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5 Homework solutions 5.2.4 Since the trace is 4 and the determinant is 5, then the xed point is an unstable spiral (alternatively this follows because the eigenvalues are 2 + i and 2 i, since they are complex it spirals and since the real part is > 0 it is unstable). The phase portrait are spirals which head out of the origin spinning clockwise (to see that it is clockwise we can, for example, plot a few points on a single trajectory to get the shape, alternatively we can see that along the positive x-axis that y < 0 which tells us how we cross the positive x-axis). 5.2.5 Since the trace is 2 and the determinant is 1 then the xed point is a degenerate node (not a star because while we have repeated eigenvalues of 1, 1 we do not have a full set of eigenvectors). The only eigenvector is 2 and is indicated below (since our 1 eigenvalue is > 0 then it moves out). To get the behavior of the remaining part we can plot a few points along a solution (i.e., such as et 2t+1 ). The phase portrait is t shown below. 1 5.2.6 Since the trace is 5 and the determinant is 4 then the xed point is a stable node (alternatively since both eigenvalues are negative then solutions tend to the origin as t ). The fast eigenvector is 2 while the slow eigenvector is 1 . Using this 1 1 we have the following phase portrait. 5.2.7 Since the trace is 0 and the determinant is 9 then the xed point corresponds to a neutrally stable center (alternatively since the eigenvalues are 3i and 3i, since the real part is 0 there will be no exponential decay or growth and the solutions will be periodic). The phase portrait will consist of ellipses centered around the origin spinning clockwise (again to see that it spins clockwise we can plot a trajectory or again look at the positive x-axis and see that y < 0 which tells us how we cross that axis which will give us the direction). 2 5.2.8 Since the trace is 0 and the determinant is 1 then the xed point is a saddle (alternatively this follows since the eigenvalues are 1 and 1, i.e., real and di er in sign so that in one direction we move towards the origin and in another direction we move away from the origin). The eigenvector associated with 1 is 1 and the eigenvector 1 associated with 1 is 2 . This gives us the following phase portrait. 1 5.2.11 The matrix A= b 0 has characteristic polynomial (x )2 so that is the only eigenvalue of A and occurs as a repeated root. Looking for the eigenvectors we see that if x = x1 is an x2 eigenvector then (A I)x = 0, i.e., 0b 00 x1 x2 = 0 0 , which reduces to the condition that b x2 = 0 and since b = 0 we must conclude that x2 = 0. So we only have a one dimensional eigenspace and it consists of all vectors of the form x = x1 . 0 To solve the system we have the obvious solution x = ce t 1 . We now need to nd 0 a second solution. From a past homework assignment (i.e., see the solution to 5.2.5 from homework 4) we know that one candidate will be x = de t (t 1 + y) where y 0 is a vector satisfying (A I)y = 1 . One such y that will work (there are in fact 0 many) is 0 1/b . So we have that the general solution is x = ce t 1 t + de t . 0 1/b When it comes to sketching the phase portrait there are several possibilities depending on whether = , 0, + and b = , +. The key to all of them is noting that we know 3 explicitly what happens on the x-axis (i.e., the direction of the eigenvector) and then we just need to gure out how curves come into/out of the degenerate node, which can be done by plotting a few points/trajectories. The di erent possibilities are shown in the phase portraits below. * Consider the following system. 03 1 x = 4 1 1 x 2 7 5 We solve this using the same approach as 2 2 matrices, i.e., nd the eigenvalues and eigenvectors. The only di erence is that there is more bookkeeping involved. First to nd the eigenvalues we nd the characteristic polynomial, in this case x 3 1 1 det(xI A) = det 4 x 1 2 7 x + 5 = x(x 1)(x + 5) + 6 28 2(x 1) + 7x 12(x + 5) = x3 + 4x2 12x + 80. To solve a cubic we hope for the best in that there will be a nice solution (i.e., rational). For cubics with integer coe cients and a 1 in front of x3 then the possible nice roots are those which divide the constant term. In our case the candidates are 1, 2, 4, 5, 8, 10, 16, 20, 40, 80. We start going through these and nd that 4 is a root. Now using that we do some long division (i.e., divide x 4 into 4 x3 + 4x2 12x + 80) and discover x3 + 4x2 12x + 80 = (x 4)(x2 + 8x + 20). Finally to get the other roots we can use the quadratic equation so that the three roots are = 4, 4 2i, 4 + 2i. The next step is to nd the eigenvectors. For = 4 we are looking for a vector x so that (A 4I)x = 0, i.e., we want 0 4 3 1 x1 4 3 1 x2 = 0 . 