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Unformatted text preview: Section 1.6 , 4
10 xvv =2V2+1; I :dv = ﬂ; ln(2v2+l) = 4lnx+lnC
2v +1 x 2y2/x2+1= Cx“; x2+2y2 = Cx6 16. The substitution v = x + y+ 1 leads to x_J dv _J2udu (v_u2)
1+J17 1+u 2u—21n(1+u)+C x = 2./x+y+ —21n(1+,/x+y+1)+C
23. v=y'“3; v'—2v/x = —l; p=x'2; y = (x+Cx2)_3 29. The substitution v = sin y yields the homogeneous equation 2xv v’ = 4x2 + v2.
Solution: sinzy = 4x2 — C x
35. F = I063 +y/x)dx = %x4 +ylnx+g(y); Fy = lnx+g’(y) = y2 +lnx = N g'(y) = y2; g(y) = %y3; %x3+;—y2+ylnx = C 36. F = I(l+ye"y)dx = x+exy+g(y); F = xe‘y+g'(y) = 2y+xexy= N y
g’(y) = 2y; g(y) = y’; x+e“’+y2 = C
51. The substitution y’ = p, y" = p p’ = p(dp/dy) in y" = 2 y( y’)3 yields , d 1
pp = 2w3 => J; = IZydy => —; = y2+C. x = Jridy = ly3Cx+D,
p 3 y3+3x+Ay+B = O 59. The substitution x u — 1, y = v — 2 yields the homogeneous equation dv u—v E _ u+v'
The substitution v = pu leads to lnu = JME— = —l[ln(p2+2p—1)—lnC]. (F2 +21) *1) 2
We thus obtain the implicit solution
u2(p2+2p—1)= C
2
u2[3—2—+211)= v2+2uv—u2 = C
u u
(y+2)2+2(x+1)(y+2)—(x+1)2 = C
y2+2xyx2+2x+6y = C.
69. With a = 100 and k = 1/10, Equation (19) in the text is
y = 50[(x/100)9/10— (x/100)‘“‘°].
The equation y’(x) = 0 then yields
(x/100)“‘° = (9/11)”, so it follows that ymix = 50[(9/11)9’2—(9/11)”’2] e 3.68 mi. 71. (a) With a = 100 and k = w/ v0 = 2/ 4 = 1/ 2, the solution given by equation (19) in the textbook is y(x) = 50[(x/100)“2 — (x/100)3/2]. The fact that y(0) = 0 means that
this trajectory goes through the origin where the tree is located. (b) With k = 4/4 = l the solution is y(x) = 50[1 — (x/100)2] and we see that the
swimmer hits the bank at a distance y(0) = 50 north of the tree. (c) With k = 6/4 = 1 the solution is y(x) = 50[(x/100)_“2 — (x/100)5/2]. This
trajectory is asymptotic to the positive x—axis, so we see that the swimmer never reaches
the west bank of the river. ...
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 Spring '07
 TERRELL,R
 Math

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