midsol1 - MAP 2302 Section 0100 Fall-2011 Midterm 1 1....

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Unformatted text preview: MAP 2302 Section 0100 Fall-2011 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 = 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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midsol1 - MAP 2302 Section 0100 Fall-2011 Midterm 1 1....

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