Exercises 5.5which is a quadratic inr2.Sor2=−23±√529−2406=−23±176.Since−20/3and−1 are negative, the roots of the characteristic equation are±iβ1and±iβ2,whereβ1=r203,β2=1.Hencex(t)=c1cosβ1t+c2sinβ1t+c3cosβ2t+c4sinβ2t.Solving the Frst equation of the system given in (5.28) fory,wegety(t14(D2+4)[x]=14±(−β21)c1cosβ1t+(−β21)c2sinβ1t+(−β22)c3cosβ2t+(−β22)c4sinβ2t².Next we substitute into the initial conditions. Settingx(0) =−1,x0(0) = 0 yields−1=c1+c3,0=c2β1+c4β2.±rom the initial conditionsy(0) = 0,y0(0) = 0, we get
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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.