20D - Homework Assignment 5

Transcription

1 Brian Bowers TA for Hui Sun MATH D Homework Assignment 5 November 8, 3 D - Homework Assignment 5 First, I present the list of all matrix row operations. We use combinations of these steps to row reduce matrices. Note that I will use r i to represent the ith row of the matrix. Swap: We are allowed to swap two entire rows of a matrix. [I will notate swapping rows r i and r j by r i r j.] Multiply: We are allowed to multiply any one row of a matrix by any nonzero scalar expression like, π, e t t + 7, etc. [I will notate multiplying row r i by k as r i kr i.] 3 Combine: Given two rows r i and r j, we are allowed to replace r i by r i + kr j for any scalar k. [I will notate this as r i r i + kr j.] For our purposes, row reduction is the process of trying to get a matrix to have s along the diagonal and have s above and below the diagonal. In other words, we re trying to get our matrix to look like this:?????? and then like this:??? Note that we can t always get our matrix to look exactly like the ones above, but our goal is to always get as close as possible. 7.3 #3,4 In each problem, determine whether the members of the given set of vectors are linearly independent for < t <. If they are linearly dependent, find the linear relation among them. [The vectors are written as row vectors to save space.] 3 x t e t, e t, x t e t, e t, x 3 t 3e t,. 4 x t sin t, sin t, x t sin t, sin t. A set of column vectors {x, x,..., x n } each with k entries is considered to be linearly dependent if and only if there exist scalars {c, c,..., c n } where at least one of these constants is not such that c x + c x c n x n. We could rewrite the above equation in matrix form as c x x x n c.., where the vector at the end of the equation has k entries. Using the above setup as motivation, we construct the following steps to determine if n vectors each with k entries are linearly independent or linearly dependent: c n

2 Augment: Write the augmented matrix below: x x x n, where represents the vector with k entries, all of which are. Row Reduce: Row reduce the above matrix. 3 Conclude: There are two conclusions to be reached here: i If the row reduction gives us a matrix that looks like, then we conclude that the vectors are linearly independent. ii Otherwise, we conclude that the vectors are linearly dependent. If we re deciding between dependent/independent, then we re done. However, sometimes we re asked to find a linear relation among the vectors if they are linearly dependent. In this case, each row of the row-reduced matrix gives us an equation involving c, c,..., c n that we can use to find such a linear relation among the vectors. For example, the matrix gives us the system of equations { c + c + c 3 c + c c 3 We can solve the equations as { c c 3 c c 3. Now all that remains is to choose a value for c 3. Recall that we want at least one of the c, c, or c 3 to be nonzero. So, as a general rule, when I get to choose a value for a constant, I just choose. Note that choosing c 3 means that c 3 is nonzero! If we choose c 3, we get c and c. Looking way back to our original equation, we can fill in our constants: c x + c x c n x n x x + x n Okay - let s do the problems now.

3 3 We follow the steps from above to make an augmented matrix and row reduce it: e t e t 3e t r e t r 3 e t e t e t e t r r e t r 3 e t 6e t r e t r 3 6 r r r 3 6 Since the left part of the matrix is not an identity matrix, this is case ii from our steps above. In other words, we determined that the vectors are linearly dependent. Moreover, we can read off two equations from the matrix we got: { c + c 3c 3 c + c + 6c 3 { c 3c 3 [Next, choose c 3 ] c 6c 3 c 3 c 6 c 3 Thus, we can write the following linear relation: 3x 6x + x 3. 4 Once again, we follow the steps from above to make an augmented matrix and row reduce it: sin t sin t r sin t r / sin t sin t sin t sin t r r sin tr / 3 sin t/ r 3 sin t r / r r r This is case i from our steps above because the left part of the matrix is an identity matrix. In other words, the vectors are linearly independent. 3

4 7.3 #6,7,8, In each problem, find all eigenvalues and eigenvectors of the given matrix /4 5 a b Quick reminder: det ad bc. c d To find the eigenvalues of a matrix A, we simply solve the equation deta λi for λ. [Note that A λi is the same as A, except we subtract λ from each entry along the diagonal of A.] We name these solutions λ, λ, etc.. Next, to find the eigenvector corresponding to λ i we call this eigenvector x, we row reduce the augmented matrix A λi I and use the resulting matrix to spit out a system of equations, just like we did to find linear relations. Let s do some problems - I think examples are clearer for this than general explanations. 4

