20 through 22 indicate other ways of finding the second

Chapter 3, Problem 22

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20 through 22 indicate other ways of finding the second solution when the characteristic equation has repeated roots.(a) If ar2 + br + c = 0 has equal roots r1, show thatL[ert] = a(ert)+ b(ert)+ cert = a(r r1)2ert. (i)Since the right side of Eq. (i) is zero when r = r1, it follows that exp(r1t) is a solution ofL[y] = ay+ by+ cy = 0.(b) Differentiate Eq. (i) with respect to r, and interchange differentiation with respect tor and with respect to t, thus showing thatrL[ert] = Lrert= L[tert] = atert(r r1)2 + 2aert(r r1). (ii)Since the right side of Eq. (ii) is zero when r = r1, conclude that t exp(r1t) is also a solutionof L[y] = 0.