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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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x (y-iv)" + x(y-,v)' + (x - —) y-,v = 0. On carrying out the
1 1 4 1
differentiations and making use of the fact that y1 is a solution, we obtain x2y1v" + (2x2y1 + xy1)v' = 0. This is a first order linear equation for v ,
v" + (2y1/y1 + 1/x)v' = 0, with solution
/
y1 1
v (x) = cexp[-J (2— + — )dx] = cexp[-2lny1 - lnx]
y1 x 1
1c 2
= c---- = ----------------------- = c csc x,
2 ,-1.2. xy1 x(x sin x)
where c is an arbitrary constant, which we will take to be
one. Then v(x) = J csc x dx = -cotx + k where again k is an
arbitrary constant which can be taken equal to zero. Thus
-1/2 -1/2 y2(x) = y1(x)v(x) = (x sinx)(-cotx) = -x cosx. The
-1/2
second solution is usually taken to be x cosx. Note that c = -1 would have given this solution.
31b. Let y2(x) = exv(x), then y2 = exv' + exv, and
y"2 = exv" + 2exv + exv. Substituting in the D.E. we
obtain xexv" + (xex-Nex)v' = 0, or v" + (1-N/x)v' = 0.
This is a first order linear D.E. for v with integrating
50
Section 3 . 5
factor m(x) = exp[J (1-N/x)dx] = x Nex. Hence
, -N x /, / _ , / N -x , . ,
(x e v ) = 0, and v = cx e which gives
v(x) = cJxNe xdx + k. On taking k = 0 we obtain as the second solution y2(x) = cexJ xNe-xdx. The integral can be evaluated by using the method of integration by parts.
At each stage let u = xN or xN —, or whatever the power of x that remains, and let dv = e x. Note that this dv is not related to the v(x) in y2(x). For N = 2 we have
-x -x
y2(x) = cexJ x2e-xdx = cex[x2—— - Ã 2x——dx] 2 -1 -1
-x -x
2 x e ñ e
= -cx + ce [2x-------- - J 2---dx]
-1 -1
22 = c(-x - 2x - 2) = -2c(1 + x + x /2!).
Choosing c = -1/2! gives the desired result. For the
general case c = - 1/N!
33. (y2/y—)' = (y—y'2 - ó1ó2)/ó12 = W(y—,y2)/y—2. Abel's identity
is W(y—,y2) = c exp[-l tp(r)dr]. Hence
Jt0
(y2/y—)' = cy—-2 exp[-lt p(r)dr]. Integrating and setting
t0
c = 1 (since a solution y2 can be multiplied by any constant) and taking the constant of integration to be zero we obtain
exp[-J sp(r)dr]
y2(t) = y—(t) Jt------------------------—- ds.
[y—(s) ] 2
35. From Problem 33 and Abel's formula we have
y2 exp[J (1/t)dt] elnt 2 2
(—) = ---------------- = ---------- = tcsc (t ). Thus
y ¦ 2 ³ .2 \ ³ 2 / , 2 \
y1 sin (t ) sin (t )
22 y2/y— = -(1/2)cot(t ) and hence we can choose y2 = cos(t )
22
since y1 = sin (t ).
r t r t
The general solution of the D.E. is y = c—e 1 + c2e 2
V2 2
b -4ac )/2a provided b - 4ac Ï 0.
2
In this case there are two possibilities. If b - 4ac > 0 2 1/2
then (b - 4ac) < b and r— and r2 are real and
rt r t
negative. Consequently e 1 ^ 0 and e 2 ^ 0; and hence
Section 3.6
51
2
y ^ 0, as t ^ • . If b - 4ac < 0 then r1 and r2 are complex conjugates with negative real part. Again
r t r t
e 1 ^ 0 and e 2 ^ 0; and hence y ^ 0, as t ^ • .
2 r t r t
Finally, if b - 4ac = 0, then y = c1e 1 + c2te 1 where
r1 = -b/2a < 0. Hence, again y ^ 0 as t ^ • . This
conclusion does not hold if either b = 0 (since y(t) = c1coswt + c2sinwt) or c = 0 (since y^t) = c1).
42. Substituting z = lnt into the D.E. gives
d2y dy
--- + — + 0.25y = 0, which has the solution
dz2 dz
I \ -z/2 -z/2 . ... 4.-1/2 4.-1/2-,
y(z) = c1e + c2ze so that y(t) = c1t + c2t lnt
Section 3.6, Page 178
1. First we find the solution of the homogeneous D.E., which
2
has the characteristic equation r -2r-3 = (r-3)(r+1) = 0. 3t -t 2t Hence yc = c1e + c2e and we can assume Y = Ae for the
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