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Unformatted text preview: MAP2302 Name: Test 1 No calculators are allowed. Each problem is worth 25 points. Show all work for full credit. 1. Solve the initial value problem. dy dx = x 2 (1 + y ) , y (0) = 3 . Solution: The equation is separable. So, dy 1 + y = x 2 dx Z dy 1 + y = Z x 2 dx log (1 + y ) = 1 3 x 3 + C 1 + y = Ke x 3 / 3 (where K = e C ) y = Ke x 3 / 3 1 . Applying the initial condition, we find 3 = y (0) = Ke 1 = K 1 , so K = 4 and y = 4 e x 3 / 3 1 2,3. Solve the equation for y by two different methods. ( x 2 + 1) dy dx + 2 xy x = 0 Method 1: The equation is linear. Rewrite in standard form: dy dx + 2 x 1 + x 2 y = x 1 + x 2 , so P ( x ) = 2 x (1 + x 2 ) 1 and Q ( x ) = x (1 + x 2 ) 1 . The integrating factor is ( x ) = exp Z P ( x ) dx = exp Z 2 x 1 + x 2 dx = exp(log (1 + x 2 )) = 1+ x 2 . So the general solution is y ( x ) = 1 1 + x 2 Z xdx + C which integrates to y = 1 1 + x 2 x 2 2 + C (1) Method 2: The equation is exact. Rewrite in differential form: (2 xy x ) dx + (1 + x 2 ) dy...
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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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