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

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23 3

t

40

Section 3 . 2

25.

27.

30.

32.

34.

2

For y1 = x, we have x (0) - x(x+2)(1) + (x+2)(x) = 0 and

for y2 = xex we have x2(x+2)ex-x(x+2)(x+1)ex+(x+2)xex = 0.

x 2 x

From Problem 4, W(x,xe ) = x e Ï 0 for x > 0, so y1 and y2 form a fundamental set of solutions.

Suppose that

P(x)y" + Q(x)y' + R(x)y = [P(x)y']' + [f(x)y]'. On

expanding the right side and equating coefficients, we find f'(x) = R(x) and P'(x) + f(x) = Q(x). These two conditions on f can be satisfied if

R(x) = Q'(x) - P"(x) which gives the necessary condition

P"(x) - Q'(x) + R(x) = 0.

We have P(x) = x, Q(x) = - cosx, and R(x) = sinx and the condition for exactness is satisfied. Also, from Problem 27, f(x) = Q(x) - P'(x) = - cosx -1, so the D.E. becomes (xy')' - [(1 + cosx)y]' = 0. Hence

xy' - (1 + cosx)y = c1. This is a first order linear

D.E. and the integrating factor (after dividing by x) is m(x) = exp[-| x-1(1 + cosx)dx]. The general solution is

x

y = [m(x)] -1[c1l t-1 m(t)dt + c2]. Jx0

We want to choose m(x) and f(x) so that m(x)P(x)y" + m(x)Q(x)y' + m (x)R(x)y = [m(x)P(x)y']' + [f(x)y]'.

Expand the right side and equate coefficients of y", y' and y. This gives m'(x)P(x) + m(x)P'(x) + f(x) = m(x)Q(x) and f'(x) = m(x)R(x). Differentiate the first equation and then eliminate f'(x) to obtain the adjoint equation Pm" + (2P' - Q)m' + (P" - Q' + R)m = 0.

2

P = 1-x , Q = -2x and R = a(a+1). Thus

2P' - Q = -4x + 2x = -2x = Q and

P" - Q' + R = -2 + 2 + a(a+1) = a(a+1) = R.

Write the adjoint D.E. given in Problem 32 as

ë ë ë ë ë

Pm" + Qm' + Rm = 0 where P = P, Q = 2P' - Q, and

ë

R = P" - Q' + R. The adjoint of this equation, namely the adjoint of the adjoint, is

ë ë ë ë ë ë

Py" + (2P' - Q)y' + (P" - Q' + R)y = 0. After

ë ë ë

substituting for P, Q, and R and simplifying, we obtain

Py" + Qy' + Ry = 0. This is the same as the original

equation.

Section 3.3

41

37. From Problem 32 the adjoint of Py" + Qy' + Ry = 0 is

Pm" + (2P' - Q)m' + (P" - Q' + R)m = 0. The two equations

are the same if 2P' - Q = Q and P" - Q' + R = R. This

will be true if P' = Q. Hence the original D.E. is self-

2

adjoint if P' = Q. For Problem 33, P(x) = x so P'(x) = 2x and Q(x) = x. Hence the Bessel equation of order v is not self-adjoint. In a similar manner we find that Problems 34 and 35 are self-adjoint.

Section 3.3, Page 152

2. Since cos30 = 4cos30 - 3cos0 we have

cos30 - (4cos30-3cos0) = 0 for all 0. From Eq.(1) we have k1 = 1 and k2 = -1 and thus cos30 and 4cos30 - 3cos0 are linearly dependent.

= -2/t Ï 0.

6. W(t,t-1) =

t t-1

1 -t-2

7. For t>0 g(t) = t and hence f(t) - 3g(t) = 0 for all t.

Therefore f and g are linearly dependent on 0<t. For t<0 g(t) = -t and f(t) + 3g(t) = 0, so again f and g are linearly dependent on t<0. For any interval that includes the origin, such as -1<t<2, there is no c for which f(t) +cg(t) = 0 for all t, and hence f and g are linearly independent on this interval.

12. The D.E. is linear and homogeneous. Hence, if y1 and y2 are solutions, then y3 = y1 + y2 and y4 = y1 - y2 are

solutions. W(y3,y4) = y3y4 - y3y4 = (y— + y2)(y— - y’2) -

(Ó— + Ó2)(Ó³ - y2) = -2(y1y/2 - y—y2) = -2W(y1,y2), is not zero since y— and y2 are linearly independent solutions. Hence y3 and y4 form a fundamental set of solutions. Conversely, solving the first two equations for y— and y2, we have y— = (y3+y4)/2 and y2 = (y3-y4)/2, so y— and y2 are solutions. Finally, from above we have W(y—y2) = -W(y3,y4)/2.

15. Writing the D.E. in the form of Eq.(7), we have p(t) = -(t+2)/t. Thus Eq.(8) yields

42

Section 3 . 3

f-(t+2) 2 t

W(t) = cexp[-J ---------- dt] = ct e .

t

20. From Eq.(8) we have W(y1,y2) = cexp[-| p(t)dt], where

2

p(t) = 2/t from the D.E. Thus W(y1,y2) = c/t . Since W(y1,y2)(1) = 2 we find c = 2 and thus W(y1,y2)(5) = 2/25.

24. Let c be the point in I at which both y1 and y2 vanish. Then W(y1,y2)(c) = y1(c)y/2(c) - y1 (c)y2 (c) = 0. Hence, by

Theorem 3.3.3 the functions y1 and y2 cannot form a fundamental set.

26. Suppose that y1 and y2 have a point of inflection at t0

//

and either p(t0) Ï 0 or q(t0) Ï 0. Since y1(t0) = 0 and

ó 2 (t0) = 0 it follows from the D.E. that p(t0)y1(t0) +

q(t0)y1(t0) = 0 and p(t0)y'2(t0) + q(t0)y2(t0) = 0. If p(t0) = 0 and q(t0) Ï 0 then y1(t0) = y2(t0) = 0, and W(y1,y2)(t0) = 0 so the solutions cannot form a fundamental set. If p(t0) Ï 0 and q(t0) = 0 then

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