Math 25B Midterm III Information Fall 28 SOLUTIONS TO PRACTICE PROBLEMS Problem Determine whether the following matrix is diagonalizable or not If it is, find an invertible matrix S and a diagonal matrix D such that S AS = D A := 2 2 2 SOLUTION: Since this matrix is upper triangular, the eigenvalues simply appear on the main diagonal λ = 2,,, So to decide if A is diagonalizable, we only need to determine whether the repeated eigenvalue λ = carries its weight, so we consider nullspace(a I) = nullspace 2 2 2 If we label the variables as (x, y, z, w), then we see that w =, 2z + w =, and x + 2w = From this, we have w = z = x = Only y is a free parameter, so we only obtain one basis vector for the eigenspace, which is not enough for λ = Hence, A is not diagonalizable Problem 2 Without a calculator, find A if [ A = 2 SOLUTION: We will seek to diagonalize A: The characteristic polynomial is det(a λi) = ( λ)( λ) 2 = λ 2 λ 2 = (λ 2)(λ + ), so the eigenvalues are λ = 2 and λ = Next we find each eigenspace: [ We have E λ= = nullspace(a + I) = nullspace, which clearly produces the 2 2 eigenvector (, ) for its basis
[ 2 We have E λ=2 = nullspace(a 2I) = nullspace 2 the eigenvector (, 2) for its basis, which clearly produces We may use the two eigenvectors we have found to form an invertible matrix S, by placing the eigenvectors into the columns of S: [ S = 2 Using the formula for the inverse of a 2 2 matrix (or just computing it directly, we have [ 2 S = Moreover, we know that S AS = D, where [ D = 2 Thus, A = (SDS ) = SD S = S [ 2 S You can use the fact that 2 = 24 and complete the matrix multiplication to get a final answer if you wish, but if you ve taken it this far, that is definitely good enough for this problem! Problem (a): Write the DE y + 4xy 6x 2 y = x 2 sin x as an operator equation and give the associated homogeneous DE SOLUTION: The operator form is and the associated homogeneous DE is (D 2 + 4xD 6x 2 )y = x 2 sin x, (D 2 + 4xD 6x 2 )y =
(b): Write the DE (D 2 + )(D + )(y) = e 4x + ln(2x + ) in terms of the expressions y, y, y, etc Is this differential equation linear? Why or why not SOLUTION: We can rewrite the differential equation as y + y + y + y = e 4x + ln(2x + ), and this differential equation IS LINEAR, since it has the form (a D + a D 2 + a 2 D + a )(y) = F (x) Problem 4 Find the complementary function y c (x) and derive an appropriate trial solution for y p (x) for the given DE Do not solve for the constants that arise in your trial solution (a): (D + )(D 2 + )y = 4xe x SOLUTION: The roots of the characteristic equation (r +)(r 2 +) = are r = and r = ±i Thus, we have y c (x) = c e x + c 2 cos x + c sin x Now an annihilator for xe x is A(D) = (D ) 2, so applying this to both sides of the DE, we obtain (D ) 2 (D + )(D 2 + )y =, whose general solution is c e x + c 2 cos x + c sin x + Ae x + Bxe x The last two terms constitute the trial solution: y p (x) = Ae x + Bxe x (b): (D 2 2D + 2) (D 2) 2 (D + 4)y = e x cos x e 2x SOLUTION: The characteristic equation is (r 2 2r + 2) (r 2) 2 (r + 4) =,
with roots Hence, r = ± i, ± i, ± i, 2, 2, 4 y c (x) = c e 4x + c 2 e 2x + c xe 2x + c 4 e x cos x + c 5 e x sin x + c 6 xe x cos x + c 7 xe x sin x + c 8 x 2 e x cos x + c 9 x 2 e x sin x Now the appropriate annihilator of e x cos x is D 2 2D + 2 and the appropriate annihilator of e 2x is D 2 Hence, we apply A(D) = (D 2 2D + 2)(D 2) to both sides of the DE The resulting homogeneous DE is 2th-order, with 9 of the terms of its general solution already comprising y c (x) The three additional terms constitute our trial solution for y p (x): y p (x) = Ax e x cos x + Bx e x sin x + Cx 2 e 2x Problem 5 Find a basis for the solution set of each differential equation below In the case of part (b), use the Wronskian to verify that the basis solutions are indeed linearly independent (a): (D 2 + 4) 2 (D + )y = SOLUTION: The roots of the characteristic equation are r =, ±2i, ±2i Therefore, a basis for the solution set of this DE is {e x, cos 2x, sin 2x, x cos 2x, x sin 2x} (b): y + y + y + y = SOLUTION: The characteristic equation is r + r 2 + r + = (r + ),
