Practice Problems - Week #7 Laplace - Step Functions, DE Solutions Solutions

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1 For Quetion -6, rewrite the piecewie function uing tep function, ketch their graph, and find F () = Lf(t). 0 0 < t < 2. f(t) = (t 2 4) 2 < t In tep-function form, f(t) = u 2 (t 2 4) The graph i the olid line in the diagram below: Practice Problem - Week #7 Laplace - Step Function, DE Solution Solution Note the dotted line: thi i what y = t 2 4 = (t 2)(t + 2) look like over the entire interval, but it i only turned on at t = f(t) = 0 0 < t < π in(t) π < t In tep-function form, f(t) = u π in(t) The graph i the olid line in the diagram below: Lu 2 (t 2 4) = e 2 L(t + 2) 2 4 = e 2 Lt 2 + 4t = e 2 L(t 2 + 4t = e 2 ( ) Note the dotted line: thi i in(t) look like over the entire interval, but it i only turned on at t = π. Noting in(t + π) = in(t), Lu π in(t) = e π Lin(t + π) = e π L in(t) ( ) = e π 2 +

2 3. f(t) = 0 < t < 2 t 2 < t In tep-function form, we want the function to turn on at t = 0 and off at t = 2: (u 0 u 2 ). When then want t to turn on at t = 2: tu 2. f(t) = (u 0 u 2 ) + tu 2 The graph i the olid line in the diagram below. Note the dicontinuity at t = 2. L(u 0 u 2 + tu 2 ) = e0 e 2 = e 2 + e 2 L(t + 2) ) + e 2 ( Simplifying, = + e 2 + e < t < 4. f(t) = e t < t < < t Step-function form: f(t) = e t (u u 2 ) The graph i the olid line in the diagram below. Taking the Laplace tranform thi time, don t get confued between exponential in time (tranform to exponential in, which come from tep function in time. t 2 0 < t < 3 5. f(t) = 9 3 < t < < t L(e t u e t u 2 ) = e L(e (t+) ) e 2 Le (t+2) = e Le e t e 2 Le 2 e t = e e + e 2 e 2 + Step-function form: f(t) = t 2 (u 0 u 3 ) + 9(u 3 u 5 ) The graph i the olid line in the diagram below. ), and a 2

3 Lt 2 u 0 t 2 u 3 + 9u 3 9u 5 = e 0 L(t + 0) 2 e 3 L(t + 3) 2 + 9e 3 9e 5 = Lt 2 e 3 Lt 2 + 6t e 3 9e 5 = 2 ( 2 3 e ) + 9e 3 9e 5 Canceling the 9 e 3 / term, = 2 3 e e 3 2 9e 5 t 0 < t < 2 t < t < 3 6. f(t) = t 4 3 < t < < t Step-function form: f(t) = t(u 0 u ) + (2 t)(u u 3 ) + (t 4)(u 3 u 4 ) The graph i the olid line in the diagram below. L(t u 0 t u ) + (2u tu 2u 3 + t u 3 ) + (t u 3 4u 3 t u 4 + 4u 4 ) =Lt u 0 2t u + 2u 6u 3 + 2t u 3 t u 4 + 4u 4 = 2 2e Lt + + 2e 6e 3 + 2e 3 Lt + 3 e 4 Lt e 4 = ( 2 2e 2 + ) + 2e ( 6e 3 + 2e ) ( e 4 ) + 4e 4 = 2 2e 2 + 2e 3 2 e 4 2 Notice how all the contant tranform (/ term) diappear after the hift? That a reult of the function being continuou, and o having no jump dicontinuitie: you ee imilar effect in earlier example a well. For Quetion 7-2, find the invere Laplace tranform of the given function, write the t-domain verion in piecewie form, and ketch the graph of f(t) = L F () 7. F () = e 2 Thee function are in the table: f(t) = L F () = u 2 0 < t < 2 = 0 2 < t 3

