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

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x (y-iv)" + x(y-,v)' + (x - —) y-,v = 0. On carrying out the

1 1 4 1

differentiations and making use of the fact that y1 is a solution, we obtain x2y1v" + (2x2y1 + xy1)v' = 0. This is a first order linear equation for v ,

v" + (2y1/y1 + 1/x)v' = 0, with solution

/

y1 1

v (x) = cexp[-J (2— + — )dx] = cexp[-2lny1 - lnx]

y1 x 1

1c 2

= c---- = ----------------------- = c csc x,

2 ,-1.2. xy1 x(x sin x)

where c is an arbitrary constant, which we will take to be

one. Then v(x) = J csc x dx = -cotx + k where again k is an

arbitrary constant which can be taken equal to zero. Thus

-1/2 -1/2 y2(x) = y1(x)v(x) = (x sinx)(-cotx) = -x cosx. The

-1/2

second solution is usually taken to be x cosx. Note that c = -1 would have given this solution.

31b. Let y2(x) = exv(x), then y2 = exv' + exv, and

y"2 = exv" + 2exv + exv. Substituting in the D.E. we

obtain xexv" + (xex-Nex)v' = 0, or v" + (1-N/x)v' = 0.

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 —, 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—— - Ã 2x——dx] 2 -1 -1

-x -x

2 x e ñ 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/y—)' = (y—y'2 - ó1ó2)/ó12 = W(y—,y2)/y—2. Abel's identity

is W(y—,y2) = c exp[-l tp(r)dr]. Hence

Jt0

(y2/y—)' = cy—-2 exp[-lt p(r)dr]. Integrating and setting

t0

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) = y—(t) Jt------------------------—- ds.

[y—(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 ³ .2 \ ³ 2 / , 2 \

y1 sin (t ) sin (t )

22 y2/y— = -(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 = c—e 1 + c2e 2

V2 2

b -4ac )/2a provided b - 4ac Ï 0.

2

In this case there are two possibilities. If b - 4ac > 0 2 1/2

then (b - 4ac) < b and r— and r2 are real and

rt 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-,

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

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