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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.
Download (direct link): elementarydifferentialequations2001.pdf
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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
-. 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 + (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) 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 ) 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 1. Solving, we find
2 t + 1 2t Likewise, c1 and c2 satisfy
1 = ---e --- e
13 3
c1 + c2 = 0 and c1 - 2c2 = 1 for 2, so that
1 t - 1 2t
2 = _e --- e
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