5 6 We solve the equation deta λi : deta λi 5 λ det 3 λ 5 λ λ 3 λ 6λ + 8 λ 4λ λ 4, Thus, we have the eigenvalues λ 4 and λ. Now we row reduce the appropriate augmented matrices. i λ 4: r r 3r Thus, the first row gives us the equation x x, which we can solve to get x x. I will select x, which yields x. Thus, we get the eigenvector x x. x ii λ : r r r 3 Thus, the first row gives us the equation 3x x, which we can solve to get x 3x. I will select x, which yields x 3. Thus, we get the eigenvector x x. x 3 5

6 7 We solve the equation deta λi : deta λi 3 λ det 4 λ 3 λ λ 4 λ λ + 5 λ ± 45 λ ± i Thus, we have the eigenvalues λ + i and λ i. Now we row reduce the appropriate augmented matrices. i λ + i: 3 + i i 4 + i 4 i r i r i [We need to rationalize that denominator] 4 i i +i +i 4 i i 4 i r r 4r i Thus, the first row gives us the equation x + i x, which we can solve to get x +i x. +i I will select x, which yields x +i. Thus, we get the eigenvector x x. x ii λ i: 3 i + i 4 i 4 + i r +i r +i [We need to rationalize that denominator] 4 + i +i i i 4 + i +i 4 + i r r 4r +i Thus, the first row gives us the equation x + +i x, which we can solve to get x i x. i I will select x, which yields x i. Thus, we get the eigenvector x x. x 6

7 8 We solve the equation deta λi : deta λi λ det λ λ λ λ + 4λ + 3 λ + 3λ + λ 3, Thus, we have the eigenvalues λ 3 and λ. Now we row reduce the appropriate augmented matrices. i λ 3: r r r Thus, the first row gives us the equation x + x, which we can solve to get x x. I will select x, which yields x. Thus, we get the eigenvector x x. x ii λ : + + r r r r r Thus, the first row gives us the equation x x, which we can solve to get x x. I will select x, which yields x. Thus, we get the eigenvector x x. x 7

8 We solve the equation deta λi : deta λi 3 λ 3/4 det 5 λ 3 λ λ 3/4 5 λ + λ + 3/4 λ + 3/λ + / λ 3/, / Thus, we have the eigenvalues λ 3/ and λ /. Now we row reduce the appropriate augmented matrices. i λ 3/: 3 + 3/ 3/4 3/ 3/ / 5 5/ r 3 r / 5 5/ r r +5r / Thus, the first row gives us the equation x x, which we can solve to get x x. I will select x, which yields x. Thus, we get the eigenvector x / x. ii λ /: 3 + / 3/4 5/ 3/4 5 + / 5 3/ r 5 r 3/ 5 3/ r r +5r 3/ x Thus, the first row gives us the equation x 3 x, which we can solve to get x 3 x. I will select x, which yields x 3. Thus, we get the eigenvector x 3/ x. x 8

9 7.4 #4 If x y and x y, the the second order equation y + pty + qty corresponds to the system { x x, x qtx ptx. Show that if x and x are a fundamental set of solutions of Equations, and if y and y are a fundamental set of solutions of Equation, then W [y, y ] cw [x, x ], where c is a nonzero constant. [Hint: y t and y t must be linear combinations of x t and x t.] First note that to transform from the single differential equation to the system of differential equations, we use the transformation y x. Now suppose x and x are a fundamental set of solutions of Equations, and suppose y and y are a fundamental set of solutions of Equation. Since x know that any solution y could be written as and any different solution y could be written as y c x + c x, y c 3 x + c 4 x. x and x x x are solutions, we x Next, note that [ x ] W [x, x x ] W, x x x x det x x x x x x Finally, we piece everything together: W [y, y ] y y y y [c x + c x ][c 3 x + c 4 x ] [c x + c x ] [c 3 x + c 4 x ] [c x + c x ][c 3 x + c 4 x ] [c x + c x ][c 3 x + c 4 x ] [c x + c x ][c 3 x + c 4 x ] [c x + c x ][c 3 x + c 4 x ] since x x and x x c c 3 x x + c c 4 x x + c c 3 x x + c c 4 x x c c 3 x x c c 4 x x c c 3 x x c c 4 x x c c 3 c c 3 x x + c c 4 c c 3 x x + c c 3 c c 4 x x + c c 4 c c 4 x x c c 4 c c 3 x x + c c 3 c c 4 x x c c 4 c c 3 x x c c 4 c c 3 x x c c 4 c c 3 x x x x c c 4 c c 3 W [x, x ] cw [x, x ] Moreover, we know that c is nonzero because we know that W [y, y ] is not zero since y and y are a fundamental set of solutions and the Wronskian of any fundamental set of solutions is nonzero. 9

