Exercises 4.5Thus a general solution on (−π/(2V),π/(2V)) isy1(t)=yh(t)+yp(tC1cost+C2sint+11−V2cosVt.(4.3)Sincey(t)≡0fort≤−(2V), the initial conditions for the above solution arey1±−π2V²=y01±−π2V²=0.From (4.3) we obtainy1±−π2V²=C1cos±−π2V²+C2sin±−π2V²y01±−π2V²=−C1sin±−π2V²+C2cos±−π2V²+V1−V2.Solving the system yieldsC1=VV2−1sinπ2V,C2=VV2−1cosπ2V,andy1(tVV2−1sinπ2Vcost+VV2−1cosπ2Vsint+11−V2cosVt=VV2−1sin±t+π2V²−1V2−1cosVt,−π2V<t<π2V.Fort>π/(2V) given equation is homogeneous, and its general solution,
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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.