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plqsolns - 642:527 1. Does ∞ ENTRANCE QUIZ: SOLUTIONS...

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Unformatted text preview: 642:527 1. Does ∞ ENTRANCE QUIZ: SOLUTIONS FALL 2007 x2 + 1 dx converge or diverge? Justify your answer. x4 + 4 0 Solution: The integrand is well behaved everywhere so the only question is what happens as ∞ x2 + 1 dx x2 1 x → ∞. When x is very large, 4 ≈ 4 = 2 so the convergence is the same as (with x +4 x x x2 a ∞ ∞ 1 1 dx =− = , so the integral converges. a > 0 arbitrary). But 2 x xa a a 2. Suppose that f (x, y ) = xy 2 sin(x2 y ). Find fx and fxy . Solution: From the product rule and the chain rule, fx = fxy ∂f = y 2 sin(x2 y ) + (xy 2 )(2xy ) cos(x2 y ) = y 2 sin(x2 y ) + 2x2 y 3 cos(x2 y ), ∂x ∂ ∂f = = 2y sin(x2 y ) + y 2 (x2 cos(x2 y )) + 6x2 y 2 cos(x2 y ) + 2x2 y 2 (−x2 sin(x2 y )) ∂y ∂x = 2y (1 − x4 y ) sin(x2 y ) + 7x2 y 2 cos(x2 y ). 3. Find the general solution y (x) of the equation y ′′ = 3y . Solution: This is a linear, second order, homogeneous, constant coefficient equation, so we look for a solution√ = erx . Plugging this in we find that it is a solution if r 2 = 3, so the general solution y √ is y (x) = c1 e 3x + c2 e− 3x , where c1 and c2 are arbitrary constants. 4. Find the general solution y (x) of the equation y ′ = xy 3 . dy dy Solution: We separeate variables: = xy 3 , so 3 = x dx and, integrating both sides of this dx y x2 1 + C , where C is the integration constant (which must be put in at this equation, − 2 = 2y 2 √ stage). Solving for y gives y (x) = ±1/ D − x2 , where D = −2C is just a different arbitrary constant. 5. Find the eigenvalues of the matrix A = 34 , and find an eigenvector for one of them. 32 3−λ 4 = λ2 − 5λ − 6 = 0: λ1 = 6, 3 2−λ u1 3−6 4 4 . = 0, finding u = u2 3 2−6 3 Solution: The eigenvalues are the roots of the equation λ2 = −1. To find an eigenvector u for λ1 we solve Similarly, and eigenvector v for λ2 is v = 1 . −1 6. Find the radius of convergence of the power series Solution: We try the ratio test: if bn = n3 x lim n 2n ∞ n=0 n3n x2n . is a typical term of the series then n→∞ (n + 1)3n+1 x2(n+1) (n + 1) bn+1 = lim 3|x|2 = 3|x|2 . = lim n x2n n→∞ n→∞ bn n3 n √ √ The series converges if this limit is less than 1, i.e., if |x| < 1 3; the radius of convergence is 1/ 3. 1 ...
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