10 7.4 #6,7 Given vectors x t and x t, complete the following: a Compute the Wronskian of x and x. b In what intervals are x and x linearly independent? c What conclusion can be drawn about the coefficients in the system of homogemeous differential equations satisfied by x and x? d Find this system of equations and verify the conclusions of part c. Complete the above for the following sets of vectors: t t 6 x t and x t t t 7 x e t and x t t t e t. 6 a W [x, x ] det x x t t det tt t t t b We know that two solutions are linearly independent if and only if their Wronskian is nonzero. But the Wronskian, t, is only zero when t. Thus, the solutions are linearly independent everywhere else, namely, and,. c According to Theorem 7.4.3, at least one of the coefficients must be discontinuous at t. If it weren t then the Wronskian wouldn t vanish. d Essentially, this is asking us to find a matrix A such that x Ax for any solution x. Note that any solution x could be written as x c x + c x. So let s simplify from our basic equation: x Ax [ c x + c x ] a a [ c a a x + c x ] [ ] t t [ ] a a c + c t t c t a a + c t c t + c t a a c t + c t c + c t a a c + c t c + c t a [c t + c t ] + a [c + c t] c a [c t + c t ] + a [c + c t] c + tc a t + a c + a t + a tc c + c a t + a c + a t + a tc Looking at the coefficients of c and c from the top entry, we see that { a t + a t a t + a t. We can solve the first equation to get a a t. Substituting this into the second equation, we get t a t + a tt a t, which we can solve as a t t.

11 Substituting this back into the expression for a, we get a t t t t t t t 3t t Looking at the coefficients of c and c from the bottom entry, we see that { a t + a a t + a t. Solving the first equation, we get a a t. Substituting this into the second equation, we get a t + a tt a t, which we can solve to get a t. Plugging this back into the expression for a, we get a t t t. Now we have all entries from the matrix A. So, we rewrite the original matrix equation x Ax as t x 3t t t x. t t This confirms our assessment from part c that at least one of these coefficients is discontinuous at t. In fact, all of them are discontinuous at t! 7 a W [x, x ] det x x t e det t t e t t e t te t t te t tt e t b We know that two solutions are linearly independent if and only if their Wronskian is nonzero. But the Wronskian, tt e t, is only zero when t or t. Thus, the solutions are linearly independent everywhere else, namely,,, and,. c According to Theorem 7.4.3, at least one of the coefficients must be discontinuous at t, and at least one of the coefficients must be discontinuous at t. If it weren t then the Wronskian wouldn t vanish. d Essentially, this is asking us to find a matrix A such that x Ax for any solution x. Note that any solution x could be written as x c x + c x. So let s simplify from our basic equation: x Ax [ c x + c x ] [ a a c a a x + c x ] [ ] t e t [ ] a a c + c t e t t e t c a a + c t e t c t + c e t a a c t + c e t c t + c e t a a c t + c e t c t + c e t a [c c + c e t t + c e t ] + a [c t + c e t ] a [c t + c e t ] + a [c t + c e t ] tc + e t c a t c + e t + a tc + a e t + a e t c c a t + a tc + a e t + a e t c

