0. 4 Copy right reserved by Yingwei Wang

5 A = 1, C = 0, B = 2, D = 1, y p (t) = tcost+2cost. Moreover, the the complete solution to (1.10) is ( ) ( ) 3 3 y(t) = c 1 e t/2 cos 2 t +c 2 e t/2 sin 2 t tcost+2cost. See Examples 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and Figure in Chapter Variation of parameters method to find a particular solution Consider the second order, non-homogeneous, linear differential equation with variable coefficients y +p(t)y +q(t)y = f(t). (2.1) Suppose we have already known the general solution to the homogeneous equation y +p(t)y +q(t)y = 0 is y c (t) = c 1 y 1 (t)+c 2 y 2 (t). (2.2) Now we can assume that the particular solution takes the following form The first derivative of y p (t) is y p (t) = c 1 (t)y 1 (t)+c 2 (t)y 2 (t). (2.3) y p (t) = c 1y 1 +c 2y 2 +c 1 y 1 +c 2 y 2. (2.4) We assume that Then we arrive at c 1 y 1 +c 2 y 2 = 0. (2.5) y p(t) = c 1 y 1 +c 2 y 2, (2.6) y p(t) = c 1 y 1 +c 2 y 2 +c 1y 1 +c 2y 2. (2.7) 5 Copy right reserved by Yingwei Wang

6 Plugging (2.3), (2.6) and (2.7) into the equation (2.1) yields c 1 y 1 +c 2 y 2 = f(t). (2.8) Combining the (2.5) and (2.8) gives us the linear system with unknowns c 1,c 2: [ ][ ] [ ] y1 y 2 c 1 0 y 1 y 2 c =. (2.9) 2 f(t) We can use the Cramer s rule to solve c 1,c 2 from (2.9), i.e., c 1 (t) = y 2f W[y 1,y 2 ], c 2 (t) = y 1 f W[y 1,y 2 ], (2.10) where W[y 1,y 2 ] = y 1 y 2 y 1 y 2 is the Wronskian determinant defined in Lecture note 15. In short, the particular solution to the problem (2.1) is y 2 f y p (t) = y 1 (t) W[y 1,y 2 ] dt+y y 1 f 2(t) dt (2.11) W[y 1,y 2 ] Remark 2.1. If you substitute y,y,y in (2.1) by the y p,y p,y p defined by (2.3), (2.6) and (2.7), respectively, you will have many, many terms on the left hand side. Fortunately, if you use the relation (2.5) and also y 1 +p(t)y 1 +q(t)y 1 = 0,y 2 +p(t)y 2 +q(t)y 2 = 0, you will find that only two terms survive in the left hand side, as shown in (2.8). See the discussion on pages in your textbook. Example 3. Consider the problem y t+2 y + t+2 y = t. (2.12) t t 2 Solution: From the Example 6 in Lecture note 16, we have already known that two linearly independent solutions to the homogeneous equation are And the Wronskian determinant is y 1 (t) = t, y 2 (t) = te t. W[y 1,y 2 ] = t 2 e t. Let us assume the particular solution to be y p (t) = c 1 (t)y 1 (t)+c 2 (t)y 2 (t). 6 Copy right reserved by Yingwei Wang

7 Then in the linear system (2.9), we have y 1 (t) = t, y 2 (t) = te t, f(t) = t. Then we can obtain the c 1,c 2 by Cramer s rule It follows that c 1(t) = tet ( t) t 2 e t = 1, c t( t) 2 (t) = = e t. t 2 e t c 1 (t) = t, c 2 (t) = e t. So the particular solution to the problem (2.12) is y p (t) = c 1 (t)y 1 (t)+c 2 (t)y 2 (t), = t 2 +t. Example 4. Find the complete solution to the problem Solution: The characteristic polynomial is y y 2y = 2e t. (2.13) P(λ) = λ 2 λ 2λ = (λ 2)(λ+1). (2.14) It has two distinct real roots: λ 1 = 2,λ 2 = 1. Thus, the two linearly independant solutions to the homogeneous equation y y 2y = 0 are Now we have two ways to find the particular solution. 1. Undetermined coefficients method. y 1 (t) = e 2t, y 2 (t) = e t. (2.15) The right hand side is f(t) = 2e t. And we know that 1 is one of the roots to the characteristic polynomial (2.14), i.e., P( 1) = 0. So the particular solution should takes the following form y p (t) = Yte t. (2.16) 7 Copy right reserved by Yingwei Wang

