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# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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Substituting in the D.E. we have
2
P(1/X)[X4dy + 2X3~y ] + Q(1/X)[-X^^] + R(1/X)y = 0,
dX2 dX dX
X4P(1/X)^-y + [2X3p(1/X) - X2Q(1/X)]^y + R(1/X)y = 0. dX2 dX
The result then follows from the theory of singular points at X = 0.
oo
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/S)/S2]/P(1/S) = 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 Sf(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 - è ,
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 ³ -1/2 ³ -.1-3/2 -
y = c1|x+1| + c2|x+1| , x Ô -1.
r2
4. If y = x then r(r-1) + 3r + 5 = 0. So r + 2r + 5 = 0
and r = (-2 ± ä/4-20 )/2 = -1 ± 2i. Thus the general
solution of the D.E. is
y = c1x-1cos(2ln|x|) + c2x-1sin(2ln|x|), x Ô 0.
9. Again let y = xr to obtain r(r-1) - 5r + 9 = 0, or
2
(r-3) = 0. Thus the roots are x = 3,3 and
y = cj_x3 + c2x3ln|x|, x Ô 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 + c2x-1 and
3 1/2 -2
y = — c^x - c,x . Setting x = 1 in y and y we obtain
2 1 2
3
c1 + 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
ri,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 â > 0, since the \j (a-1) 2-40 term must be less than |a-1|.
22. Assume that y = v(x)xr1. Then y' = v(x)r1xr1 1 + v'(x)xr1
and y" = v(x)r1(r1-1)xr1 2 + 2v'(x)r1xr1 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
r,+2 n r,+1 / 0 „ , . . , , r-,+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
88
Section 5.6
2z 2z
u(z) = c1e + c2ze + (1/4)z + 1/4. Hence 22
y(x) = c^x + c2x lnx + (1/4)lnx + 1/4.
31. If x > 0, then |x| = x and |x|r1 = xr1 so we can choose c1 = k1. If x < 0, then |x| = -x and
ri ' --4 ri ' -4 r1-ri ------- ----- = (-1) ããö,
|x| 1 = (-x) 1 = (-1) 1x 1 and we can choose c1 = (-1) 1k1,
or k1 = (-1) r1c1 = (-1) r1c1. In both cases we have
c2 = k2.
Section 5.6, Page 271
2. If the D.E. is put in the standard form y" + p(x)y +
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