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

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-16L2

--------- + LtanA + 3 > H.

22 u cos A

-161.98 sinA

31e. Setting L = 350 and H = 10 we get --------------- + 350------ > 7

cos2A cosA

2

or 7cos A - 350cosAsinA + 161.98 < 0. This can be solved numerically or by plotting the left side as a function of A and finding where the zero crossings are.

31f. Setting L = 350, and H = 10 in the answer to part d

18

Section 2.4

2

-16(350) 2

yields ------------ + 350tanA = 7, where we have chosen the

22 u cos A

equality sign since we want to just clear the wall.

2 2 1,960,000

Solving for u we get u = ----------------------. Now u will

175sin2A-7cos2A

have a minimum when the denominator has a maximum. Thus

350cos2A + 7sin2A = 0, or tan2A = -50, which yields

A = .7954 rad. and u = 106.89 ft./sec.

Section 2.4, Page 72

I. If the equation is written in the form of Eq.(1), then p(t) = (lnt)/(t-3) and g(t) = 2t/(t-3). These are defined and continuous on the intervals (0,3) and (3,<^>), but since the initial point is t = 1, the solution will be continuous on 0 < t < 3.

2

4. p(t) = 2t/(2-t)(2+t) and g(t) = 3t /(2-t)(2+t).

8. Theorem 2.4.2 guarantees a unique solution to the D.E.

22

through any point (t0,y0) such that t0 + y0 < 1 since df 2 2 1/2

— = -y(1-t -y ) is defined and continuous only for dy

1-t2-y2 > 0. Note also that f = (1-t2-y2)1/2 is defined

and continuous in this region as well as on the boundary 22

t +y = 1. The boundary can't be included in the final

df

region due to the discontinuity of — there.

dy

1+t2 df 1+t2 1+t2

II. In this case f = ----------- and — = ---------- - ---------,

y(3-y) dy y(3-y)2 y2 (3-y)

which are both continuous everywhere except for y = 0 and y = 3.

13. The D.E. may be written as ydy = -4tdt so that 2

y 2 2 2

— = -2t +c, or y = c-4t . The I.C. then yields 2

2 2 2 2 2 2 y0 = c, so that y = y0 - 4t or y = ±v y0-4t , which is

defined for 4t2 < y0 or |t| < |y0|/2. Note that y0 Ï 0 since Theorem 2.4.2 does not hold there.

17. From the direction field and the given D.E. it is noted

that for t > 0 and y < 0 that y' < 0, so y ^ -• for

y0 < 0. Likewise, for 0 < y0 < 3, y' > 0 and y' ^ 0 as

Section 2.4

19

y ^ 3, so y ^ 3 for 0 < y0 < 3 and for y0 > 3, y' < 0 and again y' ^ 0 as y^ 3, so y^3 for y0 > 3. For y0 = 3, y' = 0 and y = 3 for all t and for y0 = 0, y' = 0 and y = 0 for all t.

22a. For y1 = 1-t, y1 = -1 =

-t+[t2+4(1-t) ] 1/2 2

-t+[(t-2)2] 1/2 2

-t+It-2I

= -1 if 2

(t-2) > 0, by the definition of absolute value. Setting t = 2 in y1 we get Ó³(2) = -1, as required.

22b. By Theorem 2.4.2 we are guaranteed a unique solution only

-t+(t2+4y)1/2 2 _!/2

where f(t,y) = ---------------- and fy(t,y) = (t +4y) ' are

2 y

continuous. In this case the initial point (2,-1) lies

2 df in the region t + 4y < 0, in which case — is not

Ýó

continuous and hence the theorem is not applicable and there is no contradiction.

22

22c. If ó = y2(t) then we must have ct + c = -t /4, which is

not possible since c must be a constant.

23a. To show that f(t) = e2t is a solution of the D.E., take

its derivative and substitute into the D.E.

24. [c f(t)]' + p(t)[cf(t)] = c[f'(t) + p(t)f(t)] = 0 since f(t) satisfies the given D.E.

25. [y1(t) + Ó2(t)]' + p(t)[y1(t) + Ó2(t)] =

y1(t) + p(t)y^(t) + ó2 (t) + p(t)y2(t) = 0 + g(t).

27a. For n = 0,1, the D.E. is linear and Eqs.(3) and (4) apply.

_ T ^ 1-n _ dv , -ndy dy 1 ndv

27b. Let v = ó then -------- = (1-n)y --- so --------- = -ó ---,

dt dt dt 1-n dt

which makes sense when n Ï 0,1. Substituting into the n

y dv n

D.E. yields ------ ---- + p(t)y = q(t)y or

1-n dt

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