whose roots are r =,, Hence, a basis for the solution set of this DE is {e x, xe x, x 2 e x } The Wronskian is not identically zero (calculation omitted), so the solutions are indeed linearly independent Problem 6 Problem 5 SOLUTION: (a): (b): Write the general solution to each differential equation in y(x) = c e x + c 2 cos 2x + c sin 2x + c 4 x cos 2x + c 5 x sin 2x Problem 7 Solve the IVP y(x) = c e x + c 2 xe x + c x 2 e x (D + 2D 2 4D 8)y = subject to the initial conditions y() =, y () = 6, y () = 8 SOLUTION: The characteristic equation is r + 2r 2 4r 8 = (r 2)(r 2 + 4r + 4) = (r 2)(r + 2) 2, with roots r = 2, 2, 2 Hence, the general solution to the differential equation is y(x) = c e 2x + c 2 e 2x + c xe 2x In order to apply the initial conditions, we need to find y (x) and y (x), which we get from differentiation: and y (x) = 2c e 2x 2c 2 e 2x + c [e 2x 2xe 2x y (x) = 4c e 2x + 4c 2 e 2x + c [ 2e 2x 2e 2x + 4xe 2x Now we evaluate each of these at x = : = y() = c + c 2
6 = y () = 2c 2c 2 + c 8 = y () = 4c + 4c 2 4c The corresponding augmented matrix for the linear system consisting of these three equations is 2 2 6, 4 4 4 8 which quickly reduces (with elementary row operations) to 4 6 4 8 From the last row, we see that c = 2 Thus, we have c 2 = 2 and c = 2 Hence, the solution is y(x) = 2e 2x 2e 2x 2xe 2x Problem 8 Let v be an eigenvector of A corresponding to the eigenvalue λ Prove that if B = S AS, then S v is an eigenvector of B corresponding to the eigenvalue λ SOLUTION: We have B(S v) = (BS )v = (S A)v = S (Av) = S (λv) = λ(s v), which shows that S v is an eigenvector of B corresponding to λ (Note that S v since v (since it is an eigenvector of A) and S is invertible) Problem 9 Find the general solution to the differential equation under the assumption that x > y + 4y + 4y = e 2x x 2 SOLUTION: Setting y + 4y + 4y = = r 2 + 4r + 4 = (r + 2) 2 = = r { 2, 2} = y c (x) = c e 2x +c 2 xe 2x Let y (x) = e 2x and y 2 (x) = xe 2x We have W [y, y 2 (x) = e 4x Then a particular solution to the given differential equation is
y p (x) = y u + y 2 u 2, where the formulas for u and u 2 are given in Equation (677) in the text We have x (te 2t )e 2t x u = dt = dt = ln x t 2 e 4t t and Consequently, Hence, x (e 2t )e 2t u 2 = dt = t 2 e 4t x y p (x) = e 2x ln x e 2x y(x) = e 2x (c + c 2 x ln x ) Problem Find the general solution to x (t) := 2 x(t), and use the Wronskian to check your answer SOLUTION: We call the matrix in this system A We must find the eigenvalues: λ det(a λi) = det 2 λ = ( λ)[(2 λ)( λ) + = λ = ( λ)(λ 2 2λ + ) = ( λ)(λ ) 2 Thus, the eigenvalues NS( are λ =,, Now we search for eigenvectors, by computing ) NS(A I) = Let z = t, y = t, x = s So we get two eigenvectors:, Therefore, we obtain two linearly independent solutions to the system as follows: x (t) = e t and x 2 (t) =
To get a third solution, we seek a solution of the form x (t) = e t (w + tw 2 ), where (A I) 2 w = and w 2 = (A I)w Since (A I) 2 is the zero matrix, any vector w will do, except that we d best not let w be an eigenvector of A or else we ll end up with w 2 = Let us choose (there are many choices here) w = Then w 2 = (A I)w = Hence, x (t) = e t + t Provided x, x 2, x are linearly independent (see below), we can form the general solution to this linear system: x(t) = c e t + c 2 e t + c e t + t We confirm the answer by placing the three solutions into the Wronskian to check their linear independence: Factoring the e t from each column of the Wronskian, it suffices to consider the determinant of which is required Problem (a): Verify that + t t, Thus, the solutions obtained above are linearly independent, as x p (t) := [ t 4 5t + 6