0 2 7 9 x3 This is equivalent to the following three equations: 4x1 + 3x2 + x3 = 0 4x1 3x2 x3 = 0 2x1 + 7x2 9x3 = 0 If we now add twice the third equation to the rst equation (eliminating the x1 term) we have 17x2 17x3 = 0 showing that x2 = x3 . Putting this last relationship into any of the above equations we can conclude that x1 = x2 . So we have that the eigenvector in this case is 1 x = 1 . 1 For = 4 + 2i we are looking for a vector x so that (A ( 4 + 2i)I)x = 0, i.e., we want x1 4 2i 3 1 0 x2 = 0 . 4 5 2i 1 2 7 1 2i 0 x3 This is equivalent to the three following equations: (4 2i)x1 + 3x2 + x3 = 0 4x1 + (5 2i)x2 x3 = 0 2x1 + 7x2 + ( 1 2i)x3 = 0 If we now add the rst and second equation together (eliminating the x3 term) we have (8 2i)x1 + (8 2i)x2 = 0 showing that x2 = x1 . Putting this last relationship into any of the above equations we can conclude that x3 = ( 1 + 2i)x1 . So we have that the eigenvector in this case is 1 x = 1 . 1 + 2i 5 For = 4 2i we can repeat the above process or just use the fact that it will be the complex conjugate of the eigenvector we just found so that the eigenvector in this case is 1 x = 1 . 1 2i We now have that the general solution will be 1 1 1 x = c1 e4t 1 + c2 e( 4+2i)t 1 + c3 e( 4 2i)t 1 . 1 1 + 2i 1 2i However this solution is somewhat unsatisfying since it involves imaginary numbers. So we now work to rewrite it. One approach (a bit lengthy) is to do what we did in the previous homework (of course other approaches also work). Doing that we get the following: 1 1 c2 e( 4+2i)t 1 + c3 e( 4 2i)t 1 1 2i 1 + 2i c2 (cos(2t) + i sin(2t)) + c3 (cos(2t) i sin(2t)) c2 (cos(2t) + i sin(2t)) c3 (cos(2t) i sin(2t)) = e 4t c2 ( 1 + 2i)(cos(2t) + i sin(2t)) + c3 ( 1 2i)(cos(2t) i sin(2t)) (c2 + c3 ) cos(2t) + i(c2 c3 ) sin(2t) (c2 + c3 ) cos(2t) i(c2 c3 ) sin(2t) = e 4t (c2 + c3 )( cos(2t) 2 sin(2t)) + i(c2 c3 )(2 cos(2t) sin(2t)) cos(2t) sin(2t) +i(c2 c3 ) e 4t cos(2t) sin(2t) = (c2 + c3 ) e 4t cos(2t) 2 sin(2t) 2 cos(2t) sin(2t) =C2 =C3 So that we have that the general solution is 1 cos(2t) sin(2t) + C3 e 4t . cos(2t) sin(2t) x = C1 e4t 1 + C2 e 4t 1 cos(2t) 2 sin(2t) 2 cos(2t) sin(2t) To nd the particular solution we 1 0 = C1 0 need to nd C1 , C2 , C3 so that 1 1 0 + C2 1 + C3 0 . 1 1 1 2 6 In other words we need to have C1 + C2 =1 C1 C2 =0 C1 C2 + 2C3 = 0 Subtracting the second from the third shows that we need to have C3 = 0. Adding the rst two shows that 2C1 = 1 so that C1 = 1/2 and the second shows that C2 = C1 = 1/2. So we have that the particular solution is 1 cos(2t) 1 4t 1 4t 1 +e cos(2t) x= e 2 2 1 cos(2t) 2 sin(2t) 1 4t 1 4t cos(2t) 2e + 2e 1 4t 1 4t cos(2t) = . 2e 2e 1 4t 2e 1 2 e 4t cos(2t) e 4t sin(2t) Now consider the following system. 1 0 2 x = 0 1 2 x 2 2 0 First to nd the eigenvalues we nd the characteristic polynomial, in this case x 1 0 2 0 x+1 2 det(xI A) = det 2 2 x (x 1)(x + 1)x + 0 + 0 4(x + 1) 4(x 1) = x3 9x = x(x 3)(x + 3). So we have that the eigenvalues are = 3, 0, 3. The next step is to nd the eigenvectors. For = 3 we are looking for a vector x so that (A + 3I)x = 0, i.e., we want 4 0 2 x1 0 0 x2 = 0 . 2 2 2 2 3 x3 0 7 This is equivalent to the following three equations: 4x1 2x1 + 2x3 = 0 2x2 2x3 = 0 2x2 + 3x3 = 0 The second equation tells us that x2 = x3 while the rst equation tells us that x3 = 2x1 . So we have that the eigenvector in this case is 1 x = 2 . 2 For = 0 we are looking for a vector x so that Ax = 0, i.e., we want 1 0 2 x1 0 0 1 2 x2 = 0 . 2 2 0 0 x3 This is equivalent to the following three equations: x1 x2 2x1 2x2 + 2x3 = 0 2x3 = 0 =0 The rst equation tells us that x1 = 2x3 while the second equation tells us that x2 = 2x3 . So we have that the eigenvector in this case is 2 x = 2 . 1 For = 3 we are looking for a vector x so that (A 3I)x = 0, i.e., we want 2 0 2 x1 0 0 4 2 x2 = 0 . 2 2 3 x3 0 This is equivalent to the following three equations: 2x1 2x1 4x2 2x2 + 2x3 = 0 2x3 = 0 3x3 = 0 The rst equation tells us that x1 = x3 while the second equation tells us that x3 = 2x2 . So we have that the eigenvector in this case is 2 x = 1 . 2 8 We now have that the general solution will be 1 2 2 x = c1 e 3t 2 + c2 2 + c3 e3t 1 . 2 1 2 To nd the particular solution we need to nd c1 , c2 , c3 so that 1 1 2 2 0 = c1 2 + c2 2 + c3 1 0 2 1 2 In other words we need to have c1 2c2 2c3 = 1 2c1 2c2 + c3 = 0 2c1 + c2 2c3 = 0 If we now double the rst equation becomes c1 and add it to the second and third equations this 2c2 2c3 = 1 6c2 3c3 = 2 3c2 6c3 = 2 If we now multiply the second equation by 2 and add it to the third equation this becomes c1 2c2 2c3 = 1 6c2 3c3 = 2 9c2 = 2 It immediately follows that c2 = 2/9 substituting this into the second equation we see that c3 = 2/9 then substituting these values into the rst equation we see that c1 = 1/9. So we have that the particular solution is 1 2 2 1 3t 2 2 2 2 e3t 1 x= e 9 9 9 1 2 2 3t + 4 + 4e3t e 1 2e 3t + 4 2e3t . = 9 2e 3t 2 + 4e3t 9

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