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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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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) +
³ 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(ó 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: 2t
Y = (At+B)e . Thus Y' = Ae2t + 2(At+B)e2t and
Y" = 4Ae2t+4(At+B)e2t. Substituting into the D.E. we find -3At = 3t and 2A - 3B = 0, yielding A = -1 and B = -2/3. Since the characteristic equation is
2
r - 2r - 3 = 0, the general solution is
3t -t 2 2t 2t y = c.e +c2e - —e - te .
1 2 3
Section 3.6
53
-3t
19a. The solution of the homogeneous D.E. is yc = c—e + c2.
4
After inspection of the nonhomogeneous term, for 2t we
2 -3t
must assume a fourth order polynominial, for t e we must assume a quadratic polynomial times the exponential, and for sin3t we must assume Csin3t + Dcos3t. Thus
4 3 2 2 -3t
Y(t) = (A0t +A—t +A2t +A3t+A4) + (B0t +B—t+B2)e +Csin3t+Dcos3t.
However,since e-3t and a constant are solutions of the
homogeneous D.E., we must multiply the coefficient of e 3t and the polynomial by t. The correct form is
432
Y(t) = t(A0t + A—t + A2t + A3t + A4) +
2 -3t
t(B0t + B—t + B2)e + Csin3t + Dcos3t.
-t
22a. The solution of the homgeneous D.E. is yc = e [c—cost + c2sint]. After inspection of the nonhomogeneous term, we
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