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

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solution -v + t/2 = c or v = y' = -2/(c1 - t ) where

c1 = 2c. We must consider separately the cases c1 = 0,

2

c1 > 0 and c1 < 0. If c1 = 0, then y' = 2/t or

y = -2/t + c2. If c1 > 0, let c1 = k . Then

y' = -2/(k-t ) = -(1/k)[1/(k-t) + 1/(k+t)], so that y = (1/k)ln|(k-t)/(k+t)l+c2. If c1 < 0, let c1 = -k2.

2 2 -1 Then y' = 2/(k + t ) so that y = (2/k)tan (t/k) + c2.

Finally, we note that y = constant is also a solution of the D.E.

Following the procedure outlined, let v = dy/dt and y" = dv/dt = v dv/dy. Thus the D.E. becomes

yvdv/dy + v = 0, which is a separable equation with the solution v = c1/y. Next let v = dy/dt = c/y, which again

2

separates to give the solution y = c1t + c2.

Again let v = y' and v' = vdv/dy to obtain

22 2y v dv/dy + 2yv = 1. This is an exact equation with

-1 1/2

solution v = ± y (y + c^ . To solve this equation,

38

Section 3 . 2

1/2

we write it in the form ± ydy/(y+c1) = dt. On

observing that the left side of the equation can be

1/2

written as ± [(y+c1) - c1]dy/(y+c1) we integrate and

find ± (2/3)(y-2c1)(y+c1) 1/2 = t + c2.

39. If v = y', then v' = vdv/dy and the D.E. becomes v dv/dy + v2 = 2e y. Dividing by v we obtain

dv/dy + v = 2v 1e y, which is a Bernoulli equation (see Prob.27, Section 2.4). Let w(y) = v , then dw/dy = 2v dv/dy and the D.E. then becomes dw/dy + 2w = 4e y, which

is linear in w. Its solution is w = v2 = ce-2y + 4e-y.

Setting v = dy/dt, we obtain a separable equation in y and t, which is solved to yield the solution.

40. Since both t and y are missing, either approach used above will work. In this case it's easier to use the approach of Problems 28-33, so let v = y' and thus v' = y" and the D.E. becomes vdv/dt = 2.

43. The variable y is missing. Let v = y', then v' = y" and the D.E. becomes vv' - t = 0. The solution of the

22

separable equation is v = t + c1. Substituting v = y'

and applying the I.C. y'(1) = 1, we obtain y' = t. The positive square root was chosen because y' > 0 at t = 1. Solving this last equation and applying the I.C. y(1) = 2, 2

we obtain y = t /2 + 3/2.

Section 3.2, Page 14 5

2. W(cost,sint) =

cost sint -sint cost

22 = cos t + sin t = 1

4. W(x,xe ) =

xe

-Ë. JS

1 e + xe

x 2 x x 2 x

= xe + x e - xe = x e .

Dividing by (t-1) we have p(t) = -3t/(t-1), q(t) = 4/(t-1) and g(t) = sint/(t-1), so the only point of discontinuity is t = 1. By Theorem 3.2.1, the largest interval is -• < t < 1, since the initial point is t0 = -2.

Section 3.2

39

12.

14.

15.

18.

21.

p(x) = 1/(x-2) and q(x) = tanx, so x = p/2, 2, 3p/2... are points of discontinuity. Since t0 = 3, the interval specified by Theorem 3.2.1 is 2 < x < 3p/2.

For y = t1/2, y' = —1-1/2 and y" = - —1-3/2. Thus

24

2 1 -1 1 -1 ÓÓ + (y) = -_ t + —t = 0. Similarly y = 1 is also

44

1/2

a solution. If y = c1(1) + c2t is substituted in the

3/2

D.E. you will get -c1c2/4^ , which is zero only if

c1 = 0 or c2 = 0. Thus the linear combination of two

solutions is not, in general, a solution. Theorem 3.2.2 is not contradicted however, since the D.E. is not linear.

y = f(t) is a solution of the D.E. so L[f](t) = g(t). Since L is a linear operator,

L[cf](t) = cL[f](t) = cg(t). But, since g(t) Ï 0, cg(t) = g(t) if and only if c = 1. This is not a contradiction of Theorem 3.2.2 since the linear D.E. is not homogeneous.

W(f,g) =

t g 1 g'

2 t t = tg - g = te, or g - —g = te . This

1 t

---g = tet

t

1 ' 1

---g - ---

t t2

1t

or ( —g) = e . Integrating and multiplying by t we

t

obtain g(t) = tet + ct.

t -2t

From Section 3.1, e and e are two solutions, and t -2t

since W(e ,e ) Ï 0 they form a fundamental set of solutions. To find the fundamental set specified by

Theorem 3 .2 .5, let y(t) = c1e + c2e -2t and c2

, where c1

satisfy

c1 + c2 = 1 and c1 - 2c2 = 0 for Ó1. Solving, we find

2 t + 1 2t Likewise, c1 and c2 satisfy

Ó1 = ---e --- e

13 3

c1 + c2 = 0 and c1 - 2c2 = 1 for Ó 2, so that

1 t - 1 2t

Ó2 = _e --- e

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