12 Looking at the coefficients of c and c from the top entry, we see that { t a t + a t e t a e t + a e t. We can solve the first equation to get a t a t. Substituting this into the second equation, we get e t t a t e t + a e t, which we can solve as a t t t t. Substituting this back into the expression for a, we get t t t a t. t Looking at the coefficients of c and c from the bottom entry, we see that { a t + a t e t a e t + a e t. Solving the first equation, we get a a t. t Substituting this into the second equation, we get e t a e t + to get a t t t. Plugging this back into the expression for a, we get a a t t t t t t t t t t t t t t a t t 4t t t 3 +t t t t e t, which we can solve t t t Now we have all entries from the matrix A. So, we rewrite the original matrix equation x Ax as x t t t t t t t t t t This confirms our assessment from part c that at least one of these coefficients is discontinuous at t, and at least one of these coefficients is discontinuous at t. In fact, all of them are discontinuous at t! x.

13 HOW TO SOLVE EQUATIONS OF THE FORM x Ax if A has distinct, real eigenvalues: i Find the eigenvalues of A. We will call the eigenvalues r, r,..., r n ii Find the eigenvectors corresponding to r, r,..., r n ; we will call these eigenvectors ξ, ξ,..., ξ n, respectively. iii The general solution is x c ξ e rt + c ξ e rt c n ξ n e rnt. Note that equations of the form x Ax are referred to as homogeneous linear systems [as opposed to nonhomogeneous linear systems, which look like x Ax + gt] 7.5 #a,a,3a For each problem, find the general solution of the given system of equations and describe the behavior of the solution as t. 3 a x x a x x 3 4 3a x x 3 3

14 a I will follow the steps outlined above: i We first find the eigenvalues by solving the equation below: deta λi 3 λ det λ 3 λ λ λ λ λ λ + Thus, we get eigenvalues r and r. ii We find the eigenvectors, starting with ξ, which corresponds to eigenvalue r. To do this, we row reduce the augmented matrix below: A r I 3 4 r r r The first row gives us the equation x x, which we can solve to get x x. If we select x, we get x. Thus, we have the eigenvector ξ x. Next, we find ξ, the eigenvector associated with eigenvalue r. To do this, we row reduce the augmented matrix below: A r I 3 4 / r 4 r r r r / x The first row gives us the equation x x, which we can solve to get x x. If we select x, we get x. Thus, we have the eigenvector ξ x. x iii Thus, the general solution is x c ξ e rt + c ξ e rt c e t + c e t By creating a phase portrait, we can see that x approaches the line formed along the vector ξ, i.e. x x. 4

15 a I will follow the steps outlined above: i We first find the eigenvalues by solving the equation below: Thus, we get eigenvalues r and r. deta λi λ det 3 4 λ λ 4 λ 3 λ + 3λ + λ + λ + ii We find the eigenvectors, starting with ξ, which corresponds to eigenvalue r. To do this, we row reduce the augmented matrix below: A r I /3 r 3 r 3 r r 3r /3 The first row gives us the equation x 3 x, which we can solve to get x 3 x. If we select x, we get x 3. Thus, we have the eigenvector x /3 ξ. x Next, we find ξ, the eigenvector associated with eigenvalue r. To do this, we row reduce the augmented matrix below: A r I r r r r 3r The first row gives us the equation x x, which we can solve to get x x. If we select x, we get x. Thus, we have the eigenvector ξ x. iii Thus, the general solution is x c ξ e rt + c ξ e rt /3 c e t + c By creating a phase portrait, we can see that x approaches the line formed along the vector ξ, i.e. x x. x e t 5

16 3a I will follow the steps outlined above: i We first find the eigenvalues by solving the equation below: Thus, we get eigenvalues r and r. deta λi λ det 3 λ λ λ 3 λ λ λ + ii We find the eigenvectors, starting with ξ, which corresponds to eigenvalue r. To do this, we row reduce the augmented matrix below: A r I r r 3r The first row gives us the equation x x, which we can solve to get x x. If we select x, we get x. Thus, we have the eigenvector ξ x. Next, we find ξ, the eigenvector associated with eigenvalue r. To do this, we row reduce the augmented matrix below: A r I /3 r 3 r 3 r r 3r /3 x The first row gives us the equation x 3 x, which we can solve to get x 3 x. If we select x, we get x 3. Thus, we have the eigenvector x /3 ξ. x iii Thus, the general solution is x c ξ e rt + c ξ e rt c e t /3 + c e t By creating a phase portrait, we can see that x approaches the line formed along the vector ξ, i.e. x 3x. 6

