MATH201 Midterm 2013 F Solution

MATH201 Midterm 2013 F Solution - MATH 201Midterm...

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MATH 201—Midterm Examination Solutions 1. (a) Use the definition of the Laplace transform to compute the Laplace transform F ( s ) of f ( t ) = ( 0 if 0 t < 1 1 2 ( e 2 t + 1 ) if 1 t < . [6] Solution: F ( s ) = L{ f ( t ) } ( s ) = Z 1 1 2 ( e 2 t + 1 ) e - st dt = 1 2 e (2 - s ) t (2 - s ) 1 - e - st 2 s 1 = e (2 - s ) 2( s - 2) + e - s 2 s provided that s > 2 (so that the limits converge). (b) What is the domain of F ( s ) in part (a)? [3] Solution: The domain is all s > 2. For s < 2, the limits in the last step of the calculation above do not converge. For s = 2, then R 1 1 2 ( e 2 t + 1 ) e - st dt in the above calculation becomes R 1 1 2 ( 1 + e - 2 t ) dt > R 1 1 2 dt , which obviously does not converge. (c) Find the inverse Laplace transform of F ( s ) = 2 ( s - 2)( s 2 - 2 s +2) . [6] Solution: Use partial fractions to find that 2 ( s - 2)( s 2 - 2 s + 2) = 2 ( s - 2) (( s - 1) 2 + 1) = 1 s - 2 - ( s - 1) (( s - 1) 2 + 1) - 1 (( s - 1) 2 + 1) Take the Laplace transform of each term on the right to get L - 1 2 ( s - 2)( s 2 - 2 s + 2) ( t ) = L - 1 1 s - 2 ( t ) - L - 1 ( s - 1) (( s - 1) 2 + 1) ( t ) - L - 1 1 (( s - 1) 2 + 1) ( t ) = e 2 t - e t (cos t + sin t ) .
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ID#: Name: 2. (a) Find the solution of y 0 = 1 2 x y - y 3 , x > 0, such that y (1) = - 1 2 . [6] Solution: This equation has the Bernoulli form. Multiply by 1 /y 3 and substitute with z = 1 /y 2 to get z 0 + 1 x z = 2 . This is a linear equation and is easy to solve. The initial condition y (1) = - 1 / 2 becomes z (1) = 4. Solving the linear equation and applying the initial condition, we get z ( x ) = x + 3 /x .
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