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**Unformatted text preview: **Signature Printed Name My signature above certifies that I have complied with the University of Pennsylvania’s Code of Academic Integrity in completing this examination. Math 425 Exam 2 Jerry L. Kazdan April 26, 2011 12:00 – 1:20 Directions This exam has three parts, Part A, short answer, has 1 problem (10 points). Part B has 4 shorter problems (9 points each, so 36 points). Part C has 3 traditional problems (15 points each so 45 points). Total is 91 points. Closed book, no calculators or computers– but you may use one 3 00 × 5 00 card with notes on both sides. Part A: Short Answer (1 problem, 10 points). 1. Let S and T be linear spaces and A : S → T be a linear map. Say V and W are particular solutions of the equations A V = Y 1 and A W = Y 2 , respectively, while Z 6 = 0 is a solution of the homogeneous equation A Z = 0. Answer the following in terms of V , W , and Z . a) Find some solution of A X = 3 Y 1 . Solution : X = 3 V b) Find some solution of A X =- 5 Y 2 . Solution : X =- 5 W c) Find some solution of A X = 3 Y 1- 5 Y 2 . Solution : X = 3 V- 5 W d) Find another solution (other than Z and 0) of the homogeneous equation A X = 0. Solution : X = 2 Z e) Find another solution of A X = 3 Y 1- 5 Y 2 . Solution : X = 3 V- 5 Y 2 + Z Part B: Short Problems (4 problems, 9 points each so 36 points) B–1. Suppose f is a function of one variable that has a continuous second derivative. Show that for any constants a and b , the function u ( x,y ) = f ( ax + by ) is a solution of the nonlinear PDE u xx u yy- u 2 xy = 0 ....

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