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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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\2/3
v4pz
V
2/3
dv
2/3
Thus ------ = - kV , for k > 0.
dt
21.
22.
24.
Section 1.2 Page 14
2
3
1b. dy/dt = -2y+5 can be rewritten as ------------- = -2dt. Thus
y-5/2
ln I y-5/2 I = -2t + c1, or y-5/2 = ce-2t. y(0) = y0 yields
c = y0 - 5/2, so y = 5/2 + (y0-5/2)e-2t.
If y0 > 5/2, the solution starts above the equilibrium solution and decreases exponentially and approaches 5/2 as t^ro. Conversely, if y < 5/2, the solution starts below 5/2 and grows exponentially and approaches 5/2 from
below as tςε.
3
dy/dt ³ ³
4a. Rewrite Eq.(ii) as ------------- = a and thus ln | y | = at+ci; or
y
y = ceat.
dy dyi
4b. If y = y1(t) + k, then --------- = ----. Substituting both
dt dt
dyi
these into Eq.(i) we get --- = a(y1+k) - b. Since
dt
dy1
--- = ay1, this leaves ak - b = 0 and thus k = b/a.
dt
Hence y = y1(t) + b/a is the solution to Eq(i).
6b. From Eq.(11) we have p = 900 + cet/2. If p(0) = p0, then
c = p0 - 900 and thus p = 900 + (p0 - 900)et/2. If p0 < 900, this decreases, so if we set p = 0 and solve for T (the time of extinction) we obtain eT/2 = 900/(900-p0), or T = 2ln[900/(900-p0)].
8a. Use Eq.(26).
8b. Use Eq.(29).
dQ dQ/dt ³ ³
10a.  = -rQ yields ------------ = -r, or ln | Q | = -rt + c1. Thus
dt Q 1 Q = ce-rt and Q(0) = 100 yields c = 100. Hence Q = 100e-rt. Setting t = 1, we have 82.04 = 100e-r, which yields r = .1980/wk or r = .02828/day.
dQ/dt -1
13a. Rewrite the D.E. as ----------- = , thus upon integrating and
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(t1) = CV. The solution of this
dt C 1
D.E. is Q(t) = Ee-t/CR, so Q(t1) = 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 tf(s)ds, then u'(t) = f(t).
J0
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 32u1 -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
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