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Ordinary Diff Eq Exam Review Solutions 96

# Ordinary Diff Eq Exam Review Solutions 96 - 98 1 Solutions...

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98 1 Solutions 30. ( 1 ; 1 ) 31. (3 ; 1 ) 32. ( 1 ; 2) , ( 2 ; 0) , (0 2) , (2 ; 1 ) 33. The initial condition occurs at t = 0 which is precisely where a 2 ( t ) = t 2 has a zero. Theorem 6 does not apply. 34. In this case y ( t 0 ) = 0 and y 0 ( t 0 ) = 0 . The function y ( t ) = 0 , t 2 I is a solution to the initial value problem. By the uniqueness part of Theorem 6 y = 0 is the only solution. 35. The assumptions say that y 1 ( t 0 ) = y 2 ( t 0 ) and y 0 1 ( t 0 ) = y 0 2 ( t 0 ) . Both y 1 and y 2 therefore satisfies the same initial conditions. By the uniqueness part of Theorem 6 y 1 = y 2 . Section 4.2 1. dependent; 2 t and 5 t are multiples of each other. 2. independent 3. independent 4. dependent; e 2 t + 1 = e 1 e 2 t and e 2 t 3 = e 3 e 2 t , they are multiples of each other. 5. independent 6. dependent; ln t 2 = 2 ln t and ln t 5 = 5 ln t , they are multiples of each other. 7. dependent;
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