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Solution for Homework 7 1 [KA] Sec. 8.5, page 468 4. (a) The linearization of the equation at x = π is x0 = x − π. (c) The general solution of the linearized equation is x = Aet + π. The initial condition x(0) = 3 gives A = 3 − π. So the solution is x = (3 − π)et + π. (d) The general solution of the linearized equation is Aet = csc x + cot x, i.e., x = 2 tan−1 (A−1 e−t ). The initial condition x(0) = 3 gives A = (tan(3/2))−1 . So the solution is x = 2 tan−1 (e−t tan(3/2)). (e) This question is impossible to answer precisely. 5. (a) The fixed points (equilibriums) are x = kπ for k integers; linearize the equation at x = kπ: x0 = (−1)k (x − kπ); x = kπ is stable if k is odd and unstable if k is even. (b) The fixed points are y = kπ + π/2 for k integers; linearize the equation at y = kπ + π/2: y 0 = (−1)k+1 (y − kπ − π/2); y = kπ + π/2 is stable if k is even and unstable if k is odd. (c) The fixed points are x = 1 and x = 2; linearize the equation at x = 1: x0 = −(x − 1); linearize the equation at x = 2: x0 = x − 2; x = 1 is stable and x = 2 is unstable. (d) The fixed points are y = 1 and y = 2; linearize the equation at y = 1: y 0 = y − 1; linearize the equation at y = 2: y 0 = −(y − 2); y = 1 is unstable and the solution y = 2 is stable. [KA] Sec. 3.1, page 137 1. (a) The solution is x = ( 12 − t)−1 . The interval of existence is (−∞, 12 ). (b) The solution is x = (− 12 − t)−1 . The interval of existence is (− 12 , ∞). (c) No such solution exists. (d) The initial condition is x(0) = 10 and the corresponding solution is 1 http://www.math.ucsb.edu/˜xichen/math3c01s/hw7key.pdf 1 1 x = ( 10 − t)−1 . −1 −1 (e) The solution is x = (x−1 0 − t) . The interval of existence is (−∞, x0 ). −1 −1 (f) The solution is x = (−x−1 0 − t) . The interval of existence is (−x0 , ∞). 3. (b) Since x2 + t2 is increasing in both x and t when x > 0 and t > 0, the Euler’s method underestimates the true value of x(t) for t > 0. (c) The general solution is x = (C −t)−1 . By the initial condition x(0.9) = 14, C = 34/35. (d) Example 4 shows that the solution has a singularity greater than π/4 and (c) shows it has a singularity less than 34/35. 4. Solve x0 = x2 with x(0) = 2 and we obtain x = (1/2−t)−1 ; solve x0 = x2 +1 with x(0) = 2 and we obtain x = tan(t + tan−1 (2)). So the solution has an singularity between π/2 − tan−1 (2) and 1/2. p 10. The general solution is x = A |t2 − 1|. √ (a) The solution is x = 1 − t2 . So the interval of existence is (−1, 1). √ (b) The solution is x = − 1 − t2 . So the interval of existence is (−1, 1). (c) The solution is x = 0. So the interval of existence is (−∞, ∞). √ (d) The solution is x = 3−1/2 t2 − 1. So the interval of existence is (1, ∞). √ (d) The solution is x = −3−1/2 t2 − 1. So the interval of existence is (1, ∞). 11. The general solution is x = (C − (p − 1)t)1/(1−p) . (a) The solution is x = (1 − (p − 1)t)1/(1−p) . So the interval of existence is (−∞, 1/(p − 1)). (b) The solution is x = (x1−p − (p − 1)t)1/(1−p) . So the interval of existence 0 1−p is (−∞, x0 /(p − 1)). [S] Chap. 2, page 4 2.2. The quadratic approximation of ln x at x = 1 is (x − 1) − 21 (x − 1)2 . 