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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.
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2
equal to M(x,y), which yields g'(x) = 2x or g(x) = x . 22
12. As long as x + y Ï 0, we can simplify the equation by
2 2 3/2
multiplying both sides by (x + y ) . This gives the
exact equation xdx + ydy = 0. The solution to this
22
equation is given implicitly by x + y = c. If you apply Theorem 2.6.1 and its construction without the
simplification, you get (x2 + y2) 1/2 = C which can be
22
written as x + y = c under the same assumption required for the simplification.
14. My = 1 and Nx = 1, so the D.E. is exact. Integrating
Section 2.6
25
M(x,y) with respect to x yields
y(x,y) = 3x3 + xy - x + h(y). Differentiating this with respect to y and setting yy(x,y) = N(x,y) yields
2
h'(y) = -4y or h(y) = - 2y . Thus the implicit solution
32
is 3x + xy - x - 2y = c. Setting x = 1 and y = 0 gives
23
c = 2 so that 2y - xy + (2+x-3x ) = 0 is the implicit solution satisfying the given I.C. Use the quadratic formula to find y(x), where the negative square root is used in order to satisfy the I.C. The solution will be
32
valid for 24x3 + x2 - 8x - 16 > 0.
2
15. We want My(x,y) = 2xy + bx to be equal to 2
Nx(x,y) = 3x + 2xy. Thus we must have b = 3. This
12 2 3
gives y(x,y) = —x y + xy + h(y) and consequently 2
h'(y) = 0. After multiplying through by 2, the solution
2 2 3
is given implicitly by x y + 2x y = c.
2 2 2
19. My(x,y) = 3x y and Nx(x,y) = 1 + y so the equation is
not exact by Theorem 2.6.1. Multiplying by the
integrating factor m(x,y) = 1/xy3 we get
(1+y2)
x + ------y' = 0, which is an exact equation since
y3
My = Nx = 0 (it is also separable). In this case
12 -3-1
y = —x + h(y) and h'(y) = y + y so that
2
x2 - y-2 + 2ln|y| = c gives the solution implicitly.
22. Multiplication of the given D.E. (which is not exact) by m(x,y) = xex yields (x2 + 2x)exsiny dx + x2excosy dy,
2x
which is exact since My(x,y) = Nx(x,y) = (x +2x)e cosy.
To solve this exact equation it's easiest to integrate N(x,y) = x2excosy with respect to y to get
2x
y(x,y) = x e siny + g(x). Solving for g(x) yields the implicit solution.
23. This problem is similar to the derivation leading up to Eq.(26). Assuming that m depends only on y, we find from Eq.(25) that m' = Qm, where Q = (Nx - My)/M must depend on y alone. Solving this last D.E. yields m(y) as given. This method provides an alternative approach to Problems
2 7 through 30.
26
Section 2.7
25. The equation is not exact so we must attempt to find an
2 2
1 3x + 2x + 3y - 2x
integrating factor. Since — (M-Nx) = ----------------------- = 3
N y x 2 2 N x + y
is a function of x alone there is an integrating factor
depending only on x, as shown in Eq.(26). Then dm/dx =
3x
3m, and the integrating factor is m(x) = e . Hence the equation can be solved as in Example 4.
26. An integrating factor can be found which is a function of x only, yielding m(x) = e x. Alternatively, you might recognize that y' - y = e2x - 1 is a linear first order equation which can be solved as in Section 2.1.
27. Using the results of Problem 23, it can be shown that m(y) = y is an integrating factor. Thus multiplying the D.E. by y gives ydx + (x - ysiny)dy = 0, which can be identified as an exact equation. Alternatively, one can rewrite the last equation as (ydx + xdy) - ysiny dy = 0. The first term is d(xy) and the last can be integrated by parts. Thus we have xy + ycosy - siny = c.
29. Multiplying by siny we obtain
exsiny dx + excosy dy + 2y dy = 0, and the first two terms are just d(exsiny). Thus, exsiny + y2 = c.
31. Using the results of Problem 24, it can be shown that
m(xy) = xy is an integrating factor. Thus, multiplying
2 3 2
by xy we have (3x y + 6x)dx + (x + 3y )dy = 0, which can be identified as an exact equation. Alternatively, we can observe that the above equation can be written as d(x3y) + d(3x2) + d(y3) = 0, so that x3y + 3x2 + y3 = c.
Section 2.7, Page 103
1d. The exact solution to this I.V.P. is y = f(t) = t + 2 -e-t.
3a. The Euler formula is yn+i = yn + h(2yn - tn + 1/2) for n = 0,1,2,3 and with t0 = 0 and y0 = 1. Thus y1 = y0 + .1(2y0 - t0 + 1/2) = 1.25,
y2 = 1.25 + .1[2(1.25) - (.1) + 1/2] = 1.54,
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