MATH
test1_solns

# Multiplying by cos m 1 2 πx and integrating from x 0

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Multiplying by cos ( ( m - 1 2 ) πx ) and integrating from x = 0 to x = 1 gives 1 0 f ( x ) cos ( ( m - 1 2 ) πx ) dx = n =1 A n 1 0 cos ( ( n - 1 2 ) πx ) cos ( ( m - 1 2 ) πx ) dx Using the given orthogonality condition gives A m = 2 1 0 f ( x ) cos ( ( m - 1 2 ) πx ) dx Page 4 of 5 Please go to the next page. . . [0.5 points] [1 point] [1 point] [0.5 points] [1 point] [0.5 points] [1 point] [1 point] [1 point] [1 point]

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Solutions to Test 1 Substituting for f ( x ) gives A m = 2 1 0 ( u 0 - 1 - b 12 (1 - x 4 ) ) cos ( ( m - 1 2 ) πx ) dx = 2( u 0 - 1) 1 0 cos ( ( m - 1 2 ) πx ) dx - b 6 1 0 (1 - x 4 ) cos ( ( m - 1 2 ) πx ) dx = ( - 1) n +1 4( u 0 - 1) (2 n - 1) π - ( - 1) n +1 16 b ( π 2 (2 n - 1) 2 - 8) (2 n - 1) 5 π 5 4. 3 points (a) Solve for u ( x, t ) = u E ( x ) + n =1 u n ( x, t ) using the earlier transformations. Write down precisely the functions u n ( x, t ). (b) Find an approximate solution good for large time and state the limit as t → ∞ . Solutoin: Reversing the earlier transformations, we have u ( x, t ) = u E ( x ) + v ( x, t ) = b 12 (1 - x 4 ) + 1 + n =1 A n cos ( ( n - 1 2 ) πx ) e - ( n - 1 2 ) 2 π 2 t thus u n ( x, t ) = A n cos ( ( n - 1 2 ) πx ) e - ( n - 1 2 ) 2 π 2 t where A n is given in Part 3. For t > 1 2 , the first term in the series gives a good approximation to v ( x, t ), thus u ( x, t ) u E ( x ) + u 1 ( x, t ) = b 12 (1 - x 4 ) + 1 + 4( u 0 - 1) π - 16 b ( π 2 - 8) π 5 cos ( 1 2 πx ) e - 1 4 π 2 t We have u ( x, t ) u E ( x ) as t → ∞ Page 5 of 5 End of exam.
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• Fall '08
• Staff
• Math, Cos, Trigraph, Partial differential equation, Solutoin

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