4 The graph i hown below: 8. F () = e 3 Thi function i alo in the table: f(t) = L F () = u < t < 3 = 3 < t The graph i hown below: 9. F () = e We have to ue two rule here: 2 L 2 = in(2t) and L e a F () = f(t a)u a + 4 The graph i hown below: f(t) = L F () put a 2 on top for ine form: = 2 L e 2 get hifted invere with tep function: = 2 u in(2(t )) 0 0 < t < = 2 in(2(t )) < t Note that the hift matche up with the tep function, o we get a imple tranlation of the ine graph, with a value of zero before it tart. 0. F () = e 2 In thi example, we ue the coine and tep function rule: 4

5 L 2 = co(2t) and L e a F () = f(t a)u a + 4 The graph i hown below: f(t) = L F () = L e = u co(2(t )) 0 0 < t < = co(2(t )) < t. F () = e In thi example, we have to complete the quare: f(t) = L F () = L e 5 ( ) + 0 = L e 5 ( + ) We know that L 3 ( + ) 2 give e t in(3t), o our tranform i + 9 Put 3 in numerator: f(t) = 3 L e 5 3 ( + ) Get hifted exp/in with tep function: = 3 u 5e (t 5) in(3(t 5)) 0 0 < t < 5 Same function in piecewie form: = 3 e (t 5) in(3(t 5)) 5 < t The graph i hown below: 2. F () = e Thi require almot the ame logic a the lat problem. The only catch i that, becaue of the + group in the 5

6 denominator, we need an + group in the numerator a well to get the correct form for e at co(bt). f(t) = L F () Create + in numerator: = L e 5 (+) ( + ) Split into ditinct co and ine part: = L e 5 + ( + ) e 5 ( + ) Add 3 in numerator for correct ine form: = L e 5 + ( + ) e 5 3 ( + ) Do invere tranform: = u 5 e (t 5) co(3(t 5)) 3 u 5e (t 5) in(3(t 5)) 0, 0 < t < 5 Same in piecewie form: = e (t 5) co(3(t 5)) 3 e (t 5) in(3(t 5)), 5 < t The graph of thee hifted damped ocillation i hown below: For Quetion 3-20, find the olution to the initial value problem by uing Laplace tranform. 3. x + 3x + 2x = 0, x(0) = 0, x (0) = 2 Taking Laplace of both ide, [ 2 X() 0 ( 2) ] +3 [X() 0] +2X() = 0 L(x ) L(x ) Group X() term on left: ( )X() = 2 2 X() = We now try to put the RHS into a form matching the table entrie. Since the denominator can be factored into linear term, we do that and then ue partial fraction to eparate the factor. 2 X() = ( + )( + 2) = A + + B + 2 Solving for A and B give = o, taking invere Laplace of both ide, L X() = L x(t) = 2e 2t 2e t Check: thi atifie x(0) = 0, and (differentiating) that x (0) = = 2. Both function, e 2t and e t, alo atify the original DE, o thi olution atifie both the equation and the initial condition given. 4. x + 2x = 4, x(0) = 0 6

7 Taking Laplace of both ide, [X() 0] +2X() = 4 L(x ) Group X() term on left: ( + 2)X() = 4 X() = 4 ( + 2) We now try to put the RHS into a form matching the table entrie. Since the denominator can be factored into linear term, we do that and then ue partial fraction to eparate the factor. 5. y + 4y =, y(0) = 0, y (0) = 0 X() = 4 ( + 2) = A + B + 2 Solving for A and B give = o, taking invere Laplace of both ide, L X() = L x(t) = 2e 2t 2 Taking Laplace of both ide, [ 2 Y () 0 0 ] +4Y () = L(y ) Group Y () term on left: ()Y () = Y () = () We now try to put the RHS into a form matching the table entrie. Since the denominator can be factored into and, we do that and then ue partial fraction to eparate the factor. Y () = Solving for A, B and C give () = A + B + C = /4 (/4) + 0 = 4 4 o, taking invere Laplace of both ide, L Y () = L 4 4 y(t) = 4 4 co(2t) 7

