Exam II Review: Selected Solutions and Answers

Transcription

1 November 9, 2011 Exam II Review: Selected Solutions and Answers NOTE: For additional worked problems see last year s review sheet and answers, the notes from class, and your text. Answers to problems from the text are in the back of the book, so I don t work most of these out here. Be sure you do all the review sheet problems in preparation for the exam! 1. Define (in the context of 2nd order, linear differential equations) the following concepts and give examples where appropriate. (a) Homogeneous and non-homogeneous 2nd order linear ordinary differential equation: General second order non-homogeneous equation Homogeneous equation: set g(t) = 0. p(t)y + q(t)y + p(t)y = g(t). (b) Linear independence: Two functions f, g defined on an interval I are called linearly dependent if there exist constants c 1, c 2 not both zero so that c 1 f(t) + c 2 g(t) = 0 holds for all t in I. If this is not the case, then the functions are called linearly dependent. (c) Fundamental set of solutions: A fundamental set of solutions for a second order homogeneous equation is a pair of functions y 1 (t), y 2 (t) which are both solutions to the ODE and which and are linearly independent. By forming the linear combination y(t) = c 1 y 1 (t) + c 2 y 2 (t) from the solutions in the fundamental set, we can obtain the solution that satisfies any prescribed initial conditions by choosing the constants c 1, c 2. It does NOT make sense to talk about a fundamental set for a non-homogeneous equation. (d) The general solution to the homogeneous equation y(t) = c 1 y 1 (t) + c 2 y 2 (t) where y 1, y 2 are a fundamental set. The general solution contains all possible solutions to the ODE in the sense that given any initial condition, we can obtain the unique solution to the initial value problem by choosing the constants c 1, c 2 correctly. General solution to the non-homogeneous equation (see text 3.6): y(t) = c 1 y 1 (t) + c 2 y 2 (t) + Y (t). The combination c 1 y 1 (t) + c 2 y 2 (t) is the general solution to corresponding homogeneous equation and Y (t) is a particular solution to the non-homogeneous equation. Again, this is a general solution since we can obtain the solution to any initial value problem corresponding to this ODE by choosing the correct c 1, c 2. (e) Natural frequency and period. Quasi-period and frequency: See text chapter 3.8. (f) Resonance and amplitude modulation: See text chapter See chapters 3.1,3.4, and Give an example of two functions which are linearly independent on (, ) and two functions which are linearly dependent on (, ). Solution. Two linearly independent functions are f(t) = sin t g(t) = cos t and two linearly dependent functions are f(t) = 51! cos t + 94 π cos t. It turns out that sets of two functions are linearly dependent if and only if one is a constant multiple of the other (why?). For larger sets of functions you need the more careful definition given in 3.3.

2 4. What does the Wronskian tell you about linear dependence/independence for solutions of a 2nd order linear homogeneous differential equation? What does it tell about the linear dependence/independence of arbitrary functions? Use examples to explain your answer. Solution. For solutions to the ODE, the functions are linearly independent if and only if the Wronskian is not zero for some point t 0 in the interval. (See Theorem and remarks on 157). For arbitrary functions (not connected to any kind of differential equation), it is more complicated (see 3.3.1). In this case, the Wronskian of linearly dependent functions must be 0 for all t but it is NOT TRUE that if the Wronskian is 0 everywhere that the functions are linearly dependent. I leave the examples to you. 5. Are the functions f(t) = t and g(t) = t linearly independent on ( 1, 1)? On (0, 1)? On ( 1, 0)? Carefully explain. Solutions. Notice that if t < 0 then t = t whereas if t > 0 then t = t. The functions f(t) = t and g(t) = t are linearly dependent on (0, 1) since if we choose c 1 = 1 and c 2 = 1, we have that c 1 t + c 2 t = c 1 t + c 2 t = 0 holds for all t. Thus the functions are linearly dependent on this interval. This also true on ( 1, 0). Let c 1 = c 2 = 1: c 1 t + c 2 t = c 1 t c 2 t = 0 and this holds for all t on ( 1, 0). So the functions are linearly dependent on ( 1, 0) also. The functions must be linearly independent on ( 1, 1) however. One way to see this is to suppose that there are two constants c 1 and c 2 (not both zero) so that must hold for all t. c 1 t + c 2 t = 0 But if this is the case, then this holds for t = 1 and t = 1. In particular, we get that c 1 + c 2 = 0 c 1 + c 2 = 0 which immediately implies that c 1 = c 2 = 0. This is a contradiction so f, g must be linearly independent on this interval. 6. Find the general solution to each of the following homogeneous equations (a) 2y 13y + 6y = 0 Solution. Characteristic equation: Roots: r 1 = 6 and r 2 = 1/2. 2r 2 13r + 6 = 0 y(t) = c 1 e 6t + c 2 e t/2 (b) y 4y + 5y = 0 Solution. Characteristic equation: Roots: r 1 = 2 + i and r 2 = 2 i. r 2 4r + 5 = 0 y(t) = e 2t (c 1 cos t + c 2 sin t). (c) y 10y + 25y = 0

