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Unformatted text preview: F ( x,y ) = Z ( x 2 + 2 xy ) dx + h ( y ) = x 3 / 3 + x 2 y + h ( y ) ∂F ∂y = x 2 + h ( y ) So we need h ( y ) = y 2 , which gives h ( y ) = y 3 / 3 + C and F ( x,y ) = x 3 / 3 + x 2 y + y 3 / 3 + C. b) The level curves of the function F ( x,y ) that you found in part a) are solutions to a ﬁrst order diﬀerential equation. Write down that diﬀerential equation. Solution . We have ω = 0 when y = y ( x ) is a solution to the diﬀerential equation. So 0 = ( x 2 + 2 xy ) dx + ( x 2 + y 2 ) dy dx dx = [( x 2 + 2 xy ) + ( x 2 + y 2 ) dy dx ] dx So the diﬀerential equation is 0 = ( x 2 + 2 xy ) + ( x 2 + y 2 ) dy dx or in normal form dy dx =x 2 + 2 xy x 2 + y 2 . Of course, you could get that just by setting ω = 0, dividing by dx and then solving for dy dx . That “dividing by dx ” step does not make sense, but the notation is designed so that it leads to the right answer!...
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This note was uploaded on 05/01/2008 for the course MATH 33B taught by Professor Staff during the Spring '07 term at UCLA.
 Spring '07
 staff
 Differential Equations, Equations

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