solC3-2sample08

solC3-2sample08 - SOLUTIONS TO CALCULUS III SAMPLE MIDTERM...

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Unformatted text preview: SOLUTIONS TO CALCULUS III SAMPLE MIDTERM 2 Problem 1. (10 points) (a) Find the indicated powers (2 3 + 2 i ) 5 ; (1 i ) 8 (b) For what value of z is e z equal to 1 i ? Solution: (a) (2 3 + 2 i ) 5 = parenleftBig 4 e i / 6 parenrightBig 5 = 4 5 e i 5 6 = 1024 parenleftBig 3 2 + 1 2 i parenrightBig = 512 3 + 512 i (1 i ) 8 = parenleftBig 2 e i / 4 parenrightBig 8 = ( 2) 8 e i 2 = 16 1 = 16 (b) | e z | = e Re( z ) and arg( e z ) = Im( z ). In order that e z = 1 i = 2 e i 5 / 4 , we need e Re( z ) = 2 , Im( z ) = 5 / 4 + 2 k for any integer k that is, for z = ln 2 + i (5 / 4 + 2 k ), its exponential e z = 1 i . Problem 2. (10 points) (a) Solve the initial-value problem 2 y + 5 y + 3 y = 0 , y (0) = 3 , y (0) = 4 (b) Solve the boundary-value problem y + 4 y + 13 y = 0 , y (0) = 2 , y ( / 2) = 1 Solution: (a) The auxiliary equation is 2 r 2 + 5 r + 3 = 0 and it has too roots 1, 3 / 2. The general solution to the differential equation is c 1 e x + c 2 e 3 x/ 2 . Plug in the initial conditions, c 1 e + c 2 e = 3 , c 1 e 3 c 2 e / 2 = 4 ....
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This note was uploaded on 11/02/2009 for the course MATH V1201 taught by Professor Perutz during the Spring '08 term at Columbia.

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solC3-2sample08 - SOLUTIONS TO CALCULUS III SAMPLE MIDTERM...

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