Department of Mathematical Sciences Examination paper for TMA11 Matematikk 3 Academic contact during examination: Eugenia Malinnikova Phone: 735557 Examination date: 6th May, 15 Examination time (from to): 9:-13: Permitted examination support material: C: Simple calculator (Casio fx-8es PLUS, Citizen SR-7X or Citizen SR-7X College, Hewlett Packard HP3S), Rottmann: Matematiske formelsamling Language: English Number of pages: 6 Number pages enclosed: Checked by: Date Signature
Page 1 of 6 Problem 1 Solve the quadratic equation z + ( + i)z + 3 =, write the solutions in normal form. Solution We use the formula for solutions of the quadratic equation z 1, = ( + i) ± ( + i) 3 = i ± i = i + i. Now we have i = e iπ/ and then one of the roots is given by i = e iπ/ = / + /i. Thus z 1 = + + ( 1 + )i, z = + ( 1 )i Problem a) Solve the initial value problem x + 6x + 8x =, x() =, x () = 8. What is the maximal value attained by this solution x(t) for t >? Solution First we find the roots of the characteristic equation λ + 6λ + 8 =, we have λ 1 = and λ =. Then the general solution to the homogeneous equation is x(t) = c 1 e t +c e t. To find the constants, we use initial conditions, clearly, x() = c 1 + c and x () = c 1 c. Solving the system, c 1 + c =, c 1 c = 8 we obtain c 1 =, c =. Thus x(t) = e t e t. To find the maximum value, we compute x (t) = 8e t + 16e t, if e t = then x is positive on (, t ) and negative on (t, + ). Therefore the maximum value of x on (, + ) is attained at t and x(t ) = e t e t = 1 = 1. b) Find the steady-state solution of the equation x + 6x + 8x = cos t. Solution We consider the complex equation z + 6z + 8z = e it, such that the real part of a solution is a solution of our initial equation. We are looking for a solution of the form z(t) = ae it. We have z + 6z + 8z = ((i) + 6(i) + 8)ae it = P (i)z(t), P (w) = w + 6w + 8. Hence z(t) = e it /P (i) and 1/P (i) = ( + 1i) 1 = (1 3i)/ and z(t) = (.1.3i)e it =.1 cos t +.3 sin t + i(.1 sin t.3 cos t),
Page of 6 x(t) =.1 cos t +.3 sin t. Alternative solution We use method of undetermined coefficients to find the particular solution of the form x(t) = a cos t + b sin t. The derivatives are: x (t) = a sin t + b cos t and x (t) = a cos t b sin t. Then x +6x+8x = a cos t b sin t+6( a sin t+b cos t)+8(a cos t+sin t) = (a + 1b) cos t + (b 1a) sin t. We are solving the equation x + 6x + 8x = cos t. Therefore we want to find a, b such that a + 1b = and b 1a =, we get a =.1, b =.3, the answer is x(t) =.1 cos t +.3 sin t. Problem 3 Find general solution of the equation (Hint (cos x) 1 dx = ln sec x + tan x.) y + y = 3x + tan(x). Solution First, solutions of the corresponding homogeneous equation are y 1 (x) = cos x and y (x) = sin x. To find the solution of the non-homogeneous equation we divide the right handside into two parts, r 1 (x) = 3x and r (x) = tan x. We can use the method of undetermined coefficients for the first part, we look for y p1 = ax + b, then y = and y + y = ax + b, thus y p1 = 3x solves the equation y + y = 3x. For the second equation we use the method of variation of parameters, W (y 1, y ) = y 1 y y 1y = cos x + sin x = 1. Thus y p = y 1 y tan(x)dx + y y1 tan(x)dx, y p (x) = cos x cos x sin x cos x dx + sin x sin xdx = (cos x) 1 dx + cos x cos xdx + sin x sin xdx = cos x ln sec x + tan x + cos x sin x sin x cos x = cos x ln sec x + tan x Summing up the terms, we get the general solution y(t) = y h (t) + y p1 (t) + y p (t) = c 1 cos x + c sin x + 3x cos x ln sec x + tan x
Page 3 of 6 Problem Let A = 1 t. t a) For which values of t does the equation Ax = b have a solution for any b in R? Solution The equation Ax = b has a solution for all b if and only if the matrix A is invertible. This happens if and only if det(a). We have det(a) = t. Thus the equation always has a solution when t ±. Alternatively, we can start by performing Gauss elimination: 1 t t R tr 1 1 t t There is no zero-rows in the final matrix if and only if t. Thus the equation Ax = b has a solution for all b if and only if t ± b) Find an LU decomposition of A (the result will depend on the parameter t). 1 t Solution We apply Gauss elimination (see above) and get U = t, to get L we remind that the only row operation in the Gauss elimination was adding ( t) 1 times the first row to the second one. Thus L =, a simple computation t 1 confirms that 1 1 t A = t 1 t. Problem 5 Given the following vectors in R 1 3 v 1 =, v = 1, v 3 3 =, v = 1, 1 1 let V = Span{v 1, v, v 3, v }. a) Are the vectors {v 1, v, v 3, v } linearly independent? Find a basis for V.
