# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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problems we find tu1(t) + tetu"2(t) = 0 and

56

Section 3 . 7

u1(t) + (t + 1)etu"2 = 2t. [Note that g(t) = 2t, since the D.E. must be put into the form of Eq.(16)]. The solution of these equations gives u1(t) = -2 and u^(t) = 2e-t.

Hence, u1(t) = -2t and u2(t) = -2e-t, and

t -t 2

Y(t) = t(-2t) + te (-2e ) = -2t - 2t. However, since t

is a solution of the homogeneous D.E. we can choose as

2

our particular solution Y(t) = -2t .

18. For this problem, and for many others, it is probably easier to rederive Eqs.(26) without using the explicit form for y1(x) and y2(x) and then to substitute for y1(x) and y2(x) in Eqs.(26). In this case if we take y1 = x-1/2 sinx and y2 = x-1/2cosx, then W(y1,y2) = -1/x.

If the D.E. is put in the form of Eq.(16), then g(x) = 3x-1/2sinx and thus u1 (x) = 3sinxcosx and

u2(x) = -3sin x = 3(-1 + cos2x)/2. Hence 2

u1(x) = (3sin x)/2 and u2(x) = -3x/2 + 3(sin2x)/4, and 2

3 sin x sinx 3x 3 sin2x cosx

Y(x) =--------------— + ( - — + ------)

2

3 sin x sinx 3x 3 sinx cosx cosx

+ ( - — + )

2 Vx 2

3 sinx ^/xcosx

2 V x 2

The first term is a multiple of y1(x) and thus can be neglected for Y(x).

22. Putting limits on the integrals of Eq.(28) and changing the integration variable to s yields

•t y2(s)g(s)ds ft y1(s)g(s)ds

fty2(s)g(s)ds , , ft

Y(t) = -y1(t) I + y2(t) I

1 W(y ,yj(s) 2 Jt,

Jt0 W(y1,y2)(s) JtQ W(y1,y2)(s)

j-t -y1(t)y2(s)g(s)ds (•

Jt0 W(y ,yj(s) Jt

It

t -y1(t)y2(s)g(s)ds ft y2(t)y1(s)g(s)ds

W(y1,y2)(s) W(y1,y2)(s)

t [y1(s)y2(t) - y1(t)y2(s)]g(s)ds

-----------------------------------. To show that

t0

y1(s)y2(s) - y 1(s)y2(s)

Y(t) satisfies L[y] = g(t) we must take the derivative of Y using Leibnitz's rule, which says that if

Y(t) = [tG(t,s)ds, then Y'(t) = G(t,t) + —(t,s)ds.

Jt0 dt

Letting G(t,s) be the above integrand, then G(t,t) = 0

2

4

2

Section 3.8

57

dG y1(s)y2(t) - y^(t)y2(s)

and — = ------------------------------ g(s). Likewise

dt W(y1,y2)(s)

3G(t,t) ft d2G

I

+ (t,s)ds

dt Jt0 at2

J‘ty1(s)y2(t) - y1 (t)y2 (s)

-----------------------------ds.

t0 W(y1,y2)(s)

Since y1 amd y2 are solutions of L[y] = 0, we have L[Y] = g(t) since all the terms involving the integral will add to zero. Clearly Y(t0) = 0 and Y'(t0) = 0.

25. Note that y1 = e1tcosmt and y2 = e1tsinmt and thus 2 It

W(yi,y2) = me . From Problem 22 we then have:

m t Is It ? ^ It ^ Is ?

e cosmse sinmt - e cosmte sinms

Y(t) = g(s)ds

jt0 me2ls

= m"1 I e1(t-s)[cosms sinmt - cosmt sinms]g(s)ds t0

= m-1 J e1(t-s) [sinm(t-s)]g(s)ds.

t0

29. First, we put the D.E. in standard form by dividing by

t2: y" - 2y'/t + 2y/t2 = 4. Assuming that y = tv(t) and

substituting in the D.E. we obtain tv" = 4. Hence v (t) = 4lnt + c2 and v(t) = 4 J lnt dt + c2t =

4(tlnt - t) + c2t. The general solution is

2 2 2 2 c1y1(t) + tv(t) = c1t + 4(t lnt - t ) + c2t . Since -4t

2

is a multiple of y2 = c2t we can write

22 y = c1t + c2t + 4t lnt.

Section 3.8, Page 197

2. From Eq.(15) we have Rcosd = -1, and Rsind = y/13 . Thus

R = V1 + 3 = 2 and 8 = tan-1(-^3) + p = 2p/3 @ 2.09440. Note

that we have to "add" p to the inverse tangent value since 8 must be a second quadrant angle. Thus u = 2cos(t-2p/3).

6. The motion is an undamped free vibration. The units are in the CGS system. The spring constant k = (100 gm)(980cm/sec2)/5cm. Hence the D.E. for the motion is 100u" + [(100 • 980)/5]u = 0 where u is measured in cm and time in sec. We obtain u" + 196u = 0

58

Section 3 . 8

so

u = Acos14t + Bsin14t. The I.C. are u(0) = 0 ^ A = 0 and u'(0) = 10 cm/sec ^ B = 10/14 = 5/7. Hence u(t) = (5/7)sin14t, which first reaches equilibrium when 14t = p, or t = p/14.

8. We use Eq.(33) without R and E(t) (there is no resistor or

impressed voltage) and with L = 1 henry and 1/C = 4x106 since

C = .25x10-6 farads. Thus the I.V.P. is Q" + 4x106 Q = 0,

Q(0) = 10-6 coulombs and Q'(0) = 0.

9. The spring constant is k = (20)(980)/5 = 3920 dyne/cm.

The I.V.P. for the motion is 20u" + 400u' + 3920u = 0 or

u" + 20u' + 196u = 0 and u(0) = 2, u'(0) = 0. Here u is

measured in cm and t in sec. The general solution of the D.E. is u = Ae-10tcos4^6t + Be-10tsin4^6t. The I.C. u(0) = 2 ^ A = 2 and u'(0) = 0 ^ -10A + ^V^B = 0. The

solution is u = e-10t[2cos4 1 + 5(sin4"\/~6t)/V~6 ]cm.

The quasi frequency is m = ^V"6, the quasi period is Td = 2pm = p/2^/6 and Td/T = 7/2 ^/"6 since T = 2p/14 = p/7. To find an upper bound for t, write u in the form of Eq.(30): u(t) = \J 4+25/6 e-10tcos(4^~6 t-S).

Now, since | cos(4^/"6t-8) | ? 1, we have | u(t) | < .05 ^

\J4+25/6 e-10t < .05, which yields t = .4046. A more precise answer can be obtained with a computer algebra system, which in this case yields t = .4045. The original estimate was unusually close for this problem since cos(4\/~6t-8) = -0.9996 for t = .4046.

12. Substituting the given values for L, C and R in Eq.(33),

we obtain the D.E. .2Q" + 3x102 Q' + 105 Q = 0. The I.C.

are Q(0) = 10-6 and Q'(0) = I(0) = 0. Assuming Q = ert, we obtain the roots of the characteristic equation as r1 = -500 and r2 = -1000. Thus Q = c1e-500t + c2e-1000t and hence Q(0) = 10-6 ^ c1 + c2 = 10-6 and Q'(0) = 0 ^ -500c1 - 1000c2 = 0. Solving for c1 and c2 yields the solution.

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