is a particular solution to the non-homogeneous system [ [ 2t x (t) = x(t) + 5 [ SOLUTION: We have x p (t) = [ 5 5 [ t 4 5t + 6 Plugging x p into the right-hand side yields + [ 2t [ When multiplied out, this does indeed give us the vector (b): Find the general solution for the system 5 SOLUTION: We must find the solution to the associated homogeneous [ equation We begin by finding the eigenvalues of the coefficient matrix A = Omitting 5 [ the computations, we find that λ = 6 is an eigenvalue with associated [ eigenvalue, and λ = 2 is an eigenvalue with associated eigenvalue Thus, the 5 general solution to the homogeneous system is [ [ x c (t) = c e 6t + c 5 2 e 2t Adding the particular solution to this yields the final answer: [ [ [ x(t) = c e 6t + c 5 2 e 2t t 4 + 5t + 6 Problem 2 Juliet is in love with Romeo, who happens to be a fickle lover The more Juliet loves him, the more he begins to dislike her But when she dislikes him, his feelings for her warm up On the other hand, her love for him grows when he loves her and withers when he dislikes her A model for their ill-fated romance is dj dt = Ar and dr dt = Bj,
where A and B are constants, r(t) represents Romeo s love for Juliet at time t, and j(t) represents Juliet s love for Romeo at time t (a): Is A positive or negative? Why? What about B? SOLUTION: A is positive and B is negative If r > (ie Romeo loves Juliet), then her feelings become warmer towards him, which means dj should be positive dt Thus, A must be positive On the other hand, if j >, then Romeo begins to dislike her, so dr < This means that B < dt (b): Derive a differential equation for r(t) which does not involve j(t) and solve it SOLUTION: Note that r (t) = Bj (t) = B(Ar(t)) = (AB)r(t) Thus, r (AB)r =, which is a second-order constant coefficient, homogeneous DE The characteristic polynomial is r 2 AB = Thus, r = ± AB i, since AB < by part (a) Thus, the general solution is r(t) = c cos( AB t) + c 2 sin( AB t) (c): Find r(t) and j(t), in terms of A and B, given that r() = and j() = SOLUTION: From our expression for r(t) and the equation dj = Ar, we can integrate to dt find j(t) = [c sin( AB t) c 2 cos( AB t) AB Now = r() = c and = j() = c 2 Thus, c 2 = Substituting these values, AB we get and r(t) = cos AB t j(t) = sin AB t AB (d): As you may have discovered, the outcome of this relationship is a never-ending cycle of love and hate! Find what fraction of the time they both love one another
SOLUTION: They will both love each other provided both the sine and cosine functions present are positive, which occurs one-fourth (or 25%) of the time Problem Let A be a square matrix Find solutions of x = Ax of the form T (t)v, where T (t) is a scalar-valued function of t and v is an eigenvector of A SOLUTION: In order for T (t)v to be a solution, we must have T (t)v = AT (t)v = T (t)av = T (t)λv Hence, we must have T (t) = λt (t), where λ is an eigenvalue of A The characteristic polynomial is r 2 λ = If λ >, then the roots are r = ± λ, and we have T (t) = c e λt + c 2 e λt If λ =, then the roots are r =, and we have T (t) = c + c 2 t If λ <, then the roots are ±i λ, and we have T (t) = c cos( λ t)+c 2 sin( λ t) So the possible solutions have T (t) of the above forms, multiplied by their corresponding eigenvectors Problem 4 Ants are crawling around in Scott s apartment between a bottle of Von s Tea with a Twist in his kitchen and some leftover Domino s pizza in his bedroom Assume that each hour, one-third of the ants that were in the bedroom move to the kitchen, and one-fourth of the ants that were in the kitchen move to the bedroom Let k(t) denote the number of ants in the kitchen at time t, and let b(t) denote the number of ants in the bedroom at time t [ k() (a): If denotes the initial distribution of ants in the kitchen and b() bedroom (at time t = hours), determine how many ants are in the kitchen and the bedroom after l hours SOLUTION: The crux of the problem is to determine the distribution of ants in each [ room from one hour[ to the next In other words, we should determine how k(t + ) k(t) is related to Note that from hour t to hour t +, /4 of the b(t + ) b(t) ants that were in the kitchen will stay in the kitchen, while / of the ants that were in the bedroom will move to the kitchen That is, we can say k(t + ) = 4 k(t) + b(t) Similarly, we see that b(t + ) = 4 k(t) + 2 b(t)
