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exam1-solutions

# exam1-solutions - MAP2302 Test 1 Name No calculators are...

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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 0 - 1 = K - 1 , so K = 4 and y = 4 e x 3 / 3 - 1

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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 μ ( 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 x dx + 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 = 0 , so M ( x, y ) = 2 xy - x and N ( x, y ) = 1 + x 2 , and M y = N x = 2 x . We now find F ( x, y ) so that F x = M and F y = N . Integrating M with respect to x , F ( x, y ) = Z (2 xy - x ) dx = x 2 y - x 2 2 + g ( y ) , while integrating N
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