17 7.5 #5,6 Solve each initial value problem, and describe the behavior of the solution as t. 5 5 x x, x 3 6 x x, x I will follow the steps outlined above: i We first find the eigenvalues by solving the equation below: deta λi 5 λ det 3 λ 5 λ λ 3 λ 6λ + 8 λ 4λ Thus, we get eigenvalues r 4 and r. ii We find the eigenvectors, starting with ξ, which corresponds to eigenvalue r 4. To do this, we row reduce the augmented matrix below: A r I r r 3r The first row gives us the equation x x, which we can solve to get x x. If we select x, we get x. Thus, we have the eigenvector ξ x. x Next, we find ξ, the eigenvector associated with eigenvalue r. To do this, we row reduce the augmented matrix below: A r I /3 r 3 r 3 r r 3r /3 The first row gives us the equation x 3 x, which we can solve to get x 3x. If we select x, we get x 3. Thus, we have the eigenvector ξ x. 3 x 7

18 iii Thus, the general solution is x c ξ e rt + c ξ e rt c e 4t + c 3 e t Next, we use the initial conditions to find c and c : x c e 4 + c e c + c 3 c + 3c By comparing the top and bottom entries, we get the system of equations { { c + c c 7/ c + 3c c 3/. Thus, our final solution is x 7 e 4t 3 e t 3 This will be dominated by the e 4t term, so it will approach the line formed by the vector is the line x x., which 8

19 6 I will follow the steps outlined above: i We first find the eigenvalues by solving the equation below: Thus, we get eigenvalues r 3 and r. deta λi λ det 5 4 λ λ4 λ 5 λ λ 3 λ 3λ + ii We find the eigenvectors, starting with ξ, which corresponds to eigenvalue r 3. To do this, we row reduce the augmented matrix below: A r I /5 r 5 r 5 r r +5r 5 The first row gives us the equation x 5x, which we can solve to get x 5x. If we select x, we get x 5. Thus, we have the eigenvector ξ x. 5 Next, we find ξ, the eigenvector associated with eigenvalue r. To do this, we row reduce the augmented matrix below: A r I r r 5 5 r r +5r x The first row gives us the equation x x, which we can solve to get x x. If we select x, we get x. Thus, we have the eigenvector ξ x. x iii Thus, the general solution is x c ξ e rt + c ξ e rt c e 3t + c 5 Next, we use the initial conditions to find c and c : x c 3 e 3 + c 5 e c + c 5c + c e t 9

20 By comparing the top and bottom entries, we get the system of equations { { c + c c / 3 5c + c c /. Thus, our final solution is x e 3t + 5 e t This will be dominated by the e 3t term, so it will approach the line formed by the vector is the line x 5x. 5, which

21 HOW TO MAKE A PHASE PORTRAIT [if A has real, distinct eigenvalues]: If we are given a matrix equation x Ax and we find that A has distinct, real eigenvalues, then we can create a phase diagram by following the steps below: i For each eigenvector, draw the line formed starting at the origin by extending the vector infinitely in both directions. ii For each line from step i, look at the corresponding r i. If r i is positive, draw arrows on the line pointing outward. If r i is negative, draw arrows onn the line pointing inward. iii Draw in sample curves, approaching the lines and following the directions given by the arrows. 7.5 #4,7 For each problem, the eigenvalues and eigenvectors of a matrix A are given. Consider the corresponding system x Ax. a Sketch a phase portrait of the system. b Sketch the trajectory passing through the initial point, 3. c For the trajectory in part b, sketch the graphs of x versus t and of x versus t on the same set of axes. [Problems below] 4 r, ξ ; r, ξ. 7 r, ξ ; r, ξ. 4 a Following the steps outlined above, we create the phase portrait below. Note that the extended eigenvectors appear in bold black, the arrows appear in blue, the axes appear in fine black, and several sample curves appear in red. b Here we sketch the trajectory passing through initial point, 3:

