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

**Download**(direct link)

**:**

**414**> 415 416 417 418 419 420 .. 609 >> Next

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 + 7e2t2/3

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

c ad-bc

ay+b Ď 0 then dx = ( +-------------)dy. Integration then

a a(ay+b)

yields the desired answer.

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

**414**> 415 416 417 418 419 420 .. 609 >> Next