H. Heaviside’sCover-upMethod

1 H. heaviside s Cover-up MethodThe cover-up method was introduced by Oliver heaviside as a fast way to do a decom-position into partial fractions. This is an essential step in using the Laplace transform tosolve differential equations, and this was more or less heaviside s original cover-up method can be used to make a partial fractions decomposition of a rationalfunctionp(x)q(x)whenever the denominator can be factored into distinct linear first show how the method works on a simple example, and thenshow why it 7(x 1)(x+ 2)into partial know the answer will have the form(1)x 7(x 1)(x+ 2)=Ax 1+Bx+ determineAby the cover-up method, on the left-hand side we mentally remove (or coverup with a finger)

2 The factorx 1 associated withA, and substitutex= 1 into what s left;this givesA:(2)x 7(x+ 2) x=1=1 71 + 2= 2 =A .Similarly,Bis found by covering up the factorx+ 2 on the left, and substitutingx= 2into what s left. This givesx 7(x 1) x= 2= 2 7 2 1= 3 =B .Thus, our answer is(3)x 7(x 1)(x+ 2)= 2x 1+3x+ does the method work? The reason is simple. The right way to determineAfromequation (1) would be to multiply both sides by (x 1); this would give(4)x 7(x+ 2)=A+Bx+ 2(x 1).Now if we substitutex= 1, what we get is exactly equation (2), since the term on the rightdisappears. The cover-up method therefore is just any easy way of doing the calculationwithout going to the fuss of writing (4) it s unnecessary towrite the term containingBsince it will become 0.

3 0H. COVER-UP METHOD1In general, if the denominator of the rational function factors into the product of distinctlinear factors:p(x)(x a1)(x a2) (x ar)=A1x a1+..+Arx ar, ai6=aj,thenAiis found by covering up the factorx aion the left, and settingx=aiin the restof the xinto partial ,x3 x=x(x2 1) =x(x 1)(x+ 1).By the cover-up method,1x(x 1)(x+ 1)= 1x+1/2x 1+1/2x+ be honest, the real difficulty in all of the partial fractions methods (the cover-upmethod being no exception) is in factoring the the programs which dosymbolic integration, like Macsyma, or Maple, can only factor polynomials whose factorshave integer coefficients, or easy coefficients like 2.

4 And therefore they can only integraterational functions with easily-factored s cover-up method also can be used even when the denominator doesn t factorinto distinct linear factors. To be sure, it gives only partial results, but these can often bea big help. We + 6(x2+ 4)(x 2). write(5)5x+ 6(x2+ 4)(x 2)=Ax+Bx2+ 4+Cx first determineCby the cover-up method, gettingC= 2 . ThenAandBcan be foundby the method of undetermined coefficients; the work is greatly reduced since we need tosolve only two simultaneous equations to findAandB, not this plan, usingC= 2, we combine terms on the right of (5) so that both sideshave the same denominator.

5 The numerators must then also be equal, which gives us(6)5x+ 6 = (Ax+B)(x 2) + 2(x2+ 4).Comparing the coefficients say ofx2and of the constant terms on both sides of (6) thengives respectively the two equations0 =A+ 2and6 = 2B+ 8,from whichA= 2 andB= 1 .In using (6), one could have instead compared the coefficientsofx, getting 5 = 2A+B;this provides a valuable check on the correctness of our values NOTESIn Example 3, an alternative to undetermined coefficients would be to substitute twonumerical values forxinto the original equation (5), sayx= 0 andx= 1 (any valuesother thanx= 2 are usable). Again one gets two simultaneous equations method requires addition of fractions, and is usually better when only one coefficientremains to be determined (as in Example 4 below).

6 Still another method would be to factor the denominator completely into linear factors,using complex coefficients, and then use the cover-up method,but with complex numbers. Atthe end, conjugate complex terms have to be combined in pairsto produce real calculations are sometimes longer, and require skill with complex cover-up method can also be used if a linear factor is repeated, but there too it givesjust partial results. It applies only to thehighest power of the linear factor. Once again, (x 1)2(x+ 2). write(7)1(x 1)2(x+ 2)=A(x 1)2+Bx 1+Cx+ findAcover up (x 1)2and setx= 1; you getA= 1/3. To findC, cover upx+ 2, andsetx= 2; you getC= 1 leavesBwhich cannot be found by the cover-up method.

7 But sinceAandCarealready known in (7),Bcan be found by substituting any numerical value (other than1 or 2) forxin equation (7). For instance, if we putx= 0 and remember thatA= 1/3 andC= 1/9, we get12=1/31+B 1+1/92,from which we see thatB= 1 also be found by applying the method of undetermined coefficients to the equation(7); note that sinceAandCare known, it is enough to get a single linear equation in orderto determineB simultaneous equations are no longer fact that the cover-up method works for just thehighest powerof the repeated linearfactor can be seen just as before. In the above example for instance, the cover-up methodfor findingAis just a short way of multiplying equation (5) through by (x 1)2and thensubstitutingx= 1 into the resulting Ordinary Differential Notes and Exercisesc Arthur Mattuck and 1988, 1992, 1996, 2003, 2007, 20111