Chapter 5⇒A=0,B=−1⇒xh(t)=−sin 2tandx(txh(t)+xp(tc1cost+c2sint+c3cos√6t+c4sin√6t−sin 2t.From the frst equation in the original system, we havey(t12(x0+5x)=2c1cost+2c2sint−12c3cos√6t−12c4sin√6t−12sin 2We determine constantsc1andc3using the initial conditionsx(0) = 0 andy(0) = 1.0=x(0) =c1+c3,1=y(0) = 2c1−c3/2⇒c3=−c1,2c1−(−c1)/2=1⇒c3=−2/5,c1/5.To fndc2andc4, computex0(t)andy0(t), evaluate these ±unctions att= 0, and use the othertwo initial conditions,x0(0) =y0(0) = 0. This yieldsx0(0) =c2+√6c4−2,y0(0) = 2c2−√6c4/2−1⇒c4=√6/5,c2=4/5.There±ore, the required solution isx(t25cost+45sint−25cos√6t+√65sin√6t−sin 2t,y(t45cost+85sint+15cos√6t−
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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.