3 Solution. Characteristic equation: Roots: r = 5 (repeated). r 2 10r + 25 = 0 y(t) = c 1 e 5t + c 2 te 5t. 7. Describe how each of the solutions to the previous problems behave as t. Assume that y(0) = 0 and y (0) = 1 for each of the previous problems. Make a sketch of the solutions with these initial conditions. Proof. For the first problem, and y(0) = c 1 + c 2 = 0. Compute Solve the equations c 1 + c 2 = 0 and 6c 1 + c2 2 y(t) = c 1 e 6t + c 2 e t/2 y (t) = 6c 1 e 6t + c 2 2 et/2. y (0) = 6c 1 + c 2 2 = 1 = 1 to find that c 1 = 2/11 and c 2 = 2/11. Thus and y(t) as t. For the second problem, and y(0) = c 1 = 0. Compute Thus which oscillates and grows as t. For the third problem, and y(0) = c 1 = 0. Compute so that c 2 = 1/5. So and we note that y(t) as t. y(t) = 2 11 e6t 2 11 et/2 y(t) = e 2t (c 1 cos t + c 2 sin t) y (t) = 2c 2 e 2t sin t + c2e 2t cos t y (0) = c 2 = 1. y(t) = e 2t sin t y(t) = c 1 e 5t + c 2 te 5t y (t) = 5c 2 e 5t + c 2 te 5t y (0) = 5c 2 = 1 y(t) = t 5 e5t 8. Additional practice problems for homogeneous constant coefficient case: (a) 3.1: 9-11 (b) 3.4: 7-10 (c) 3.5: You must be able to use the reduction of order method. In this approach, you are given one solution y 1 (t) and must construct a second solution y 2 (t) by setting y 2 (t) = y 1 (t)v(t). You plug this back in to the original equation and find a first order equation for v. Some good practice problems are: 3.5:

4 3.6: #3 10. You must know how to use the method of undetermined coefficients: 3.6: 1-10 are good practice. I will certainly ask you to use this method for some easy non-homogeneous terms like g(t) = 3 cos 2t or g(t) = e t. 11. You must know how to use the variation of parameters formula. 3.7: 1-10 are good practice. 12. Spring-mass: , 3.9: 5,6,9. I will give you a spring mass problem (or part of one) to solve and interpret. It will be very similar to one of these problems. 13. Circuits: 3.8: 12, Some proofs; (a) Easy proofs: i. Suppose that y 1 and y 2 are a fundamental set for y + p(t)y + q(t)y = 0. Prove that c 1 y 1 (t) and c 2 y 2 (t) are a fundamental set. Proof. You must check that c 1 y 1, c 2 y 2 are both solutions to the differential equation and that they are linearly independent. Use the operator notation Now and by the linear properties of L, we have L[y] = y + p(t)y + q(t)y L[y 1 ] = L[y 2 ] = 0 L[c 1 y 1 ] = c 1 L[y 1 ] = 0. The same proof works for c 2 y 2. Therefore c 1 y 1, c 2 y 2 are both solutions to the differential equation. To verify that c 1 y 1, c 2 y 2 are linearly independent, you should compute the Wronskian. W (c 1 y 1, c 2 y 2 )(t) = c 1y 1 c 2 y 2 c 1 y 1 c 2 y 2 = c 1c 2 W (y 1, y 2 )(t) Now W (y 1, y 2 )(t) 0 since y 1, y 2 are linearly independent. Provided that c 1 and c 2 are not 0, then the solutions are linearly independent. Therefore, c 1 y 2, c 2 y 2 form a fundamental set. ii. Suppose p(t) and q(t) are continuous for all t. Find the unique solution to y + p(t)y + q(t)y = 0, y(0) = 0, y (0) = 0. Proof. We know that solutions to 2nd order linear ODE initial value problems with continuous coefficient functions are unique (theorem 3.2.1). It clear that y(t) = 0 satisfies the ODE and also the initial conditions. Therefore, it is the unique solution. iii. Prove that the sum of any two solutions to a 2nd order linear homogeneous equation is a solution to the same equation. Proof. General homogeneous linear 2nd order ODE: y + p(t)y + q(t)y = 0. Assume y 1, y 2 are solutions. Then prove this result by plugging the sum y = y 1 + y 2 into the differential equation and rearranging. iv. Suppose y 1 (t), y 2 (t) are solutions to a 2nd order linear homogeneous equation and that Y (t) is a solution to the corresponding non-homogeneous equation. Prove that is a solution the non-homogeneous equation. y = c 1 y 1 (t) + c 2 y 2 (t) + Y (t) Proof. General non-homogeneous linear 2nd order ODE: y + p(t)y + q(t)y = g(t). Prove this result by substituting y(t) into the differential equation and rearranging.

5 v. Suppose y 1 (t) is a solution to y + p(t)y + q(t)y = 0. Prove that y 2 (t) = cy 1 (t) is also solution for any constant c. Does y 1 (t), y 2 (t) constitute a fundamental set? Explain. Proof. Plug y 2 (t) into the equation to verify that it is a solution. y 1 (t) and y 2 (t) are NOT linearly independent however because one is a multiple of the other! So these are not a fundamental set. vi. The difference of two solutions to a 2nd order linear non-homogeneous equation is a solution to the corresponding homogeneous equation. Proof. Consider the non-homogeneous equation: If Y 1 (t) and Y 2 (t) are solutions to this equation y + p(t)y + q(t)y = g(t). Y 1 + p(t)y 1y + q(t)y = g(t) Y 2 + p(t)y 2 + q(t)y 2 = g(t). Subtract these two equations and simplify to obtain (Y 2 Y 1 ) + p(t)(y 2 Y 1 ) + q(t)(y 2 Y 1 ) = 0. Therefore the difference of two solutions to the non-homogeneous equation is a solution the corresponding homogeneous equation. (b) Harder proofs: i. Prove Abel s formula (Theorem 3.3.2): In text and notes. ii. Prove the variation of parameters formula: In text (Theorem 3.7.1) and notes. iii. Consider the equation ay + by + cy = 0. Suppose that the corresponding characteristic equation has one repeated root r. Prove y 1 (t) = e rt is a solution to the differential equation. Then use variation of parameters to construct a second solution: Done in class notes. iv. Use the series for e x to prove that Proof. Recall from calculus II that e (λ+µi)t = e λt (cos µt + i sin µt) e x = and that this series converges for all values of x. We will assume two facts which require some more advanced mathematics to carefully establish: A. That e (λ+iµ)t = e λt e iµt B. That we can replace x in the power series with iµt and still have a convergent series. We have e (λ+µi)t = e λt e iµt ( ) = e λt (iµt) n n! ( ) = e λt ( 1) n (µt) 2n ( 1) n 1 (µt) 2n 1 + i (2n)! (2n 1)! = e λt (cos µt + i sin µt) It might have been helpful in doing this proof to look up the cosine and sine power series. Also, its a good idea to write out the first few terms of each of the series appearing above if your not quite sure how to get the general pattern for the terms. x n n! n=1