Putting this together, we can write the matrix system [ [ [ k(t + ) /4 / k(t) = b(t + ) /4 2/ b(t) Let us denote A = [ /4 / /4 2/ [ k(t) Each time we multiply by the matrix A, the effect on the vector is to determine the number of ants in each room exactly one hour later Thus, to find the b(t) number of ants in the kitchen and the bedroom after l hours, we must determine [ [ k(l) a() = A l b(l) b() This necessitates an expression for A l, which we will obtain by diagonalizing the matrix A Note that [ /4 λ / det(a λi) = det = ( /4 2/ λ 4 λ)(2 λ) 2 = λ2 7 2 λ+ 5 2 = (λ )(λ 5 2 ) Hence, the eigenvalues of A are λ = and λ = 5 2 Next, we must find corresponding eigenvectors for these two eigenvalues For λ =, we look at [ /4 / nullspace(a I) = nullspace /4 / and it is easy to see that a vector that is in this nullspace is v = For λ = 5, we look at 2 nullspace(a 5 2 I) = nullspace [ / / /4 /4 and it is easy to see that a vector that is in this nullspace is v 2 =,, [ 4 [
[ 4 Now we can diagonalize A by using the matrices S = and D = Using the formula for the inverse of a 2 2 matrix S, we have S = [ 7 4 Now since S AS = D, we have A = SDS, = A l = (SDS ) l = SD l S = 7 [ 4 [ 5/2 [ (5/2) l [ 4 Multiplying out the matrices on the right, we obtain A l = [ 4 + (5/2) l 4 4(5/2) l 7 (5/2) l + 4(5/2) l The number of ants in each room at time l hours is therefore, [ [ k(l) a() = A l = [ 4 + (5/2) l 4 4(5/2) l b(l) b() 7 (5/2) l + 4(5/2) l [ k() b() (b): As time approaches infinity, what fraction of the ants will be in each room? Does the initial distribution of ants affect the answer here? Why or why not? SOLUTION: As l, the expressions (5/2) l in the expression above tend to zero, leaving [ [ [ [ [ k( ) k() 4 4 k() 4 = A (k() + b()) = 7 b( ) b() b() ((k() + b()) = 7 which implies that, in the long run, 4/7 of the initial ants will be in the kitchen, and /7 of them will be in the bedroom, regardless of the initial distribution of the ants between the rooms In other words, as we can see, the values of k() and b() are not important here; we only need to focus on the total number of ants in the problem, which stays constant at k() + b() [ Problem 5 Solve the initial value problem x = Ax, where x() = [ 2 and A = 4 7
2 λ SOLUTION: We have det(a λi) = if and only if = if and 4 λ only if λ 2 +6λ+9 = if and only if (λ+) 2 = This means that the only eigenvalue of A is λ = (with multiplicity 2) [ Eigenvectors for λ = : We have A + I =, which gives rise to corresponding eigenvectors of the form v = r(, ), where r R Therefore, [ we obtain one solution to the linear system of differential equations as x (t) = e t We next seek a second linearly independent solution to this linear system of the form x(t) = e t (v + tv ), where v and v are determined from (A + I) 2 v =, (A + I)v, v = (A + I)v [ In this case, we have A + I = and (A + I) 2 = 2 Hence, we may choose [ v to be any vector such that (A + I)v Let us choose, say, v = Then v = (A+I)v = equations is [ Therefore, our second solution to this system of differential x 2 (t) = e t ([ [ + t ) Hence, the general solution to this system is [ ([ [ x(t) = c e t + c 2 e t + t [ We have been given that x() =, so that [ [ c + c = 2 c ) Therefore, c + c 2 = and c = Therefore, c 2 = Hence, the solution to this initial-value problem is [ ([ [ ) [ x(t) = e t + e t + t = e t t t