Math 54: Mock Final. December 11, y y 2y = cos(x) sin(2x). The auxiliary equation for the corresponding homogeneous problem is

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1 Name: Solutions Math 54: Mock Final December, 25 Find the general solution of y y 2y = cos(x) sin(2x) The auxiliary equation for the corresponding homogeneous problem is r 2 r 2 = (r 2)(r + ) = r = 2, r = The general solution to the homogeneous problem is therefore y h = C e 2x + C 2 e x For finding a particular solution, we use the method of undetermined coefficients Assume a particular solution is of the form y p = (A cos(x)+a 2 sin(x))+(b cos(2x)+ B 2 sin(2x)) We then get y p = ( A sin(x) + A 2 cos(x)) + 2( B sin(2x) + B 2 cos(2x)) y p = (A cos(x) + A 2 sin(x)) 4(B cos(2x) + B 2 sin(2x)) Plugging these in the differential equations yields y p y p 2y p = [ (A cos(x) + A 2 sin(x)) 4(B cos(2x) + B 2 sin(2x))] [( A sin(x) + A 2 cos(x)) + 2( B sin(2x) + B 2 cos(2x))] 2[(A cos(x) + A 2 sin(x)) + (B cos(2x) + B 2 sin(2x))] = ( 3A A 2 ) cos(x) + (A 3A 2 ) sin(x) + ( 6B 2B 2 ) cos(2x) + (2B 6B 2 ) sin(2x) Since we want this to equal cos(x) sin(2x), we require 3A A 2 = A 3A 2 =, 6B 2B 2 = 2B 6B 2 = We therefore get A = 3, A =, B =, B 2 2 = 3 We conclude that the general 2 solution of the non-homogeneous problem is y = y h + y p = C e 2x + C 2 e x (3 cos(x) + sin(x)) + 2 ( cos(2x) + 3 sin(2x))

2 2 Define f(x) = cos(x) on [, π] (a) Find the Fourier sine series of f (You might find the formula sin(a) cos(b) = (sin(a + B) + sin(a B))/2 useful) The Fourier sine series of f is n= b n sin(nx) where b n = 2 π = 2 π = π π π π f(x) sin(nx) dx cos(x) sin(nx) dx sin((n + )x) + sin((n )x) dx Note that π sin(kx) dx = whenever k is even so we only need to consider even n in the expression above (so (n ± ) is odd) We have b n = [ ] π cos((n + )x) cos((n )x) π n + n = [ ] cos((n + )π) cos((n )π) + π n + n = [ 2 π n ] n where we used the fact that cos(rπ) = for odd r To sum up, the required Fourier series is b 2k sin((2k)x) = ( 2 π 2k ) sin((2k)x) 2k k= k= (b) What does the series in (a) converge to at x = and x = π/4? Note that the series in (a) is actually the Fourier series corresponding to the odd extension f o of f and therefore converges pointwise to it wherever f o is continuous Thus, at x = π/4, the series converges to f o (π/4) = f(π/4) = cos(π/4) = / 2 At x =, since f o is discontinuous, the series converges to (f o ( + ) + f o ( ))/2 = (f() f())/2 = 2

3 3 (a) Show that for any a, b, c, θ (a cos(θ) + b sin(θ) + c) 2 2(a 2 + b 2 + c 2 ) Define u = a b and v = cos(θ) sin(θ) Then, u u = a 2 + b 2 + c 2, v v = c cos 2 (θ) + sin 2 (θ) + = and u v = a cos(θ) + b sin(θ) + c so by the Cauchy- Schwartz inequality, we have (u v) 2 (u u)(v v) (a cos(θ) + b sin(θ) + c) 2 2(a 2 + b 2 + c 2 ) (b) Let a, b be non-zero vectors in R n Define T (v) = (v a)b Find the rank and nullity of T Note that all the elements in the range of T are scalar multiples of b and, since a and b are non-zero, all the multiples are included in the range Thus, rank(t ) = By the rank-nullity theorem, it follows that rank(a)+nullity(t ) = n nullity(t ) = n Another way of seeing that nullity(t ) = n is that T sends precisely those elements to that are orthogonal to a, ie, the kernel of T is precisely the orthogonal complement of span{a} Since span{a} is one-dimensional, its orthogonal complement has dimension (n ) 3

4 4 Find the solution of the wave equation 2 u t 2 = 64 2 u x 2, < x < π, t > satisfying the boundary conditions u(, t) =, u(π, t) = (for t > ) and initial conditions u(x, ) = 2 and u (x, ) = 2 sin(7x) 4 sin(x) (for < x < π) t For the given boundary conditions, the general solution of the wave equation is (note that α = 64 = 8) u(x, t) = Plugging in t =, we get [a n cos(8nt) + b n sin(8nt)] sin(nx) () n= u(x, ) = a n sin(nx) The a n s are just the Fourier sine coefficients of f(x) = 2 (on < x < π) so that n= π a n = 2 (2) sin(nx) dx π = 4 [ cos(nx) ] π π n = 4 { n even nπ ( ( )n ) = 8/nπ n odd (2) Next, differentiating () with respect to t and plugging in t = gives u t (x, ) = (8nb n ) sin(nx) n= Compare this with g(x) = 2 sin(7x) 4 sin(x) to infer that 8(7)b 7 = 2 b 7 = 8()b = 4 b = with all the other b 2 n s equal to zero We conclude that the solution is () with the coefficients given by (2) and the last statement 28, 4