Page of 6 Solutuon We consider the matrix with column vectors v 1, v, v 3 and v and perform the Gauss elimination to find out if the columns are linearly dependent or not. 1 3 3 C = 1 1 1 1 1 3 1 1 5 3 1 1 R 3 R 1 R +R 1 3 3 1 1 5 1 1 1 3 1 1 5 3 6 SW AP (R,R 3 ) R R 3 1 3 1 1 5 3 Since there is no pivot element in the last column, the columns are linearly dependent. There are pivot elements in the first three columns, thus {v 1, v, v 3 } form a basis for V. b) Find an orthogonal basis for V. Solution We use the Gram-Schmidt algorithm to find an orthogonal basis: ũ = v v u 1 u 1 u 1 u 1 = u 3 = v 3 v 3 u 1 u 1 u 1 u 1 v 3 u u u u = 1 u 1 = v 1 =, 1 1 5 = 1 5 1, u = 1 1 5 5 1 3 5 3 3 1 = 3 1 5 1 c) Does there exist a vector u in R which is orthogonal to v 1, v, v 3, v? Yes, since according to part (a) the dimension of V is three and V is a subspace of R, there exists a non-zero vector u in V. One can find such vector by solving the system C T u = ;for example u = 1 1 T is such a vector.
Page 5 of 6 Problem 6 a) Find (complex) eigenvalues and (complex) eigenvectors of the matrix 1 1 3 Solution First, the characteristic polynomial is det(b λi) = (1 λ)(3 λ)+ = λ λ + 5 = (λ ) + 1. The roots are λ 1 = + i, λ = λ 1 = i. We find an eigenvector corresponding to λ 1, 1 i B ( + i)i =, v 1 1 i 1 = 1 i For λ = λ 1, we obtain v = v 1 =. b) Find the solution of the system x 1 = x 1 x x = x 1 + 3x that satisfies the initial conditions x 1 () = 1 and x () = 1. Write down the answer using real-valued functions. Solution We know that v 1 e λ 1t and v e λ t are (complex conjugate) solutions of this system, we find two real solutions: ( ) 1 e (+i)t + e ( i)t 1 i ( ) = et (cos t + i sin t) + (cos t i sin t) = e t cos t 1 i cos t + sin t ( ) 1 e (+i)t e ( i)t = e t sin t i 1 i cos t sin t General solution is a linear combination of these two, x(t) = a 1 e t cos t + a cos t + sin t e t sin t. cos t sin t a Then x() = 1 and the initial conditions give a a 1 a 1 =.5 and a = 1.5. Finally cos t 3 sin t x(t) = cos t + sin t
Page 6 of 6 Alternative solution General solution has the form x(t) = c 1 v 1 exp(λ 1 t) + c v exp(λ t), where λ 1, are eigenvalues, v 1, are corresponding eigenvectors and c 1, are some constants. The initial conditions imply 1 = c 1 1 + c 1 i, and solving the linear system, we obtain c 1 = (1 + 3i)/, c = (1 3i)/. Then we have x(t) = et ( (1 + 3i)(cos t + i sin t) + (1 3i)(cos t i sin t) 1 i = et ( (cos t 3 sin t + i(sin t + 3 cos t)) 1 i ) + (cos t 3 sin t i(sin t + 3 cos t)) = e t cos t 3 sin t cos t + sin t ) Problem 7 Suppose that A is an m n-matrix with real entries. Prove that x A T Ax for each x in R n and therefore each real eigenvalue of the matrix A T A is non-negative. Solution First, note that x A T Ax = x T A T Ax = (Ax) T Ax = Ax. If A T Av = λv for some real λ and v then v A T Av = v λv = λ v and since v > we conclude that λ.