ESE 318-02, Fall 2014Lecture 2, Aug. 28, 2014PFE continued, ODEs with Laplace, Unit StepSlight Review. An inverse transform problem that does not require PFE. 45524!4!4tebsbasbatassOr simply use Line 17 of Table in Zill.Example where PFE not required (recall the property for f(t)eat.                     261223612224!31324324242424212!32221212122121222232221221ttteetettesssssssssssssssssttttPartial Fraction Expansion (PFE), slight review. Can you “decompose” the following fraction into a sum of fractions (Step 2 of “the process”)?   52439738222322358sssssssssHere is the answer:   52433322322ssIHssGFssEsDsCsBsAPFE continued. We were doing Partial Fraction Expansion (PFE) as a method to find theInverse Laplace Transform, and were up to “Step 3” of the process.Step 3: Determine the numerators. The overall method is to equate numerators. There is aquicker way in some cases, called the “cover-up” method.3a, Equating numerators. The individual fractions from Step 2 are multiplied out to get a common denominator that matches the original. Then the resulting numerator (with the TBD constants A, B, etc.) are equated to the original numerator by matching like powers of s. The result will be multiple equations in multiple unknowns.ESE 318-02, Fall 2014        tteesssssBAABAsBAsBBsAAssBsAssBsAssssF32121212121210131113121232:1:313231312Yes, this can get tedious. At least there is a short-cut for some cases.3b, Cover-Up Method. In the case of distinct (non-repeated) real roots, we can use the “cover up” method. Consider the same case as above.    ttsseesssssssBssAsBsAsss32121212131311131221211323122131213231312We find the value by “covering up” the term in the original F(s) that corresponds to that denominator, and then plugging in for s the value that makes that denominator equal zero.Thus in the above, in finding A we cover up the (s+1) term of F(s) and evaluate what’s left at s = -1. To find B we cover up the (s+3) term and evaluate at s = -3.We can also use the cover-up method for the highest power term of a repeated linear factor, as in the following example.      211229124322132331312322122222ssssCssAsCsBsAsssHowever, we still need to equate numerators to find B in this example.The “cover up” method does not work with quadratic/complex factors or with lower-powered terms of real roots that are repeated, such as (s+3) in the above example.Step 4, Handling the Quadratic/Complex Terms. We use a complete-the-square method to manipulate quadratic terms into terms that match those of The Table of Laplace transforms.ESE 318-02, Fall 2014   ktekaskktekasasTableatatsincos:2222Consider the term below (which could be just one term of an expanded fraction resulting from PFE).      tetesssssssssstt2sin2cos241223411241312415252522322222Steps: (a) Complete the square in the denominator. (b) Manipulate the s term in the numerator so that part of the numerator matches the squared term in the denominator. (c) Manipulate the constant remainder so that it is a factor of the square root of the constant term of the denominator. (d) Inverse transform into eatcos(kt) and eatsin(kt) forms as seen in the Table.Zill 4.2.20. We’ll use the “cover-up” method in this case because all roots are real and distinct.  ttsseesssssBsAsBsAssss5914912542519141912019141915154541201Slight review, PFD. Here is a problem that could be approached 3 different ways.     323233332sCsBsAssFirst, the cover-up method can be used, but only to find C (associated with highest power of repeated real root).111291232ssCSecond method, equate numerators, where we could still use the cover-up method for C alone.   116123906123962332222CBACBABAAsCBBsAAsAssCsBsAThird: repeated break-down of the numerator to put in terms of s + 3.           323232323113631311363376332sssssssssssESE 318-02, Fall 2014In my experience, equating numerators is the easier method if not all roots are real and distinct.Transforms of derivatives. Consider the Laplace transform of the derivative of a function. Use the definition of the transform, and integration by parts.                                   0lim00lim000000tfeiffssFtftfsftfedttfestfedtsetftfedttfetfdttfdvdtsedutfveudttdftfdttfetfsttsttststststdvuststststLLLLWe had to assume that f(t)e-st goes to zero as t goes to infinity; a piecewise continuous function of exponential order satisfies that. We could continue along this line to get the general formula for higher derivatives as well (assuming all the lower order derivatives and f itself are piecewise continuous and of exponential order).                    00000)1(21)(2nnnnnnnffsfssFstftfdtdfsfsFstfLLLSolving differential equations using Laplace transforms. We can now solve linear differential equations with constant coefficients and intial conditions. There are three general steps:(1) Transform