5 5 Solve the initial value problem x = Ax for 4 A =, x() = 5 We first find the eigenvalues of A The characteristic polynomial is λ 2 6λ+9 = (λ 3) 2 so ( the only ) eigenvalue is λ = 3 with a multiplicity of two Note that (A 3I) = 2 4 is not the zero matrix so A does not have two eigenvectors corresponding to 2 it lone eigenvalue We therefore need to turn to the matrix exponential method By the Cayley-Hamilton Theorem, we get (A 3I) 2 = so that e At = e 3It+(A 3I)t = e 3It ( e (A 3I)t ) = e 3t I + (A 3I)t + (A 3I) 2 t2 t3 + (A 3I)3 2! 3! + = e 3t (I + (A 3I)t) 2t 4t = e 3t t + 2t The general solution of the given problem is therefore x(t) = e At c where c is ( a) column matrix of constants The initial condition x() = gives e A() c = c = so we conclude that the solution to the initial value problem is 2t 4t + 2t x(t) = e 3t = e 3t t + 2t + t 5

6 6 Let V be the space of vectors in R 4 such that x + x 2 + x 3 + x 4 = and let W be the set of vectors in V such that x = x 4 (a) Find an orthogonal basis {v, v 2, v 3 } for V with v = (,,, ) and such that {v, v 2 } is a basis for W Observe that if we choose v 2 = (,,, ), then v 2 W as well as being orthogonal to v Since W is two-dimensional, {v, v 2 } is a basis for it Next, let v 3 = (,,, ) Then, v 3 is orthogonal to both v and v 2 and also belongs to V Since the latter has dimension 3, {v, v 2, v 3 } is an orthogonal basis for it (b) Find the orthogonal projection of v = (2,, 3, 6) on W We have proj W (v) = v v v v v + v v 2 v 2 v 2 v 2 = 3 + v v 2 = v 2v 2 = ( 2,, 3, 2) (c) Find the distance from v to W The distance from v to W is given by v proj W (v) = (4,,, 4) = ( ) /2 = 4(2) /2 6

7 7 Show that {e 3x, e x, e 4x } is a fundamental solution set for y + 2y y 2y = Let y = e 3x, y 2 = e x, y 3 = e 4x Then, we have y + 2y y 2y = e 3x ( ) = y 2 + 2y 2 y 2 2y 2 = e x ( ) = y 3 + 2y 3 y 3 2y 3 = e 4x ( ) = so y, y 2, y 3 are solutions of the given differential equation To show that {y, y 2, y 3 } is a fundamental solution set, we only need to prove that these functions are linearly independent The Wronskian for these functions is e 3x e x e 4x W (x) = 3e 3x e x 4e 4x 9e 3x e x 6e 4x = (e 3x )(e x )(e 4x ) = e 2x [( 6 + 4) ( ) + (3 + 9)] = e 2x [ ] = 84e 2x which is never zero Hence, {y, y 2, y 3 } is a linearly independent collection of functions and therefore a fundamental solution set 7

8 8 Consider the matrix A = 2 2 (a) Find a fundamental matrix for z = Az We first find the eigenvalues of A We have = λ 2 4λ + 5 so( λ = [4 ± ) i 6 2]/2 = 2 ± i Corresponding to λ = 2 + i, we have A λi = i i which has v = in its null space Thus, two linearly independent solutions of z = Az are sin(t) z (t) = e 2t cos(t) e 2t sin(t) = e 2t cos(t) cos(t) z 2 (t) = e 2t sin(t) + e 2t cos(t) = e 2t sin(t) ( cos(t) A fundamental matrix is therefore Z(t) = e 2t sin(t) ) sin(t) cos(t) e (b) Find a particular solution of z 2t = Az + f(t) where f(t) = e 2t By variation of parameters, a particular solution is z p (t) = Z(t)v(t) where v(t) = Z(t) f(t) dt cos(t) sin(t) e = e 2t 2t sin(t) cos(t) e 2t dt cos(t) sin(t) = dt sin(t) cos(t) sin(t) + cos(t) = cos(t) sin(t) Thus, cos(t) sin(t) sin(t) + cos(t) z p (t) = e 2t sin(t) cos(t) cos(t) sin(t) = e 2t ( cos 2 (t) + sin 2 (t) cos 2 (t) + sin 2 (t) ) = e 2t 8

9 6 2 9 Let A = 5 (a) Diagonalize A We first find the eigenvalues of A We have λ 2 5λ + 4 = (λ 4)(λ ) = λ =, λ = Corresponding to λ = 4, we have A λi = which has v 5 5 = ( ) 5 2 in its null space Similarly, corresponding to λ =, we get A λi = which has v 2 = in its null space Define 5 P = ( ) 2 5 Then we should have A = P DP, D = 4 (b) By using (a) or otherwise, find a matrix B such that B 2 = A Observe that if we have a matrix E such that E 2 = D, then (P EP ) 2 = P E 2 P = P DP = ( A ) Since D is a diagonal matrix with positive entries 2 only, we can set E = to ensure that E 2 = D (the diagonal entries of E can also be taken to be negative) Defining B = P EP then gives the required matrix: B = ( ) = =