22 The point, 3 is marked as a red circle. c To sketch x versus t or x versus t, it is useful to think about the general solution to our differential equation. The general solution is x x c e rt ξ + c e rt ξ c e t + c e t c e t + c e t c e t + c e t. x Next we use our initial condition [ x : 3] c e x + c e c + c c e + c e c + c This gives us the system of equations { c + c c + c 3 { c /4. c 7/4. 3 Thus, we see that our particular solution is x c e x t + c e t c e t + c e t 4 e t e t e t + 7. e t x Now we can graph x and x versus t. 7 a Following the steps outlined above, we create the phase portrait below.

23 Note that the extended eigenvectors appear in bold black, the arrows appear in blue, the axes appear in fine black, and several sample curves appear in red. b Here we sketch the trajectory passing through initial point, 3: The point, 3 is marked as a red circle. c To sketch x versus t or x versus t, it is useful to think about the general solution to our differential equation. The general solution is x x c x e rt ξ + c e rt ξ c e t + c e t c e t + c e t c e t c e t. Next we use our initial condition [ x ] : 3 c e x + c e c e c e This gives us the system of equations { c + c c c 3 c + c. c c 3 { c 7/4 c /4. Thus, we see that our particular solution is x c e x t + c e t 7 c e t c e t 4 et + 4 et. x Now we can graph x and x versus t. 7 et et 3

24 HOW TO SOLVE EQUATIONS OF THE FORM x Ax if A has two complex eigenvalues: i Find the eigenvalues of A. We will call the eigenvalues r λ + µi and r λ µi. ii Find the eigenvectors corresponding to r, r ; we will call these eigenvectors ξ a + bi and ξ a bi respectively. iii The general solution is x c u + c v, where we define u e λt a cosµt b sinµt and v e λt a sinµt + b cosµt. 7.6 #a,3a For each problem, express the general solution of the given system of equations in terms of real-valued functions. 3 a x x 4 5 3a x x a We follow the steps outlined above: i We solve the equation deta λi : 3 λ det 4 λ 3 λ λ 4 λ λ + 5 λ ± 45 λ ± i So, we have eigenvalues r + i and r i. In particular, we have λ and µ. 4

25 ii Next, we find the ξ by row reducing the augmented matrix A r I : A r I 3 + i 4 + i i 4 i r i r +i 4 i +i i i [rationalizing the denominator] 4 i i 4 i i r r 4r Now we can read off the equation from our matrix: We can solve this equation to get x + i x. x + i x. Then we arbitrarily select x, and we get x +i. Thus, we have the eigenvector ξ x + i/ / / + i. x / / In other words, we have a and b. SHORTCUT: Once we have one eigenvector from a matrix with complex eigenvalues, we get for free that the other eigenvector is the complex conjugate of that eigenvector. In other words, we now know that ξ / / i. iii Finally, we use our formula to conclude that the general solution is x c u + c v c e λt a cosµt b sinµt + c e λt a sinµt + b cosµt [ ] [ c e t / / cost sint + c e t / sint + c e [ t cost sint] + c e [ t sint + cost] c e t cost + c e t sint ] / cost 3a We follow the steps outlined above: 5

26 i We solve the equation deta λi : λ 5 det λ λ λ 5 λ + λ ±i So, we have eigenvalues r i and r i. In particular, we have λ and µ. ii Next, we find the ξ by row reducing the augmented matrix A r I : A r I i 5 i 5 r i r i i 5 i +i +i [rationalizing the denominator] i i i r r r i Now we can read off the equation from our matrix: We can solve this equation to get x + ix. x + ix. Then we arbitrarily select x, and we get x + i. Thus, we have the eigenvector ξ x + i + i. x In other words, we have a and b. SHORTCUT: Once we have one eigenvector from a matrix with complex eigenvalues, we get for free that the other eigenvector is the complex conjugate of that eigenvector. In other words, we now know that ξ i. iii Finally, we use our formula to conclude that the general solution is x c u + c v c e λt a cosµt b sinµt + c e λt a sinµt + b cosµt [ ] [ ] c cos t sin t + c sin t + cos t c [ cos t sin t] + c [ sin t + cos t] c cos t + c sin t 6

