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Unformatted text preview: Solutions to Homework 2 Section 2.2 21. Seperate the variables and integrate, using y (0) = 1 integraldisplay y 1 3 s 2 6 sds = integraldisplay x 1 + 3 s 2 ds. (1) Yielding y 3 3 y 2 = x 3 + x 2. Notice that in the differential equation y = 0 , 2 would make the denominator zero therefore the solution cannot pass through these points. The corresponding x coordinates are x = 1 , 1. So the solution is valid on the interval  x  < 1. 30. a. Eq. (i) can be rewritten as eq. (ii) by miltypling top and bottom of equation (i) by 1 /x . b. If y = vx then dy dx = x dv dx + v by the product rule. c. Direct substitution back into eq. (ii) and rearranging gives eq. (iii), x dv dx = v 2 4 1 v . (2) d. Sepearting the variables and integrating, integraldisplay 1 v v 2 4 dv = integraldisplay 1 x dx, (3) gives the solution 1 / 4ln( v 2) 3 / 4ln( v + 2) = ln( x ) + ln( c ) . (4) Which can be rewritten as 1 ( v 2) 1 / 4 ( v + 2) 3 / 4 = cx, (5) 1 or substituting y/x back in for v , e. 1 ( y x 2 ) 1 / 4 ( y x + 2 ) 3 / 4 = cx. (6) f. The direction field and integral curves seen in fig.1 are in fact symmetric about the origin....
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This note was uploaded on 10/23/2009 for the course MATH 22B taught by Professor Hunter during the Winter '08 term at UC Davis.
 Winter '08
 Hunter
 Differential Equations, Equations

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