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Differential Equations Solutions 114

# Differential Equations Solutions 114 - 124 Chapter 20...

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124 Chapter 20. Solutions: Solution of Ordinary Differential Equations CHALLENGE 20.15. Using the notation of the pointer, we let a ( t ) = 1 > 0, b ( t ) = 8 . 125 π cot((1 + t ) π/ 8), c ( t ) = π 2 > 0, and f ( t ) = 3 π 2 . These are all smooth functions on [0 , 1]. Since c ( t ) = π 2 / 2 > 0 and 1 0 [ f ( t )] 2 d t = π 4 4 , the solution exists and is unique. Since f ( t ) < 0, the Maximum Principle tells us that max t [0 , 1] u ( t ) max( 2 . 0761 , 2 . 2929 , 0) = 0 . Letting v ( t ) = 3, we see that v ( t ) + 8 . 125 π cot((1 + t ) π/ 8) v ( t ) + π 2 v ( t ) = 3 π 2 and v (0) = v (1) = 3. Therefore the Monotonicity Theorem says that u ( t ) v ( t ) for t [0 , 1]. Therefore we conclude 3 u ( t ) 0 for t [0 , 1]. Note on how I constructed the problem: The true solution to the prob- lem is u ( t ) = cos((1 + t ) π/ 8)
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