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10.1. By the ratio test, the series converges if |2n xn /2n − 1xn−1 | = |2x| < 1 and diverges if |2n xn /2n − 1xn−1 | = |2x| > 1. It follows that the radius of convergence is 1/2. 10.2. We have y 00 (x) = xy 0 (x) + y. By the Taylor’s formula, an = y n (0)/n!. Consequently, a0 = y(0) = 1, a1 = y 0 (0) = 1, y 00 (0) = 0 · y 0 (0) + y(0) = 1, hence a2 = 1/2. y 000 = 2y 0 + xy 00 , hence y 000 (0) = 2y 0 (0) = 2, and a3 = 2/6 = 1/3. y (4) = 3y 00 + xy 000 , hence y (4) (0) = 3y 00 (0) = 3, and a4 = 3/24 = 1/8. 10.3. We solve it in the same way as the previous problem. 1 y 00 = − y 0 − y. x a0 = y(1) = 1, a1 = y 0 (1) = 1, y 00 (1) = −y 0 (1) − y(1) = −2, so that a2 = −1. 1 0 1 00 y − y − y0 , x2 x hence y 000 (1) = y 0 (1) − y 00 (1) − y 0 (1) = −y 00 (1) = −1, and a3 = −1/6. y 000 = y (4) = − 2 2 1 + 2 y 00 − y 000 − y 00 , x3 x x hence y (4) (1) = −2 + 2y 00 (1) − y 000 (1) − y 00 (1) = −3, and a4 = −1/8. 10.4. The radii of convergence of the Taylor series of the solution of y 00 + 4x 0 1 y + y=0 3 1+x 1 + x3 4x around x0 is not less than the radius of convergence of the Taylor series of 1+x 3 1 and 1+x3 around x0 . It is equal to the distance to x0 to the closest root of the 3 polynomial 1 + x3 . √ We have 1 + x = (1 + x)(1 − x + x2 ), hence the roots of √ 3 3 1 1 3 1 + x are −1, 2 + i 2 , and 2 − i 2 . Distance from x0 = 0 to all these roots is equal to one, since q 1 2 ±i √ 3 2 = 1 4 + 34 = 1 and | − 1| = 1. Consequently, the radius of convergence of the Taylor series of y(x) around x0 = 0 is at least 1. √ Distances from x0 = 2 to the roots of 1+x3 are |−1−2| = 3, 12 ± i 23 − 2 = q √ √ 9 3 3 < 3. Consequently, the radius of convergence − 23 ± i 23 = 4 + 4 = √ around x0 = 2 is at least 3. 1 10.5. (1) Substitution of y = ∞ X P∞ n=0 an xn into the equation gives (n + 1)an+1 xn − n=0 ∞ X an−1 xn = 0 n=1 It follows that a1 = 0 and (n + 1)an+1 − an−1 = 0 for all n ≥ 1. We get hence a recurrent relation nan − an−2 = 0, an = an−2 . n for all n ≥ 2. C C , a6 = a64 = 2·4·6 , e.t.c.. In Denote a0 = C. Then a2 = C2 , a4 = a42 = 2·4 C C general, a2k = 2·4···2k = 2k ·k! . Since a1 = 0, a3 = a31 = 0, a5 = a53 = 0, e.t.c., so that all an for odd n are equal to zero. We get then a series solution y(x) = ∞ X k=1 0 y y x2 /2 (2) We have y 0 = xy, or 2 ln |y| = x /2 + c1 , or y = Ce Note that x2 /2 Ce C x2k . 2k · k! = x, hence R dy y = R x dx. Consequently, . ∞ ∞ X X (x2 /2)n Cx2n =C = ; n! 2n · n! n=1 n=1 and we see that the obtained solutions coincide. 10.6. We are looking for solutions in the form y = xr . Substitution into the equation gives r(r − 1)xr + 4rxr + 2xr = 0, hence r(r −1)+4r +2 = 0, or r2 +3r +2 = 0. The roots are r = −1 and r = −2. It follows that the general solution of the equation is y = c1 x−1 + c2 x−2 . 2