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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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Q-CV CR
simplifying we get Q = De-t/CR + CV. Q(0) = 0 ^ D = -CV and thus Q(t) = CV(1 - e-t/CR).
13b. limQ(t) = CV since lim e t/CR = 0.
t—t—
dQ Q
13c. In this case R— + — = 0, Q(tO = CV. The solution of this
dt C 1
D.E. is Q(t) = Ee-t/CR, so Q(ti) = Ee-t1/CR = CV, or E = CVet1/CR. Thus Q(t) = CVet1/CRe-t/CR = CVe"(t"t1>/CR.
4
Section 1.3
Section 1.3, Page 22
2. The D.E. is second order since there is a second
derivative of y appearing in the equation. The equation
2 2 is nonlinear due to the y term (as well as due to the y
term multiplying the y" term).
6. This is a third order D.E. since the highest derivative is y''' and it is linear since y and all its derivatives
22
appear to the first power only. The terms t and cos t do not affect the linearity of the D.E.
-3t ' -3t -3t
8. For y (t) = e we have y (t) = -3e and y" (t) = 9e .
1 11
Substitution of these into the D.E. yields
-3t -3t -3t -3t
9e + 2(-3e ) - 3(e ) = (9-6-3)e = 0.
14. Recall that if u(t) = I f(s)ds, then u'(t) = f(t).
Jo
rt
16. Differentiating e twice and substituting into the D.E.
2 rt rt 2 rt rt
yields re - e = (r -1)e . If y = e is to be a solution of the D.E. then the last quantity must be zero 2 rt
for all t. Thus r -1 = 0 since e is never zero.
r
19. Differentiating t twice and substituting into the D.E.
2 r-2 r-1 r 2 r
yields t [r(r-1)t ] + 4t[rt ] + 2t = [r +3r+2]t . If
r
y = t is to be a solution of the D.E., then the last term
2
must be zero for all t and thus r + 3r + 2 = 0.
22. The D.E. is second order since there are second partial derivatives of u(x,y) appearing. The D.E. is nonlinear due to the product of u(x,y) times ux(or uy).
5
du1 2 -a2t d2u1 -a2t
26. Since --------- = -a e sinx and ---------------- = -e sinx we have
dt 3x2
22 2 -a2t 2 -a2t
a [-e sinx] = -a e sinx, which is true for all t and x.
6
CHAPTER 2
Section 2.1, Page 38
— (ye3t) = te3t + et. Integration of both sides yields
1. m(t) = exp(J3dt) = e3t. Thus e3t(y'+3y) = e3t(t+e 2t) or
d dt
11
ye3t = —te3t - —e3t + et + c, and division by e3t gives 3 9
the general solution. Note that Jte3tdt is evaluated by integration by parts, with u = t and dv = e3tdt.
2. m(t) = e-2t. 3. m(t) = et.
4. m(t) = exp(J —) = elnt = t, so (ty)' = 3tcos2t, and
t
integration by parts yields the general solution.
6. The equation must be divided by t so that it is in the
form of Eq.(3): y' + (2/t)y = (sint)/t. Thus
r2dt 2 2
m(t) = exp(J = t , and (t y) = tsint. Integration
t
then yields t y = -tcost + sint + c.
t2 r4tdt 2 2
7. m(t) = e . 8. m(t) = exp(J ) = (1+t ) .
1+t2
11. m(t) = et so (ety)' = 5etsin2t. To integrate the right
side you can integrate by parts (twice), use an integral table, or use a symbolic computational software program to find ety = et(sin2t - 2cos2t) + c.
13. m(t) = e-t and y = 2(t-1)e2t + cet. To find the value
for c, set t = 0 in y and equate to 1, the initial value
of y. Thus -2+c = 1 and c = 3, which yields the solution
of the given initial value problem.
15. m(t) = exp(J dt) = t2 and y = t2/4 - t/3 + 1/2 + c/t2.
t
Setting t = 1 and y = 1/2 we have c = 1/12.
18. m(t) = t2. Thus (t2y)' = tsint and
t2y = -tcost + sint + c. Setting t = p/2 and
y = 1 yields c = p2/4 - 1.
20. m(t) = tet.
Section 2.1
7
21b. m(t) = e-t/2 so (e-t/2y) = 2e-t/2cost. Integrating (see comments in #11) and dividing by e-t/2 yields
4 8 t/2 4
y(t) = cost + —sint + cet/2. Thus y(0) = + c = a,
5 5 5
4 4 8 4 t/2
or c = a + — and y(t) = cost + —sint + (a + —)et/2.
5 5 5 5
4
If (a + —) = 0, then the solution is oscillatory for all
5
t, while if (a + — ) n 0, the solution is unbounded as 5
t ? •. Thus a
0
21a.
4
5
24a.
24b. m(t) = exp
-cost
y(t) :
4c
f 2dt 2
J — = t2, so
(t2y)' = sint and
t
2
c p
+ — . Setting t = yields
t
2
ap2 ap2/4 - cost
— = a or c = -------------- and hence y(t) = , which
p2 4 t2
is unbounded as t ^ 0 unless ap 2/4 = 1 or a0 = 4/p2.
2
24c. For a = 4/p2 y(t) =
1 - cost
. To find the limit as
t2
t ? 0 L'Hopital's Rule must be used:
sint cost 1
limy(t) = lim = lim = — .
t?0 t?0 2t t?0 2 2
28. (e ty)/ = e t + 3e tsint so
-t
e y = -e
-t -t sint+cost
- 3e (-------------
2
) + c or
3 -t t
y(t) = -1 - ( )e (sint+cost) + ce .
2
If y(t) is to remain bounded, we must have c = 0. Thus
3 5 5
y(0) = -1 - — + c = y0 or c = y0 + — = 0 and y0 = .
2 0 0 2 0 2
8
Sect ion
2 . 2
30. m(t) = eat so the D.E. can be written as
(eaty)' = beate-1t = be(a-1)t. If a n l, then integration and solution for y yields y = [b/(a-1)]e-1t + ce-at. Then
lim y is zero since both l and a are positive numbers.
x——•
If a = l, then the D.E. becomes (eaty)' = b, which yields y = (bt+c)/e1t as the solution. L'Hopital's Rule gives
(bt+c) b
lim y = lim ----------- = lim —— = 0.
t——• t——• e t——• le
32. There is no unique answer for this situation. One
—2t
possible response is to assume y(t) = ce + 3 — t, then y'(t) = —2ce—2t — 1 and thus y' + 2y = 5 — 2t.
35. This problem demonstrates the central idea of the method of variation of parameters for the simplest case. The solution (ii) of the homogeneous D.E. is extended to the corresponding nonhomogeneous D.E. by replacing the constant A by a function A(t), as shown in (iii).
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