Test #2 Math 2250 Summer 2003

Transcription

1 Test #2 Math 225 Summer 23 Name: Score: There are six problems on the front and back of the pages. Each subpart is worth 5 points. Show all of your work where appropriate for full credit. ) Show the following sets of vectors in R n are linearly dependent or linearly independent. If they are linearly dependent, find a nontrivial linear combination of the vectors which equals zero. a) u = 2 ; v = 3 ; w = 2 ; c u + c 2 v + c 3 w = Free variable is c 3. Let c 3 =, then c 2 = 2 and c = 3 gives the linear combination 3u + 2v + w =. b) v = 4 ; v 2 = ; v 3 = ; v 4 = 2 3 ; This has a 4 4 diagonal submatrix (last four rows) which has nonzero determinant (product of the diagonals is nonzero), so the vectors are linearly independent.

2 2) Show whether or not the following sets of vectors W form a subspace. a) W is the set of all vectors in R 4 such that x + 2x 2 + 3x 3 + 4x 4 =. Let (x, x 2, x 3, x 4 ) and (y, y 2, y 3, y 4 ) be two vectors in W. Then checking that a linear combination a(x, x 2, x 3, x 4 )+b(y, y 2, y 3, y 4 ) satisfies the property x +2x 2 +3x 3 +4x 4 = gives us closure under addition and scalar multiplication as necessary to show W is a subspace. a(x, x 2, x 3, x 4 ) + b(y, y 2, y 3, y 4 ) = (ax + by, ax 2 + by 2, ax 3 + by 3, ax 4 + by 4 ) Checking the condition, W is a subspace. (ax + by ) + 2(ax 2 + by 2 ) + 3(ax 3 + by 3 ) + 4(ax 4 + by 4 ) = (ax + 2ax 2 + 3ax 3 + 4ax 4 ) + (by + 2by 2 + 3by 3 + 4by 4 ) = a(x + 2x 2 + 3x 3 + 4x 4 ) + b(y + 2y 2 + 3y 3 + 4y 4 ) = a + b = = b) W is the set of all f in F (the set of all real valued functions defined on the whole real line) such that f() =. Again we check a linear combination of two vectors from W, f and g. W is a subspace, as well. (af + bg)() = af() + bg() = a + b = 2

3 3) Find a basis for the solution space of the following homogeneous equations. a) x 3x 2 5x 3 6x 4 = 2x + x 2 + 4x 3 4x 4 = x + 3x 2 + 7x 3 + x 4 = x 3 is the only free variable which yields through back substitution x 4 =, x 3 = t, x 2 = 2t, x = t. Letting t = yields the single basis element, {(, 2,, )}. b) y (4) 8y + 6y = The characteristic equation is r 4 8r 2 +6 = (r 2 4) 2 = (r 2)(r 2)(r+2)(r+2) =. Thus, r = ±2 are the repeated roots yielding the basis functions {e 2x, e 2x, xe 2x, xe 2x }. 3

4 4) a) Find two linearly independent solutions to y 2y + 2y =. () The characteristic equation r 2 2r +2 = yields the roots r = ±i, which correspond to the two solutions y = e x cos(x) and y 2 = e x sin(x). b) Verify that they are linearly independent using the Wronskian. y W = y 2 y y 2 e = x cos(x) e x sin(x) e x sin(x) + e x cos(x) e x cos(x) + e x sin(x) = e x cos(x)(e x cos(x) + e x sin(x)) e x sin(x)( e x sin(x) + e x cos(x)) = e 2x (cos 2 (x) + sin 2 (x)) = e 2x c) Find a particular solution to () with the initial conditions y() = 2 and y () = 3. The general solution is y = C e x cos(x) + C 2 e x sin(x) so that y() = C = 2. Since y (x) = C ( e x sin(x) + e x cos(x)) + C 2 (e x cos(x) + e x sin(x)), we get that y () = C + C 2 = 3, so that C 2 = and the particular solution is y(x) = 2e x cos(x) + e x sin(x). 4

5 5) Find the general solution to the forced oscillatory pendulum equation without damping Lx + gx = F cos(ωt) a) when the forcing is not resonant. )Solving the homogeneous equation Lx + gx =. We get the characteristic equation Lr 2 + g =, so that r = ± g Li. From this we get the general complementary solution ( ) ( ) g g y c (t) = C cos L t + C 2 sin L t 2)Solving for the particular solution. Since we are in the nonresonance case we know that the natural frequency g L = ω is not equal to ω, so that we take our trial particular solution to look like We now plug into the equation y p (t) = A cos(ωt) + B sin(ωt). x + ω 2 x = F /L cos(ωt) (2) to solve for A and B using the method of undetermined coefficients. We get ( Aω 2 cos(ωt) Bω 2 sin(ωt)) + ω 2 (A cos(ωt) + B sin(ωt)) = F /L cos(ωt) Equating coefficients of the sin(ωt) and cos(ωt) we get the two equations which yields 3)The general solution is then A(ω 2 ω2 ) = F /L B(ω 2 ω 2 ) = A = F /L ω 2 ω 2 B = y(t) = y c + y p = C cos(ω t) + C 2 sin(ω t) + F /L ω 2 ω 2 cos(ωt) 5

6 b) when the forcing is resonant. The homogeneous solution remains the same. Now since ω = ω, we need the trial solution multiplied by the independent variable y p (t) = t(a cos(ωt) + B sin(ωt)) Plugging this into the differential equation (2) (remembering that ω = ω) we get 2( Aω sin(ωt) + Bω cos(ωt)) = F /L cos(ωt) Again, equating coefficients and solving yields A =, B = F /L 2ω, which gives the final solution y(t) = y c + y p = C cos(ω t) + C 2 sin(ω t) + F /L 2ω t sin(ωt) 6

7 6) a) Find the eigenvalues and eigenvectors of the matrix. A = 2 To find the eigenvalues we solve for λ the characteristic equation A λi =. A λi = λ 2 λ = λ( λ) 2 = λ2 λ 2 = (λ 2)(λ + ) Therefore the eigenvalues are λ =, 2. Finding the associated eigenvectors requires solving the equation (A λi)v = for the vector v. For λ =, 2 2 v = v so that v 2 is a free variable. Letting v 2 = forces v = and gives the eigenvector (, ). For λ = 2, 2 2 v = v so that v 2 is a free variable. Letting v 2 = 2 forces v = and gives the eigenvector (, 2). b) Solve the following system of first order equations x y = y = 2x + y for the general solution. Differentiating the first equation yields the second order equation in x x = y = 2x + y = x + 2x or x x 2x = Solving the characteristic equation gives the roots r =, 2 and the general solution x(t) = C e t + C 2 e 2t. Again referring to the first equation we get the solution for y(t) = C e t + 2C 2 e 2t. 7