27 7.6 #9, For each problem, find the solution of the given initial value problem. 5 9 x x, x 3 3 x x, x 9 We follow the steps outlined above: i We solve the equation deta λi : λ 5 det 3 λ λ 3 λ 5 λ + λ + λ ± 4 λ ± i So, we have eigenvalues r i and r + i. In particular, we have λ and µ. ii Next, we find the ξ by row reducing the augmented matrix A r I : A r I + i i i 5 i 5 r i r i i 5 i +i +i [rationalizing the denominator] i i i r r r i Now we can read off the equation from our matrix: We can solve this equation to get x x + ix. x + ix. Then we arbitrarily select x, and we get x + i. Thus, we have the eigenvector ξ x + i + i. In other words, we have a and b. 7

28 SHORTCUT: Once we have one eigenvector from a matrix with complex eigenvalues, we get for free that the other eigenvector is the complex conjugate of that eigenvector. In other words, we now know that ξ i. iii Finally, we use our formula to conclude that the general solution is x c u + c v c e λt a cosµt b sinµt + c e λt a sinµt + b cosµt [ ] [ ] c e t cos t sin t + c e t sin t + cos t c e t [ cos t sin t] + c e t [ sin t + cos t] c e t cos t + c e t sin t Next, we use the initial condition: c e x [ cos sin] + c e [ sin + cos] c e cos + c e sin c + c c We can solve this to find c, c. Thus, we plug this back into our general solution to get e x t [ cos t sin t] e t [ sin t + cos t] e e t cos t e t t [cos t 3 sin t] sin t e t [cos t sin t] We can note that both terms of x tend toward because of the e t terms, so x will tend toward the origin as t. We follow the steps outlined above: i We solve the equation deta λi : 3 λ det λ 3 λ λ λ + 4λ + 5 λ 4 ± 4 45 ± i So, we have eigenvalues r + i and r i. In particular, we have λ and µ. 8

29 ii Next, we find the ξ by row reducing the augmented matrix A r I : A r I 3 + i + i i i r i r i i i +i +i [rationalizing the denominator] i + i i r r +r + i Now we can read off the equation from our matrix: We can solve this equation to get x + + ix. x ix. Then we arbitrarily select x, and we get x i. Thus, we have the eigenvector ξ x i + i. x In other words, we have a and b. SHORTCUT: Once we have one eigenvector from a matrix with complex eigenvalues, we get for free that the other eigenvector is the complex conjugate of that eigenvector. In other words, we now know that ξ + i. iii Finally, we use our formula to conclude that the general solution is x c u + c v c e λt a cosµt b sinµt + c e λt a sinµt + b cosµt [ ] [ c e t cos t sin t + c e t sin t + c e t [cos t + sin t] + c e t [sin t cos t] c e t cos t + c e t sin t cos t ] Next, we use the initial condition: c e x [cos + sin] + c e [sin cos] c e cos + c e sin c c c We can solve this to find c, c 3. Thus, we plug this back into our general solution to get e x t [cos t + sin t] + 3e t [sin t cos t] e e t cos t + 3e t t [cos t 5 sin t] sin t e t [ cos t 3 sin t] We can note that both terms of x tend toward because of the e t terms, so x will tend toward the origin as t. 9

30 7.6 #3ab x α x α a Determine the eigenvalues in terms of α. b Find the critical value or values of α where the qualitative nature of the phase portrait for the system changes. c Draw a phase portrait for a value of α slightly below, and for another value slightly above, each critical value. α a Let A. To find the eigenvalues, we solve the following equation: α deta λi α λ det α λ α λα λ λ αλ + α + λ α ± α 4α + λ α ± 4 λ α ± i b Let s think about the general solution. Since our eigenvalues are complex, we can think about phase portrait for complex eigenvalues. We know that if we have eigenvalue λ + µi, the phase portrait will depend on λ. If λ >, then the phase portrait spirals outward. If λ, the phase portrait will loop around in ellipses. If λ <, the phase portrait will spiral inward. Thus, we can see that α is the critical value of α at which the qualitative nature of the phase portrait changes. c First, consider α. We would see that the phase portrait spirals out. Moreover, we can plug x into our original differential equation to see x, so the spiral moves clockwise. Thus, the graph below shows a phase portrait for α : 3

