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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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1. First solve the homogeneous D.E. The characteristic
68
Section 4.3
3 2
equation is r- r - r + 1 = 0, and the roots are r = -1, 1, 1; hence yc(t) = c1e-t + c2et + c3tet. Using the superposition principle, we can write a particular solution as the sum of particular solutions corresponding
to the D.E. y'"-y"-y'+y = 2e-t and y/"-y"-y/+y = 3. Our initial choice for a particular solution, Y1, of the
first equation is Ae-t; but e-t is a solution of the homogeneous equation so we multiply by t. Thus,
Y1(t) = Ate-t. For the second equation we choose Y2(t) = B, and there is no need to modify this choice.
The constants are determined by substituting into the individual equations. We obtain A = 1/2, B = 3. Thus, the general solution is
y = c1e-t + c2et + c3tet + 3 + (te-t)/2.
4 2 2 2
5. The characteristic equation is r - 4r = r (r -4) = 0, so yc(t) = c1 + c2t + c3e-2t + c4e2t. For the particular
2
solution correspnding to t we assume 22
Y1 = t (At + Bt + C) and for the particular solution
corresponding to et we assume Y2 = Det. Substituting Y1, in the D.E. yields -48A = 1, B = 0 and 24A-8C = 0 and substituting Y2 yields -3D = 1. Solving for A, B, C and
D gives the desired solution.
9. The characteristic equation for the related homogeneous
D.E. is r3 + 4r = 0 with roots r = 0, +2i, -2i. Hence yc(t) = c1 + c2cos2t + c3sin2t. The initial choice for
Y(t) is At + B, but since B is a solution of the homogeneous equation we must multiply by t and assume Y(t) = t(At+B). A and B are found by substituting in the D.E., which gives A = 1/8, B = 0, and thus the general solution is y(t) = c1 + c2cos2t + c3sin2t + (1/8)t . Applying the I.C. we have y(0) = 0 ^ c1 + c2 = 0, y'(0) = 0 ^ 2c3 = 0, and y"(0) = 1 ^ -4c2 + 1/4 = 1, which have the solution c1 = 3/16, c2 = -3/16, c3 = 0.
For small t the graph will approximate 3(1ócos2t)/16 and
for large t it will be approximated by t /8.
13. The characteristic equation for the homogeneous D.E. is
32
r - 2r + r = 0 with roots r = 0,1,1. Hence the complementary solution is yc(t) = c1 + c2et + c3tet. We
Section 4.3
69
consider the differential equations y'" - 2y" + y' = t3
and y'" - 2y" + y' = 2et separately. Our initial choice for a particular solution, Y1, of the first equation is
3 2
A0t + A1t + A2t + A3; but since a constant is a solution of the homogeneous equation we must multiply by t. Thus
32
Y1(t) = t(A0t + A1t + A2t + A3). For the second equation
we first choose Y2(t) = Bet, but since both et and tet are solutions of the homogeneous equation, we multiply by
2 2 t
t to obtain Y2(t) = Bt e . Then Y(t) = Y1(t) + Y2(t) by the superposition principle and y(t) = yc(t) + Y(t).
17. The complementary solution is yc(t) = c1 + c2e-t + c3et + c4tet. The superposition principle allows us to consider separately the D.E. ylv - y'" - y" +y' = t2 + 4 and
ylv - y'" - y" + y' = tsint. For the first equation our
2
initial choice is Y1(t) = A0t + A1t + A2; but this must
be multiplied by t since a constant is a solution of the
2
homogeneous D.E. Hence Y1(t) = t(A0t + A1t + A2). For
the second equation our initial choice that
Y2 = (B0t + B1)cost + + (C0t + C1)sint does not need to be
modified. Hence 2
Y(t) = t(A0t + A1t + A2) + (B0t + B1)cost + (C0t + C1)sint.
2
20. (D-a)(D-b)f = (D-a)(Df-bf) = D f - (a+b)Df + abf and
2
(D-b)(D-a)f = (D-b)(Df-af) = D f - (b+a)Df + baf. Since
a+b = b+a and ab = ba, we find the given equation holds
for any function f.
22a. The D.E. of Problem 13 can be written as
D(D-1)2y = t3 + 2et. Since D4 annihilates t3 and (D-1)
annihilates 2et, we have D5(D-1) 3y = 0, which corresponds to Eq.(ii) of Problem 21. The solution of this equation
432
is y(x) = A1t + A2t + A3t + A4t + A5 +
2 -t -t
(B1t + B2t + B3)e . Since A5 + (B2t + B3)e are
solutions of the homogeneous equation related to the original D.E., they may be deleted and thus
4 3 2 2 -t
Y(t) = A1t4 + A2t3 + A3t2 + A4t + B1t2e .
70
Section 4.4
2 2
22b. (D+1) (D +1) annihilates the right side of the D.E. of Problem 14.
3 2 2
22e. D (D +1) annihilates the right side of the D.E. of Problem 17.
Section 4.4, Page 229
1. The complementary solution is yc = c1 + c2cost + c3sint and thus we assume a particular solution of the form
Y = u1(t) + u2(t)cost + u3(t)sint. Differentiating and
assuming Eq.(5), we obtain Y' = -u2sint + u3cost and
/ / /
uL + u2cost + u3sint = 0 (a).
Continuing this process we obtain Y" = -u2cost - u3sint,
... / /
Y = u2sint - u3cost - u2cost - u3sint and
-u^sint + u^cost = 0 (b).
Substituting Y and its derivatives, as given above, into
the D.E. we obtain the third equation:
-u^cost - u^sint = tant (c).
Equations (a), (b) and (c) constitute Eqs.(10) of the
text for this problem and may be solved to give
2
u-l = tant, u2 = -sint, and u3 = -sin t/cost. Thus
uL = -lncost, u2 = cost and u3 = sint - ln(sect + tant) and Y = -lncost + 1 - (sint)ln(sect + tant). Note that the constant 1 can be absorbed in cL.
4. Replace tant in Eq. (c) of Prob. 1 by sect and use Eqs. (a) and (b) as in Prob. 1 to obtain ul = sect, u'2 = -1 and u'3 = -sint/cost.
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