smt2 - Math 5525. April 14, 2010. Midterm Exam 2. Problems...

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Math 5525. April 14, 2010. Midterm Exam 2. Problems and Solutions. Problem 1. Find the general solution of the equation xy 00 - (2 x + 1) y 0 + ( x + 1) y = 0 . Note that one of two linearly independent solutions is y 1 ( x ) = e x . Solution. By Abel’s formula, the Wronskian W ( x ) = W ( y 1 , y 2 )( x ) satisfies W 0 = - p 1 p 0 W = 2 x + 1 x W = ± 2 + 1 x ² W, so that W ( x ) = W ( x 0 ) · xe 2 x . On the other hand, W ( x ) = y 2 1 · ± y 2 y 1 ² 0 = e 2 x · ± y 2 y 1 ² 0 . Therefore, ( y 2 /y 1 ) 0 = const · x, y 2 /y 1 = const · x 2 . We can take y 2 = x 2 y 1 = x 2 e x . Then the general solution has the form y = C 1 y 1 + C 2 y 2 = ( C 1 + C 2 x 2 ) e x . Problem 2. Let y = y ( x ) be a solution of the problem y 0 = 1 + y + y 2 2 , y (0) = 0 . Show that y ( x ) ≤ - ln(1 - x ) for 0 x < 1 . Proof. The function z = - ln(1 - x ) satisfies z 0 = 1 1 - x = e z 1 + z + z 2 2 for 0 x < 1 , z (0) = 0 . Note that the comparison theorem (Theorem 11.1) can be applied in the case of non-strict inequalities, if the function
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smt2 - Math 5525. April 14, 2010. Midterm Exam 2. Problems...

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