527plsolns

527plsolns - 642:527 PREREQUISITE QUIZ: SOLUTIONS 2n x3n ....

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 642:527 PREREQUISITE QUIZ: SOLUTIONS 2n x3n . n2 n=0 ∞ Fall 2009 1. Find the radius of convergence of the power series Solution: We use the ratio test: if bn = 2n x3n /n2 is a typical term of the series then n→∞ lim 2n+1 x3(n+1) /(n + 1)2 2|x|3 n2 bn+1 = lim = lim = 2|x|3 . n→∞ n→∞ (n + 1)2 bn 2n x3n /n2 The series converges if this limit is less than 1, i.e., if |x| < 2−1/3 , so the radius of convergence is 2−1/3 . 2. Find the eigenvalues of the matrix A = 3 6 , and find an eigenvector for one of them. −1 −4 3−λ 6 = λ2 + λ − 6 = 0: −1 −4 − λ u1 3+3 6 λ1 = −3, λ2 = 2. To find an eigenvector u for λ1 we solve = 0, finding u2 −1 −4 + 3 −1 −6 . . Similarly, an eigenvector v for λ2 is v = u= 1 1 Solution: The eigenvalues are the roots of the equation 3. Suppose that f (x, y ) = x2 y ln(xy 2 ). Find fy and fxy . Solution: By the product and chain rules: fy = fxy = ∂ ∂x ∂f 1 = x2 ln(x2 y ) + x2 y 2xy = x2 ln(x2 y ) + 2x2 , ∂y xy 2 ∂f 1 y 2 + 4x = 2x ln(xy 2 ) + 5x. = 2x ln(x2 y ) + x2 ∂y xy 2 Notice that we never have to compute fx , since always fxy = fyx . 4. Find the general solution y (x) of the equation y ′ = ye2x . Solution: We separate variables (remember that y ′ = dy/dx) to obtain dy = e2x dx. y Integrating gives ln y = 1 e2x + C or y = ee 2 2x /2+C = Kee 2x /2 , where K = eC . 5. Find the solution y (x) of the initial value problem y ′′ − 4y = 0, y (0) = 3, y ′ (0) = 2. Solution: Since this is a constant coefficient linear equation we expect solutions of the form y = erx ; substituting this form into the equation we find that r must satisfy r 2 − 4 = 0, so r = ±2. The general solution is y = c1 e2x + c2 e−2x ; imposing the initial conditions gives y = 2e2x + e−2x . 6. Does sin x dx converge or diverge? Justify your answer. x2 π Solution: The function is finite for all x in the range of integration so the only question is ∞ π ∞ dx π x2 ∞ convergence at x = ∞. Now converges: ∞ dx π x2 1 = −x = 1 . π But the sin x factor just oscillates between 1 and −1, which will not destroy this convergence; we conclude that the ∞ 1 integral converges. Technically we are using a comparison test: sin2x ≤ x2 , π dx converges, so x x2 ∞ sin x 2 dx converges x π 1 ...
View Full Document

Ask a homework question - tutors are online