2 3.2. The Taylor series of cos 3x is cos 3x = P∞ n=0 (−1)n (3x)2n . (2n)! 3.4. The 4th Taylor polynomial of M 4 − T 4 at T = M is M 4 − T 4 = M 4 − ((T − M ) + M )4 = −4M 3 (T − M ) − 6M 2 (T − M )2 − 4M (T − M )3 − (T − M )4 . 4.1. The Taylor series of (1 + x)−1 at x = 0 is ∞ X 1 = (−1)n xn . 1 + x n=0 The Taylor series of (1 − x2 )−1 at x = 0 is ∞ X 1 = x2n . 1 − x2 n=0 The Taylor series of (1 + x2 )−1 at x = 0 is ∞ X 1 = (−1)n x2n . 1 + x2 n=0 4.2. The Taylor series of cos(3x) at x = 0 is cos(3x) = ∞ X (−1)n n=0 (3x)2n . (2n)! Multiply it by x2 and we obtain 2 x cos(3x) = ∞ X nx (−1) n=0 ∞ 2 (3x)2n X (−1)n 32n 2n+2 = x . (2n)! (2n)! n=0 4.3. The Taylor series of (2 − x)−1 at x = 0 is 1 1 = 2−x 2 1 1 − (x/2) ∞ = 3 ∞ 1 X x n X x n = . n+1 2 n=0 2 2 n=0 Replace x by x3 and we obtain the Taylor series of (2 − x3 )−1 at x = 0: ∞ X x3n 1 = . 2 − x3 2n+1 n=0 4.4. The Taylor series of sinh(x) at x = 0 is 1 x e + e−x 2 ! ∞ ∞ ∞ ∞ n n X X 1 X xn X x 1 x2n n n x = + (−1) = (1 + (−1) ) = . 2 n=0 n! n=0 n! 2 n! (2n)! n=0 n=0 sinh(x) = The Taylor series of x/(x + 1) at x = 0 is ∞ ∞ X X x n n =x (−1) x = (−1)n xn+1 . 1+x n=0 n=0 The Taylor series of xe−x at x = 0 is xe −x =x ∞ X nx (−1) n=0 n n! = ∞ X (−1)n n=0 xn+1 . n! The Taylor series of x2 sin(x2 ) at x = 0 is x2 sin(x2 ) = x2 ∞ X n=0 ∞ (−1)n X (x2 )2n+1 x4n+4 = (−1)n . (2n + 1)! n=0 (2n + 1)! 4.5. The Taylor series of sin(x)/x at x = 0 is ∞ ∞ 2n+1 X X sin(x) x2n −1 n x =x (−1) = (−1)n . x (2n + 1)! n=0 (2n + 1)! n=0 The Taylor series of (ex − 1)/x at x = 0 is ! ∞ ∞ ∞ n X X ex − 1 x xn X xn−1 = x−1 − 1 = x−1 = x n! n! n! n=0 n=1 n=1 4 5.4. Since (arcsin(x))0 = (1 − x2 )−1/2 , we may first find the Taylor series of (1 − x2 )−1/2 and then integrating it. By binomial theorem, ∞ ∞ X X −1/2 2 −1/2 2 n n −1/2 (1 − x ) = (−x ) = (−1) x2n . n n n=0 n=0 So arcsin(x) = ∞ X n (−1) n=0 −1/2 x2n+1 . n 2n + 1 5.5. Since ∞ et − 1 X tn−1 = , t n! n=1 Z 0 x ∞ X et − 1 dt = t n=1 x n−1 Z t 0 n! dt = ∞ X xn . n(n!) n=1 5.6. Since ∞ n ∞ n−1 X X − ln(1 − t) t t −1 =t = , t n n n=1 n=1 Z 0 x ∞ X − ln(1 − t) dt = t n=1 Z 0 x n−1 t n dt = ∞ X xn n=1 n2 . 7.3. The Taylor series of sinh(x) is ∞ X x2n sinh(x) = . (2n)! n=0 Use ratio test: 2n+2 x /(2n + 2)! x2 = 0 < 1. lim = lim 2n n→∞ n→∞ (2n + 1)(2n + 2) x /(2n)! 5 So the Taylor series always converges and hence the radius of convergence is ∞. 7.4. The Taylor series of x2 cos(3x) is 2 x cos(3x) = ∞ X (−1)n 32n n=0 (2n)! x2n+2 . Use ratio test: (−1)n+1 32n+2 x2n+4 /(2n + 2)! 9x2 = 0 < 1. = lim lim n 2n 2n+2 n→∞ (2n + 1)(2n + 2) n→∞ (−1) 3 x /(2n)! So the Taylor series always converges and hence the radius of convergence is ∞. 7.6. Use ratio test: (n + 1)!xn+1 = lim |(n + 1)x|. lim n→∞ n→∞ n!xn Since limn→∞ |(n + 1)x| = ∞ > 1 for all x 6= 0, the power series diverges for all x 6= 0. So the radius of convergence is 0. 6