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ps2sol

# ps2sol - ACM 95b/100b Problem Set 2 Solutions Prepared by...

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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 - 4 y + 37 y = 0 , y (0) = 3 , y (0) = 3 2 b) y + y = r ( t ) , y (0) = 0 , y (0) = 0 , r ( t ) = t, 0 < t < 1 0 , 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 ) } = τ f ( t - τ ) e st dt = 0 f ( u ) e s ( u + τ ) du = e F ( s ) . Thus, if we have a Laplace transform which takes the form e 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 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 ) = 0 e - αt f ( t ) e - st dt = 0 f ( t ) e - ( s + α ) t dt = F ( s + α ) . a) To begin, take the Laplace transform of the equation given: L { 4 y - 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

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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 ) = 0 r ( t ) e - st dt = 1 0 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 ) , 0 < 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 + 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 ) = t 0 f ( τ ) g ( t - τ ) = t 0 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 ) . or Y ( s ) = G ( s ) + 1 s 2 + ω 2 = 1 ω G ( s ) ω s 2 + ω 2 + 1 ω ω s 2 + ω 2 . Recognizing L { sin( ωt ) } = ω s 2 + ω 2 , and applying the convolution theorem, we get y ( t ) = 1 ω t 0 g ( τ ) sin( ω ( t - τ )) + 1 ω sin( ωt ) = 1 ω t 0 sin( ωτ ) g ( t - τ ) + 1 ω sin( ωt ) . 2
Problem 3 (2 × 10 points) Laplace transforms of integrals. a) Suppose f ( t ) has Laplace transform F ( s ). Verify the following relationship: L t 0 f ( τ ) = 1 s F ( s ) b) The current i ( t ) in an electric circuit is governed by the following integral equation Ri ( t ) + 1 C t 0 i ( τ ) = v ( t ) where R is the resistance and C is the capacitance. Use the Laplace transform to solve for

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