15_2_17 - variable. Let y = x v( x ) . Then v + x d v dx =...

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17. All circles tangent to the y -axis at the origin intersect all circles tangent to the x -axis at the origin at right angles, so they must be the streamlines of the two-dimensional dipole. As an alternative derivation of this fact, the streamlines must satisfy dx 2 xy = dy y 2 - x 2 , or, equivalently, dy dx = y 2 - x 2 2 xy . This homogeneous DE can be solved (as was that in Exercise 10) by a change in dependent
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Unformatted text preview: variable. Let y = x v( x ) . Then v + x d v dx = dy dx = v 2 x 2-x 2 2 v x 2 x d v dx = v 2-1 2 v-v = -v 2 + 1 2 v 2 v d v v 2 + 1 = -dx x ln (v 2 + 1 ) = -ln x + ln C v 2 + 1 = C x ⇒ y 2 x 2 + 1 = C x x 2 + y 2 = Cx ( x-C ) 2 + y 2 = C 2 . These streamlines are circles tangent to the y-axis at the origin....
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This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

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