Exercises 5.2Substituting these expressions into the nonhomogeneous equation given in (5.3) yields2x0p−2xp=2(−Asint+Bcost)−2(Acost+Bsint)=(2B−2A)cost+(−2A−2B)sint=−cost.By equating coeﬃcients we obtain2B−2A=−1and−2A−2B=0.By solving these two equations simultaneously forAandB,weseethatA=14andB=−14.Thus, a particular solution to the nonhomogeneous equation given in (5.3) will bexp(t)=14cost−14sintand a general solution to the nonhomogeneous equation (5.3) will bex(txh(t)+xp(tC1et+14cost−14sint.We now must Fnd a functiony(t). To do this, we subtract the second of the two di±erential
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.