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Unformatted text preview: MAP 2302 Section 0100 Fall2011 Midterm 1 1. Solve the equation (1 x 2 ) dy dx = 2 xy + 3 . 1st SOLUTION: This equation is linear. It has the following standard form y 2 x 1 x 2 y = 3 1 x 2 . Then μ = e R P = e R 2 x 1 x 2 dx = e ln1 x 2 = 1 x 2 . Then ( y (1 x 2 )) = 3 and y (1 x 2 ) = 3 x + C . It can be solved for y : y = 3 x + C 1 x 2 . 2nd SOLUTION: This equation is exact: (2 xy + 3) dx + ( x 2 1) dy = 0 with ∂M ∂y = 2 x = ∂N ∂x . Then R Mdx = x 2 y + 3 x + g ( y ) with g ( y ) = N ∂ ( x 2 y + 3 x ) ∂y = x 2 1 x 2 = 1 . Thus, g ( y ) = y and x 2 y + 3 x y = C is the solution. Clearly that it can be written as y = 3 x + C 1 x 2 . 2. Is the equation ( x y ) dx + xdy = 0 exact? Justify your answer. Solve it? SOLUTION: No, since ∂ ( x y ) ∂y = 1 6 = 1 = ∂x ∂x . 1 It is a homogeneous equation dy dx = y x 1. With the substitution v = y x we obtain x dv dx + v = v 1. Thus, dv dx = 1 x . Therefore, v = ln  x  + C and finally, y = Cx x ln ...
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This note was uploaded on 12/15/2011 for the course MAP 2302 taught by Professor Tuncer during the Spring '08 term at University of Florida.
 Spring '08
 TUNCER

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