31 Next, consider α. We would see that the phase portrait spirals out. Moreover, we can plug x into our original differential equation to see x, so the spiral moves clockwise. Thus, the graph below shows a phase portrait for α : 7.6 #6 The electric circuit shown in the figure above is described by the system of differential equations d I L I dt V C, 3 RC V where I is the current through the inductor and V is the voltage drop across the capacitor. These differential equations were derived in Problem 9 of Section 7.. a Show that the eigenvalues of the coefficient matrix are real and different if L > 4R C; show that they are complex conjugates if L < 4R C. b Suppose that R Ω, C F, and L H. Find the general solution of the system 3 in this case. c Find It and V t if I A and V V. d For the circuit of part b determine the limiting values of It and V t as t. Do these limiting values depend on the initial condition? a Let x I, and let A L V C. Then we see that our differential equation is of the form RC 3

32 x Ax. Let s find the eigenvalues of A: λ deta λi det L C RC λ λ RC λ L C λ + λ + RC LC λ RC λ RC ± RC 4 LC ± R C 4 LC λ RC ± L 4R C LR C We see that the radical expression here is real and nonnegative if and only if L 4R C is positive. So, the eigenvalues will be real and different if and only if L 4R C >, which is equivalent to if and only if L > 4R C. Similarly, we see that the radical term will be negative yielding complex conjugate eigenvalues if and only if L 4R C is negative, which is if and only if L < 4R C. b Plugging in R, C /, and L, we use our work from part a to get that the eigenvalues are λ RC ± L 4R C LR C / ± 4 / / ± 4 ± i So, we have eigenvalues r + i and r i. In particular, we see that λ and µ. Next, we find the eigenvector corresponding to r + i. To do this, we row reduce the augmented matrix A λi : + i C RC L + i r i r i L C RC i i i i i i +i +i [rationalizing the denominator] i +i i +i r r +r Next, we read off the equation this gives us: + i x + x. We can solve this to get x i x. Then, we can choose x and solve to get x i. Thus, we have the eigenvector ξ x i/ / / + i. x 3

33 In other words, we have a Thus, the general solution is x c u + c v / and b /. c e λt a cosµt b sinµt + c e λt a sinµt + b cosµt [ ] [ c e t / / cos t sin t + c e t / sin t + c e [ t cos t + sin t] + c e [ t sin t cos t] c e t cos t + c e t sin t ] / cos t c We use our initial condition to see that I x V c e [ cos + sin] + c e [ sin cos] c e cos + c e c c sin c We can solve this system of equations to get c and c 5. Thus, we get the solution c e [ x t cos t + sin t] + c e [ t sin t cos t] c e t cos t + c e t sin t [ e t cos t + sin t] 5e [ t sin t cos t] e t cos t 5e t sin t e t [ cos t + 3 sin t] e t [cos t 5 sin t] d Looking back at our general solution for part b, we see that both I and V tend toward zero because of the e t terms, regardless of c and c. I other words, It and V t tend toward as t, and this limiting value does not depend on the initial conditions. 7.6 #8ab A mass m on a spring with constant k satisfies the differential equation mu + ku, where ut is the displacement at time t of the mass from its equilibrium position. a Let x u, x u, and show that the resulting system is x x. k/m b Find the eigenvalues of the matrix for the system in part a. a We start by transforming our differential equation into a system of first-order differential equations. Using x u and x u, we can also note that u x. Thus, we can replace the original differential equation with mx + kx. We also have the relationship x x. Let s rewrite this as a system of equations: { x x x k/mx { x x + x x k/mx + x. 33

34 x Note that if we let x, we can rewrite this system as the following matrix equation: x as desired. x x x x, k/m b We find the eigenvalues by solving the following equation: λ deta λi det k/m λ λ λ k/m λ + k/m λ ±i k/m 34