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Unformatted text preview: ′ . We must solve v (xv ′ + v ) + 1 =
We must solve
xv dv
=
dx We must solve
v√ 1 + v2. 1 + v 2 − v 2 − 1. dx
dv
=
.
x
1 + v2 − v2 − 1 Integrate both sides. Let w = 1 + v 2 . It follows that dw = 2vdv . We must solve
1
2 √ dw
= ln x + C.
w−w We have
ln x + C = 1
2 √ √
Let u = w . We have du = 1 w−1/2 dw .
2
We have
ln x + C =
= − ln 1 − 1+ dw
√.
w(1 − w) √
du
= − ln 1 − u = − ln 1 − w = − ln 1 −
1−u
y
x 2 = − ln x− x2 + y 2
= − ln x −
x 1 + v2  x2 + y 2 + ln x. 2 Subtract ln x...
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This document was uploaded on 03/23/2014 for the course MATH 242 at South Carolina.
 Fall '11
 Staff
 Differential Equations, Equations

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