# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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= x + — + ...,

6

which converges about x0 = 0 and thus xp(x) is analytic 2

at x0 = 0. xq(x), by similar steps, is also analytic at x0 = 0 and thus x0 = 0 is a regular singular point. For x0 = nn, we have

(x-nn)x (x-nn)[(x-nn) + nn]

(x-nn)p(x) =

sinx - (x-nn) 3

±(x-nn)H-------------±

6

Section 5.4

85

19.

21.

(x-nn) 2

= [(x-nn)+nn][±1 ± ------------ ± ...], which

6

converges about x0 — nn and thus (x-nn)p(x) is analytic

2

at x— nn. Similarly (x+nn)p(x) and (x±nn) q(x) are analytic and thus x0 — ±nn are regular singular points.

Substituting y — an^n into the D.E. yields

n=0

2 Y. n(n-1)anxn 1 + 3Y nanxn 1 + anxn+1 = 0. The last sum

n=2 n=1 n=0

becomes Yan-2xn-1 by rePlacing n+1 by n-1, the first term

n=2

of the middle sum is 3a1, and thus we have

3ai + {[2n(n-1)+3n]an + an-2}xn 1 = 0. Hence a1 = 0 and

n=2

-an-2

an =----------, which is the desired recurrance relation.

n = n(2n+1)

Thus all even coefficients are found in terms of a0 and all odd coefficients are zero, thereby yielding only one solution of the desired form.

If F = 1/x then

dy = dy dF = -Y dy = -.2_dy

dx dF dx x2 dF dF

d2y d ,-2dy dF dy ,-2d2y 1

^ = yi (-^^-y) ^ = (-2^ -y - F^) (—^

dx2 dF dF dx dF dF2 x2

= F4d2? + 2F3^y.

dF2 dF

Substituting in the D.E. we have 2

p(1/F)[F4dy- + 2F^-dy] + Q(1/F)[-F2^y] + R(1/F)y = 0, dF2 dF dF

F4p(1/F)— + [2F3p(1/F) - F2Q(1/F)]^y + R(1/F)y — 0. dF2 dF

The result then follows from the theory of singular points at F — 0.

00

2

23. Since P(x) = x , Q(x) = x and R(x) = -4 we have

86

Section 5.5

f(S) = [2P(1/S)/S - Q(1/^)/^2]/P(1/^) = 2/S, - 1/S = 1/S

and g(S) = R(1/S)/S4P(1/S) = -4/S2. Thus the point at

infinity is a singular point. Since both f(S) and

S2g(S) are analytic at S = 0, the point at infinity is a regular singular point.

2 2 2 25. Since P(x) = x , Q(x) = x, and R(x) = x - u ,

f(S) = [2P(1/S)/S - q(1/S)/S2]/p(1/S) = 2/S - 1/S = 1/S

and g(S) = R(1/S)/S4P(1/S) = (1/S2 - u2)/S2 = 1/S4 - u2/S2.

Thus the point at infinity is a singular point. Although

Sf(S) = 1 is analytic at S = 0, S2g(S) = 1/S2 - u2 is not, so the point at infinity is an irregular singular point.

Section 5.5, Page 265

2. Comparing the D.E. to Eq.(27), we seek solutions of the form y = (x+1)r for x + 1 > 0. Substitution of y into the D.E. yields [r(r-1) + 3r + 3/4](x+1)r = 0. Thus r + 2r + 3/4 = 0, which yields r = -3/2, -1/2. The general solution of the D.E. is then

I -, I -1/2 I -,,-3/2

y = c1|x+1| + c2|x+1| , x ^ -1.

r2

4. If y = x then r(r-1) + 3r + 5 = 0. So r + 2r + 5 = 0

and r = (-2 ± "/4-20 )/2 = -1 ± 2i. Thus the general

solution of the D.E. is

y = c1x-1cos(2ln|x|) + c2x-1sin(2ln|x|), x ^ 0.

9. Again let y = xr to obtain r(r-1) - 5r + 9 = 0, or

(r-3) = 0. Thus the roots are x = 3,3 and

y = cj_x3 + c2x3ln|x|, x ^ 0, is the solution of the D.E.

r2

13. If y = x , then F(r) = 2r(r-1) + r -3 = 2r - r - 3 =

(2r-3)(r+1) = 0, so y = cj_x3/2 + c^"1 and

3 1/2 -2

y = — c-,x - c,x . Setting x = 1 in y and y we obtain

2 1 2

3

ci + c2 = 1 and —c1 - c2 = 4, which yield c1 = 2 and

3/2 -1 +

c2 = -1. Hence y = 2x - x . As x ^ 0 we have y ^ -to due to the second term.

2

16. We have F(r) = r(r-1) + 3r + 5 = r + 2r + 5 = 0. Thus

Section 5.5

87

r1,r2 = -1 ± 2i and y = x 1[c1cos(2lnx) + c2sin(2lnx)].

-2

Then y(1) = c1 = 1 and y' = -x [cos(2lnx) + c2sin(2lnx)]

+ x 1[-sin(2lnx)2/x + c2cos(2lnx)2/x] so that y'(1) = -1-2c2 = -1, or c2 = 0.

17. Substituting y = xr, we find that r(r-1) + ar + 5/2 = 0

or r + (a-1)r + 5/2 = 0. Thus

r1,r2 = [-(a-1) ± V (a-1)2-10] /2. In order for solutions

to approach zero as x^0 it is necessary that the real

parts of r1 and r2 be positive. Suppose that a > 1, then

]f (a-1)2-10 is either imaginary or real and less than a - 1; hence the real parts of r1 and r2 will be

negative. Suppose that a = 1, then r1,r2 = ±iV10 and the solutions are oscillatory. Suppose that a < 1, then

\j (a-1)2-10 is either imaginary or real and less than |a-1| = 1 - a; hence the real parts of r1 and r2 will be

positive. Thus if a < 1 the solutions of the D.E. will approach zero as x^0.

21. In all cases the roots of F(r) = 0 are given by Eq.(5) and the forms of the solution are given in Theorem 5.5.1.

21a. The real part of the root must be positive so, from

Eq.(5), a < 0. Also P > 0, since the \j (a-1) 2-4p term must be less than |a-1I.

22. Assume that y = v(x)xr1. Then y' = v(x)rj_xr1 1 + v'(x)xr1 and y" = v(x)r1(r1-1)xr1 2 + 2v'(x)rj_xr1 1 + v"(x)xr1. Substituting in the D.E. and collecting terms yields xr1+2 v" + (a + 2r1)xr1+1 v' + [r1(r1-1) + ar1 + P]xr1 v = 0.

Now we make use of the fact that r1 is a double root of f(r) = r(r-1) + ar + p. This means that f(r1) = 0 and f'(r1) = 2r1 - 1 + a = 0. Hence the D.E. for v reduces

r1+2 r1+1 r1+1

to x 1 v + x 1 v . Since x > 0 we may divide by x 1

to obtain xv" + v' = 0. Thus v(x) = lnx and a second solution is y = xr1lnx.

25. The change of variable x = ez transforms the D.E. into u" - 4u' + 4u = z, which has the solution

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