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

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-t 2 -t 2 assume Y(t) = Ae + (B0t + B—t + B2)e cost + (C0t + C—t

-t -t -t + C2)e sint. Since e cost and e sint are solutions of

the homogeneous D.E., it is necessary to multiply both the last two terms by t. Hence the correct form is -t 2 -t

Y(t) = Ae + t(B0t + B—t + B2)e cost +

2 -t t(C0t + C—t + C2)e sint.

28. First solve the I.V.P. y" + y = t, y(0) = 0, y'(0) = 1 for 0 < t < p. The solution of the homogeneous D.E. is yc(t) = c—cost + c2sint. The correct form for Y(t) is

y(t) = A0t + A—. Substituting in the D.E. we find A0 = 1

and A— = 0. Hence, y = c—cost + c2sint + t. Applying the

I.C., we obtain y = t. For t > p we have y" + y = peP t

P—t

so the form for Y(t) is Y(t) = Ee . Substituting Y(t)

P P P

in the D.E., we obtain Ee + Ee = pe so E = p/2.

Hence the general solution for t > p is Y = D—cost +

D2sint + (p/2)ep-t. If y and y are to be continuous at

t = p, then the solutions and their derivatives for t < p

and t > p must have the same value at t = p. These conditions require p = -D— + p/2 and 1 = -D2 - p/2.

Hence D— = -p/2, D2 = -(1 + p/2), and

54

Section 3 . 7

I t, 0 ? t ? p

y = f(t) = \ p-t

[ -(p/2)cost - (1 + p/2)sint + (p/2)e , t > p.

The graphs of the nonhomogeneous term and f follow.

30. According to Theorem 3.6.1, the difference of any two

solutions of the linear second order nonhomogeneous D.E. is a solution of the corresponding homogeneous D.E.

Hence Y - Y is a solution of ay" + by' + cy = 0. In

1 2

Problem 38 of Section 3.5 we showed that if a > 0, b > 0, and c > 0 then every solution of this D.E. goes to zero as t ^ • . If b = 0, then yc involves only sines and

cosines, so Y1 - Y2 does not approach zero as t ^ • .

2t

33. From Problem 32 we write the D.E. as (D-4)(D+1)y = 3e .

2t

Thus let (D+1)y = u and then (D-4)u = 3e . This last

2t

equation is the same as du/dt - 4u = 3e , which may be

-4t

solved by multiplying both sides by e and integrating

2t 4t

(see section 2.1). This yields u = (-3/2)e + Ce .

Substituting this form of u into (D+1)y = u we obtain

2t 4t t

dy/dt + y = (-3/2)e + Ce . Again, multiplying by e

2t 4t -t

and integrating gives y = (-1/2)e + C1e + C2e , where

C1 = C/5.

Section 3.7, Page 183

2. Two linearly independent solutions of the homogeneous

2t -t 2t D.E. are y1(t) = e and y2(t) = e . Assume Y = u1(t)e

+ u2(t)e t, then Y'(t) = [2u1(t)e2t - u2(t)e t] + [u1(t)e2t

Section 3.7

55

+ u"2(t)e t]. We set u1(t)e2t + u^(t)e t = 0. Then

„// . 2t -t „ ' 2t ' -t , 4- ¦ 4- 4- ¦

Y = 4u1e + u2e + 2u1e - u2e and substituting in

the D.E. gives 2u1(t)e2t - u2(t)e-t = 2e-t. Thus we have

solution u’1(t) = 2e 3t/3 and u2(t) = -2/3. Hence u,(t) = -2e-3t/9 and u2(t) = -2t/3. Substituting in the

two algebraic equations for u1(t) and u2(t) with the /1(t) = 2e-3t/3 and u2 ^1(t) = -2e-3t/9 and u2 formula for Y(t) we obtain Y(t) = (-2e 3t/9)e2t +

(-2t/3)e-t = (-2e-t/9) - (2te-t/3). Since e-t is a solution of the homogeneous D.E., we can choose Y(t) = -2te-t/3.

5. Since cost and sint are solutions of the homogeneous D.E., we assume Y = u1(t)cost + u2(t)sint. Thus

Y' = -u1(t)sint + u2(t)cost, after setting

/ / ..

u1(t)cost + u2(t)sint = 0. Finding Y and substituting

/ /

into the D.E. then yields -u1(t)sint + u2(t)cost = tant.

/ /

The two equations for u1(t) and u2(t) have the solution: 2

u1(t) = -sint/cost = -sect + cost and /

u2(t) = sint. Thus u1(t) = sint - ln(tant + sect) and

u2(t) = -cost, which when substituted into the assumed

form for Y, simplified, and added to the homogeneous solution yields

y = c1cost + c2sint - (cost)ln(tant + sect).

11. Two linearly independent solutions of the homogeneous

3t 2t

D.E. are y^ (t) = e and y2(t) = e . Applying Theorem 3.7.1 with W(y1,y2)(t) = -e5t, we obtain

>*2s/\ ë 3s / \

Y(t) = -e3tfe_g^ ds + e2tf^^t(sldS

J 5s J 5s

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