College of Engineering and Computer ScienceMechanical Engineering DepartmentMechanical Engineering 501ASeminar in Engineering AnalysisFall 2004 Number: 17472 Instructor: Larry CarettoOctober 26 Homework SolutionsPage 257, problem 15 – Find the Laplace transform of a plot which is a straight line from (0,1) to (1,0.5). Show the details of your work.The line has a slope of -0.5 and an intercept of 1 so its equation is f(t) = -t/2 + 1. If we plug this function into the definition of the Laplace transform we obtain the following (after using an integraltable for the integral of te-stdt.)     10)(11)(2111)(2121)0(21)()(221021010101100sssesesstsesedttedtedtedtetdtetfsFssstststststststRearranging to collect common terms gives the following result.2222221222122121221)(sesessesesssesesessFsssssssPage 257, problem 19 – Given F(s) = L[f(t)] = (-s-10)/(s2 – s – 2), find f(t). Show the details.We can factor the denominator of F(s) and rearrange it into two fractions using the two components in the numerator to obtain transforms that are in the transform table.)1)(2(110)1)(2(210)(2sssssssssFThe fractions on the right-hand side are found as transforms 10 and 11 in the table on page 297 of Kreyszig. Using these transforms, with a = 2 and b = -1 give the following result for f(t),   tttttttteeeeeeeetf2222433912)1(2110)1(2)1(21)( Page 257, problem31 – Find the Laplace transform of 5e2tsinh 2t. (Show the details.)Here we can use the first shifting theorem which lets us write the Laplace transform of eat f(t) in terms of the Laplace transform, F(s), of f(t). This theorem tells us that L[eatf(t)] = F(s – a). That is we take the original Laplace transform of f(t) and replace s by s – a everywhere in that transform to get the Lapalce transform of eatf(t). Since the Laplace transform of sinh at is a2/(s2 – a2), the Laplace transform of 5e2tsinh 2t is  )4(10222522sssPage 257, problem 44. If L[f(t)] = F(s) and c is any positive constant, show that L[f(ct)} = F(s/c)/c. Use this result to obtain L(cos wt) from L(cos t).We can start with the definition of the Laplace transform and use this definition to write the transform of f(ct).Engineering Building Room 1333 Mail Code Phone: 818.677.6448Email: [email protected] 8348 Fax: 818.677.706200)()]([)()]([)( dtectfctfdtetftfsFststWe can integrate this equation by the substitution y = ct so that t = y/c and dt = dy/c. Because c is a positive constant the limits of t = 0 and t =  are equivalent to y = 0 and y = . Making these substitutions in the integral and defining a new parameter k = s/c gives.00/0)(1)/()()()]([ dyeyfccydeyfdtectfctfkycsystThe final integral in this step is simply the definition of the Laplace transform except the dummy variable of integration is changed from t to y and the parameter s is replaced by the parameter k = s/c. Thus this integral gives the Laplace transform F(k) = F(s/c). Since the integral is divided by c, we have the desired result that L[f(ct)] = F(s/c)/c.To apply this result to obtain L(cos wt) from L(cos t), we start with the transform in the table for cos t and set  = 1. As we see below, this gives the correct result for L(cos t).2222111)(cos1)()(cossssssFtsssFtPage 273, problem 23 – Solve the following differential equation y’’ + 9y = 8 sin t if 0 < t <  and 0 if t > . y(0) = 0, y’(0) = 4. Use Laplace transforms. (Show the details.)Here we can use the Heaviside function (the unit step function), u(t – a) to solve one equation that includes the discontinuity on the right-hand side.ttuydtydsin8)](1[922Taking Laplace transforms of this differential equation gives]sin)([sin8)(9)0(')0()(2ttutsYysysYs It looks like we can use the second shifting theorem for the product of the sine and the unit step function, but we can only do this if the argument of the sine function is the same as the argument of the unit step function, t – p. We can use the periodicity of the sine to write sin t as –sin(t – ). Doing this, using the second shifting theorem to get the Laplace transform of the right side and substituting the initial conditions, y(0) = 0, y’(0) = 4, to get an following equation for Y(s).22222221)1(81)1(8)]sin()([81)1(8)(94)(sesttussYsYss18911891941818491)(22222222sesssssesssYssTo get the inverse, we first notice that the term 4/(s2+9) has the form of the Laplace transform of the sine, /(s2 +  2) with  = 3. So the inverse of the first term is (4/3) sin 3t. The second term can be treated by the method of partial fractions. Here we have repeated complex factors so we write the partial fractions as follows.Engineering Building Room 1333 Mail Code Phone: 818.677.6448Email: [email protected] 8348 Fax: 818.677.70621918912222 sDCssBAsssMultiplying this equation by (s2 + 9)(s2 + 1) gives.    DCsDsCsBAsBsAssDCssBAs99918232322Equating coefficients of like terms in this equation gives.8909000123DBsCAsDBsCAsThe first and third equations are satisfied only if A = C = 0. The second and fourth equations are satisfied if D = –B = 1. We then have the following result and inverse transform for the second term.ttsssssin3sin3111911891221221We see that the third term in our equation for Y(s) is simply the second term multiplied by the same term whose inverse we just found above. We can obtain the inverse of such a term from the second shifting theorem, shown below, where F(s) is the Laplace transform of f(t).)()()]([1atfatusFeasApplying this theorem to the final term in the Y(s) equation