8 6. y 2y = 4, y(0) = 0, y (0) = 0 Taking Laplace of both ide, [ 2 Y () 0 0 ] L(y ) Group Y () term on left: 2 [Y () 0] = 4 L(y ) ( 2 2)Y () = 4 Y () = 4 ( 2 2) = 4 2 ( 2) = A + B 2 + C Solving for A, B and C give 2 = o, taking invere Laplace of both ide, L Y () = L y(t) = + 2t e 2t Note: if we were to have olved thi problem uing the y c and y p approach, we would have a cae where y c = c (contant olution), o our aumed form for y p would have needed a t multiplier to avoid the overlap. Uing Laplace tranform avoid the need for thi pecial-cae logic when you are building the olution. 7. x 3x = 39 in(2t), x(0) = 2 Taking Laplace of both ide, [X() 2] 3X() = 39 L(x ) Group X() term on left: ( 3)X() = X() = ( 3)() 39 Looking jut at the more complicated right-hand term, ( 3)() = A 3 + B + C Solving for A, B and C give = Combining with the other term give ( ) ( 2 3 X() = ) = o x(t) = 5e 3t 3 co(2t) 9 2 in(2t) 8. d 4 x dt 4 x 4d3 dt 3 + x 6d2 dt 2 4dx dt + x = 0 where x(0) = 0, x (0) =, x (0) = 0, x (0) =. 8

9 Let X() := Lx(t)(). The linearity and time differentiation propertie of the Laplace tranform give 0 = Lx (4) (t)() 4Lx (t)() + 6Lx (t)() 4Lx (t)() + Lx(t)() 0 = ( 4 X() 3 x(0) 2 x (0) x (0) x (0) ) 4 ( 3 X() 2 x(0) x (0) x (0) ) +6 ( 2 X() x(0) x (0) ) 4 ( X() x(0) ) + X() = ( )X() X() = ( ) 4. To find the invere Laplace tranform, we ue a partial fraction decompoition: ( ) 4 = A + B ( ) 2 + C ( ) 3 + D ( ) 4 = A( )3 + B( ) 2 + C( ) + D ( ) 4 = A3 + ( 3A + B) 2 + (3A 2B + C) + ( A + B C + D) ( ) 4 o A = 0, B =, C = 2, and D = 4. Uing the Laplace tranform of the n-th power and the frequency hift property, we obtain X() = ( ) 2 2 ( ) ( ) 3! 3 ( ) 4 and x(t) = te t t 2 e t t3 e t. 9. y + y = e t + t +, y(0) = y (0) = y (0) = 0. Let Y () := Ly(t)(). The linearity and time differentiation propertie of the Laplace tranform together with the Laplace tranform of the exponential function and n-th power give Ly (t)() + Ly (t)() = Le t () + Lt() + L() ( 3 Y () 2 y(0) y (0) y (0) ) + ( 2 Y () y(0) y (0) ) = = 22 2 ( ) Y () = To find the invere Laplace tranform, we ue a partial fraction decompoition: ( )( + ) = A + B 2 + C 3 + D 4 + E + F ( )( + ) = (A + E + F )5 + (B + E F ) 4 + ( A + C) 3 + ( B + D) 2 C D 4 ( )( + ) o D =, C = 0, B =, A = 0, E = 2, and F = 2. Uing the Laplace tranform of the n-th power and exponential function, we obtain Y () = 2 + ( ) 3! ( ) ( ) and y(t) = t t3 + 2 et 2 e t. 20. f (t) + 2f (t) + f(t) = 4e t, f(0) = 2, f (0) =. Let F () := Lf(t)(). The linearity and time differentiation propertie of the Laplace tranform combined with the Laplace tranform of the exponential function give 4Le t () = Lf (t)() + 2Lf (t)() + Lf(t)() 4 + = ( 2 F () f(0) f (0) ) + 2 ( F () f(0) ) + F () = ( )F () 2 3 F () = ( + ) 2 ( ) = ( + ) 3. 9

10 To find the invere Laplace tranform, we ue a partial fraction decompoition: ( + ) 3 = A + + B ( + ) 2 + C ( + ) 3 = A( + )2 + B( + ) + C ( + ) 3 = A2 + (2A + B) + (A + B + C) ( + ) 3 o A = 2, B =, and C = 4. Uing the Laplace tranform of the n-th power and and the frequency hift property, we obtain ( ) ( ) F () = ( + ) ( + ) 3 and f(t) = 2e t + te t + 2t 2 e t. 0