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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.
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2
dy 8 + 4x
 = ------------, so the minimum value is attained at
dx (2-4x-x2) 2
x = -2. Note that the solution is defined for -2 - yf! < x < -2 + \[~6 (by finding the zeros of the denominator) and has vertical asymptotes at the end points of the interval.
2
25. Separating variables and integrating yields 3y + y = sin2x + c. y(0) = -1 gives c = -2 so that
2
y + 3y + (2-sin2x) = 0. The quadratic formula then
gives y = --- + \Jsin2x + 1/4 , which is defined for
-.126 < x < 1.697 (found by solving sin2x = -.25 for x and noting x = 0 is the initial point). Thus we have dy cos2x
 = ------------, which yields x = p/4 as the only
dx 1
(sin2x+)
4
critical point in the above interval. Using the second derivative test or graphing the solution clearly indicates the critical point is a maximum.
27a. By sketching the direction field and using the D.E.we
note that y' < 0 for y > 4 and y' approaches zero as y
approaches 4. For 0 < y < 4, y' > 0 and again approaches
zero as y approaches 4. Thus limy = 4 if y0 > 0. For
t
y0 < 0, y' < 0 for all y and hence y becomes negatively unbounded (-) as t increases. If y0 = 0, then y' = 0
for all t, so y = 0 for all t.
27b. Separating variables and using a partial fraction
1 1 4
expansion we have ( - ------)dy =  tdt. Hence
y y-4 3
ln | y | =  t2 + c1 and thus | | = ec1e2t2/3 = ce2t2/3,
y-4 3 1 y-4
where c is positive. For y0 = .5 this becomes
y 2t2/3 .5 1
---- = ce and thus c = ---------- =  . Using this value
4 -y 3.5 7
12
Section 2.2
for c and solving for y yields y(t) = ----------------------.
1 + 7e2t2/3
Setting this equal to 3.98 and solving for t yields t = 3.29527.
cy+d
29. Separating variables yields ----------- dy = dx. If a Ď 0 and
ay+b
ay+b Ď 0 then dx = ( +-------------)dy. Integration then
a a(ay+b)
dy dv
3 0c. If v = y/x then y = vx and  = v + x-------------- and thus the
dx dx
dv v-4
D.E. becomes v + x = -----------. Subtracting v from both
dx 1-v
dv v2-4
sides yields x =
dx 1-v
1-v 1
30d. The last equation in (c) separates into ------dv = dx. To
v2-4 x
integrate the left side use partial fractions to write
1-v A B ,
--- = ---- + ----, which yields A = -1/4 and B = -3/4.
v-4 v-2 v+2
13
Integration then gives - ln|v-2| - ln|v+2| = ln|x| - k,
44
ln|x4||v-2||v+2|3 = 4k after manipulations using properties of the ln function.
31a. Simplifying the right side of the D.E. gives
dy/dx = 1 + (y/x) + (y/x) so the equation is homogeneous.
31b. The substitution y = vx leads to
dv 2 dv dx
v + x = 1 + v + v or --------- = ----------. Solving, we get
dx 1 + v2 x
arctanv = ln|x| + c. Substituting for v we obtain arctan(y/x) - ln|x| = c.
33b. Dividing the numerator and denominator of the right side
dv 4v - 3
by x and substituting y = vx we get v + x = ----------
dx 2-v
2
dv v2 + 2v - 3
which can be rewritten as x = ---------------. Note that
dx 2 - v
v = 3 and v = 1 are solutions of this equation. For
v Ď 1, 3 separating variables gives
or
Section 2.3
13
2 - v 1
------------ dv = dx. Applying a partial fraction
(v+3)(v-1) x
decomposition to the left side we obtain
1 1 5 1 dx
[----------------]dv = , and upon integrating both sides
4 v-1 4 v+3 x
15 we find that ln|v-1| - ln|v+3| = ln|x| + c.
44
Substituting for v and performing some algebraic
manipulations we get the solution in the implicit form
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