Ps2sol - ACM 95b/100b Problem Set 2 Solutions Prepared by Lei Zhang Total 132 points Include grading section 2 points Problem 1(2 × 10 points Use

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Unformatted text preview: ACM 95b/100b Problem Set 2 Solutions Prepared by: Lei Zhang January 16, 2007 Total: 132 points Include grading section: 2 points Problem 1 (2 × 10 points) Use the shifting theorems to assist in solving the following initial value problems. a) 4 y 00- 4 y + 37 y = 0 , y (0) = 3 , y (0) = 3 2 b) y 00 + y = r ( t ) , y (0) = 0 , y (0) = 0 , r ( t ) = t, < t < 1 , t > 1 Solution 1 The first of the shifting theorems from Laplace Transforms deals with shifts in time: L{ f ( t- τ ) H ( t- τ ) } = Z ∞ τ f ( t- τ ) e st dt = Z ∞ f ( u ) e s ( u + τ ) du = e sτ F ( s ) . Thus, if we have a Laplace transform which takes the form e sτ F ( s ), where F ( s ) is the transform of a known function, the shifting theorem tells us that the inverse transform is L- 1 { e sτ F ( s ) } = f ( t- τ ) H ( t- τ ) . The second deals with shifts in s : L- 1 { F ( s + α ) } = e- αt f ( t ) because L e- αt f ( t ) = Z ∞ e- αt f ( t ) e- st dt = Z ∞ f ( t ) e- ( s + α ) t dt = F ( s + α ) . a) To begin, take the Laplace transform of the equation given: L{ 4 y 00- 4 y + 37 y } = 4 s 2 Y ( s )- 4 y (0)- 4 sy (0)- 4 sY ( s ) + 4 y (0) + 37 Y ( s ) = Y ( s ) ( 4 s 2- 4 s + 37 )- 12 s + 6 = 0 . Solving for Y ( s ): Y ( s ) = 12 s- 6 4 s 2- 4 s + 37 = 3 s- 1 2 ( s- 1 2 ) 2 + 9 . Applying the second shifting theorem, and noting that L{ cos( at ) } = s s 2 + a 2 , we have y ( t ) = L- 1 { Y ( s ) } = 3 e t/ 2 cos(3 t ) . 1 b) We can write r ( t ) in a more convenient form as follows: r ( t ) = t ( H ( t )- H ( t- 1)) . Using this definition, the Laplace transform of r ( t ), denoted R ( s ) is L{ r ( t ) } ≡ R ( s ) = Z ∞ r ( t ) e- st dt = Z 1 te- st dt = 1 s 2- e- s ( s + 1) s 2 . Taking the Laplace transform of the given equation, we find s 2 Y ( s ) + Y ( s ) = R ( s ) or Y ( s ) = R ( s ) s 2 + 1 = 1 s 2 ( s 2 + 1)- e- s ( s + 1) s 2 ( s 2 + 1) = 1- e- s s 2- 1- e- s 1 + s 2- e- s s + e- s s s 2 + 1 . Noting that L{ 1 } = 1 s L{ t } = 1 s 2 L{ sin( t ) } = 1 s 2 + 1 L{ cos( t ) } = s s 2 + 1 , and applying the first shifting theorem, we find that y ( t ) = t- sin( t ) + H ( t- 1) [- t + sin( t- 1) + cos( t- 1)] = t- sin( t ) , < t < 1 cos( t- 1) + sin( t- 1)- sin( t ) , t ≥ 1 . Problem 2 (10 points) Express the solution to the following initial value problem as a convolution integral. y 00 + ω 2 y = g ( t ) , y (0) = 0 , y (0) = 1 Solution 2 The convolution theorem for Laplace transforms states L- 1 { F ( s ) G ( s ) } = f ( t ) * g ( t ) = Z t f ( τ ) g ( t- τ ) dτ = Z t g ( τ ) f ( t- τ ) dτ. Taking the Laplace transform of the equation, we have s 2 Y ( s )- 1 + ω 2 Y ( s ) = G ( s ) ....
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This note was uploaded on 11/23/2009 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.

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Ps2sol - ACM 95b/100b Problem Set 2 Solutions Prepared by Lei Zhang Total 132 points Include grading section 2 points Problem 1(2 × 10 points Use

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