# | = |2x| &lt; 1 x /2n − 1x ```10.1. By the ratio test, the series converges if |2n xn /2n − 1xn−1 | = |2x| &lt; 1
and diverges if |2n xn /2n − 1xn−1 | = |2x| &gt; 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 &middot; 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
&plusmn;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 &plusmn; i 23 − 2 =
q
√
√
9
3
3 &lt; 3. Consequently, the radius of convergence
− 23 &plusmn; 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&middot;4&middot;6
, e.t.c.. In
Denote a0 = C. Then a2 = C2 , a4 = a42 = 2&middot;4
C
C
general, a2k = 2&middot;4&middot;&middot;&middot;2k = 2k &middot;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 &middot; k!
= x, hence
R
dy
y
=
R
x dx. Consequently,
.
∞
∞
X
X
(x2 /2)n
Cx2n
=C
=
;
n!
2n &middot; 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
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