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

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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 tςε.

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. tt

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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