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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
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This is a first order linear D.E. for v with integrating
50
Section 3 . 5
factor m(x) = exp[J (1-N/x)dx] = x Nex. Hence
, -N x /, / _ , / N -x , . ,
(x e v ) = 0, and v = cx e which gives v(x) = cJxNe xdx + k. On taking k = 0 we obtain as the second solution y2(x) = cexJ xNe-xdx. The integral can be evaluated by using the method of integration by parts.
At each stage let u = xN or xN 1, or whatever the power of x that remains, and let dv = e x. Note that this dv is not related to the v(x) in y2(x). For N = 2 we have
-x -x
y2(x) = cexJ x2e-xdx = cex[x2—— - J 2x——dx] 2 -1 -1
-x -x
2 x e c e
= -cx + ce [2x - J 2 dx]
-1 -1
22 = c(-x - 2x - 2) = -2c(1 + x + x /2!).
Choosing c = -1/2! gives the desired result. For the
general case c = - 1/N!
33. (y2/y1)' = (y1y,2 - y1y2)/y12 = W(y1,y2)/y12. Abel's identity
is W(y1,y2) = c exp[-l tp(r)dr]. Hence
Jt0
(y2/y1)' = cy1-2 exp[-lt p(r)dr]. Integrating and setting
c = 1 (since a solution y2 can be multiplied by any constant) and taking the constant of integration to be zero we obtain
exp[-J sp(r)dr]
y2(t) = y1(t) Jt----------—---------- ds.
[y1(s) ] 2
35. From Problem 33 and Abel's formula we have
y2 exp[J (1/t)dt] elnt 2 2
(—) = = = tcsc (t ). Thus
y ? 2/,2\ i 2 / , 2»
y1 sin (t ) sin (t )
22 y2/y1 = -(1/2)cot(t ) and hence we can choose y2 = cos(t )
22
since y1 = sin (t ).
r t r t
The general solution of the D.E. is y = c1e 1 + c2e 2
V2 2
b -4ac )/2a provided b - 4ac n 0.
2
In this case there are two possibilities. If b - 4ac > 0 2 1/2
then (b - 4ac) < b and r1 and r2 are real and
r t r t
negative. Consequently e 1 ^ 0 and e 2 ^ 0; and hence
Section 3.6 51
2
y ^ 0, as t ^ • . If b - 4ac < 0 then r1 and r2 are complex conjugates with negative real part. Again
r t r t
e 1 ^ 0 and e 2 ^ 0; and hence y ^ 0, as t ^ • .
2 r t r t
Finally, if b - 4ac = 0, then y = c1e 1 + c2te 1 where
r1 = -b/2a < 0. Hence, again y ^ 0 as t ^ • . This
conclusion does not hold if either b = 0 (since y(t) = c1coswt + c2sinwt) or c = 0 (since y^t) = c1).
42. Substituting z = lnt into the D.E. gives
d2y dy
+ — + 0.25y = 0, which has the solution
dz2 dz
I \ -z/2 -z/2 . ... 4.-1/2 4.-1/2-, 4_
y(z) = c1e + c2ze so that y(t) = c1t + c2t lnt
Section 3.6, Page 178
1. First we find the solution of the homogeneous D.E., which
2
has the characteristic equation r -2r-3 = (r-3)(r+1) = 0.
3t -t 2t
Hence yc = c1e + c2e and we can assume Y = Ae for the
2t 2t
particular solution. Thus Y = 2Ae and Y" = 4Ae and substituting into the D.E. yields
2t 2t 2t 2t
4Ae - 2(2Ae ) - 3(Ae ) = 3e . Thus -3A = 3 and
3t -t 2t
A = -1, yielding y = c e + c e - e .
12
4. Initially we assume Y = A + Bsin2t + Ccos2t. However,
since a constant is a solution of the related homogeneous
D.E. we must modify Y by multiplying the constant A by t
and thus the correct form is Y = At + Bsin2t + Ccos2t.
-t -t 2 -t
6. Since yc = c1e + c2te we must assume Y = At e , so
-t 2 -t -t -t 2 -t
that Y = 2Ate - At e and y" = 2Ae - 4Ate + At e .
2 -t
Substituting in the D.E. gives (At -4At+2A)e +
2 -t 2 -t -t
2(-At +2At)e + At e = 2e . Notice that all terms on
2
the left involving t and t add to zero and we are left
-t -t 2 -t
with 2A = 2, or A = 1. Hence y = c1e + c2te + t e .
8. The assumed form is Y = (At + B)sin2t + (Ct + D)cos2t, which is appropriate for both terms appearing on the right side of the D.E. Since none of the terms appearing
52
Section 3 . 6
in Y are solutions of the homogeneous equation, we do not need to modify Y.
rt
11. First solve the homogeneous D.E. Substituting y = e
gives r2 + r + 4 = 0. Hence yc = e [c1cos^ 15 t/2) +
I t -t
c2sin(y 15 t/2)]. We replace sinht by (e - e )/2 and
t -t t -t
then assume Y(t) = Ae + Be . Since neither e nor e are solutions of the homogeneous equation, there is no need to modify our assumption for Y. Substituting in the
t -t t -t
D.E., we obtain 6Ae + 4Be = e - e . Hence, A = 1/6
and B = -1/4. The general solution is
-t/2 ,--- , t -t
y = e [c1cos^ 15 t/2) + c2sin(y 15 t/2)] + e /6 - e /4.
[For this problem we could also have found a particular solution as a linear combination of sinht and cosht:
Y(t) = Acosht + Bsinht. Substituting this in the D.E. gives (5A + B)cosht + (A + 5B)sinht = 2sinht. The solution is A = -1/12 and B = 5/12. A simple calculation
t -t
shows that -(1/12)cosht + (5/12)sinht = e /6 - e /4.]
-2t t
13. yc = c1e + c2e so for the particular solution we
assume Y = At + B. Since neither At or B are solutions of the homogeneous equation it is not necessary to modify the original assumption. Substituting Y in the D.E. we obtain 0 + A -2(At+B) = 2t or -2A = 2 and A-2B = 0.
-2t t
Solving for A and B we obtain y = c1e + c2e - t - 1/2
as the general solution. y(0) = 0 ^ c1 + c2 - 1/2 = 0
and y'(0) = 1 ^ -2c1 + c2 - 1 = 1, which yield c1 = -1/2
t -2t
and c2 = 1. Thus y = e - (1/2)e - t - 1/2.
16. Since the nonhomogeneous term is the product of a linear polynomial and an exponential, assume Y of the same form:
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