8 It follows that Plugging y p,y p,y p y p(t) = Y(1 t)e t, y p(t) = Y(t 2)e t. into the equation (2.13) leads to 3Ye t = 2e t Y = 2 3. So the particular solution is y p (t) = 2 3 te t. (2.17) And the complete solution is where C 1,C 2 could be any two constants. y(t) = C 1 e 2t +C 2 e t 2 3 te t, (2.18) 2. Variation of parameters method. We can assume that the particular solution takes the following form: In the Eq.(2.10), we have We can get y p (t) = c 1 (t)e 2t +c 2 (t)e t. (2.19) y 1 (t) = e 2t, y 2 (t) = e t, f(t) = 2e t, W(y 1,y 2 ) = 3e t. c 1(t) = 2e 2t 3e t = 2 3 e 3t c 1 (t) = 2 9 e 3t, c 2(t) = 2et 3e t = 2 3 It follows that the particular solution is c 2 (t) = 2 3 t. y p (t) = 2 9 e t 2 3 te t. (2.20) 8 Copy right reserved by Yingwei Wang

9 So the complete solution is y(t) =C 1 e 2t +C 2 e t 2 9 e t 2 3 te t. ( =C 1 e 2t + C 2 2 ) e t te t, where C 1, C 2 could be any two constants. Example 5. Find the complete solution to the problem =C 1 e 2t + C 2 e t 2 3 te t, (2.21) Solution: See Example 11 on page 324 of your textbook. y +y = tanx. (2.22) Remark 2.2. Some comments on undetermined coefficients method and variation of parameters method. 1. Restrictions of undetermined coefficients method to find the particular solutions: (a) The differential equation must be with constant coefficients. (b) The right hand side function must be of a special form. For example, we can use the undetermined coefficients method to find a particular solution to the following equation: y 4y = te t. (2.23) However, we can not apply the undetermined coefficients method to the following problems: t 2 y 4y = te t ; (2.24) y 4y = t 1 e t. (2.25) Fortunately, the variation of parameters method works for the cases with both constant coefficients and variable coefficients. We have to use the variation of parameters method to find the particular solutions for the equations (2.24)-(2.25). 2. These two methods may give two different particular solutions, see (2.17) and (2.20). However, the complete solution should be essentially the same, see (2.18) and (2.21). 9 Copy right reserved by Yingwei Wang

10 3 The behavior of the solution as t 3.1 Various coefficients in the differential equations Consider the following second order differential equation The characteristic polynomial and its roots are y +αy +y = 0. (3.1) P(λ) = λ 2 +αλ+1, λ 1,2 = α+ α 2 4. (3.2) 2 Quesion: What are behaviors of the solutions as t for different values of α in (3.1)? 1. y y +y = 0; 2. y +y = 0; 3. y +y +y = 0; 4. y +2y +y = 0; 5. y +3y +y = Various initial values For different values of α, the solution to the following problem { y 2y 8y = 0, y(0) = α, y (0) = 2π, has different limits as t. It is easy to know that the solution to this problem is: y(t) = c 1 e 2t +c 2 e 4t, where c 1 = 2α π, c 2 = α+π. 3 3 Now we can conclude that: 1. if α > π, then c 2 > 0, so lim t y(t) = ; (3.3) 2. if α = π, then c 2 = 0, so lim t y(t) = 0; 3. if α < π, then c 2 < 0, so lim t y(t) =. 10 Copy right reserved by Yingwei Wang