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a) given thatwhere I is two cunent through finductor andv is the noltage drop accross capacitor for eigen valueslet A=([0,(1)/(L)],[(-1)/(C),(-1)/(RC)])put |A-lambda I|=0{:[|[-lambda,1//L],[-1//C,-1//RC]|=0],[=>-lambda(-1//RC-lambda)+(1)/(LC)=0],[=>lambda^(2)+lambda((1)/(RC))+(1)/(LC)=0],[=>lambda=-(((1)/(RC))+-sqrt((1)/(R^(2)C^(r))-(4)/(LC)))/(2)=((-(1)/(RC))+-sqrt((L-4R^(2)C)/(R^(2)C^(2))))/(2)],[lambda=-(2)/(RC)+-(1)/(2RC)sqrt(1-(4R^(2)C)/(L))]:}now tw eign values will be red and diftarent it1-(4R^(2)c)/(L) > 0i.e if L > 4R^(2)C twn L-4R^(2)C > 0, so lambda is real and diline And tw eigen values are Complex Conjugotive if 1-(4R^(2)C)/(L) < 0 i.e it L/_4R^(r)CHence provedb) Now when R=1ohm,c=(1)/(2) forad, L=1 henry. then the System(d)/(dt)((I)/(v))=([0,1],[-2,-2])([I],[v])-(1)we take ([I],[v])=x.from ean (1)x^(')=([0,1],[-2,-2])xletus assume that x=xixi^(n). then becan obtiain Th Alge beric eysten.([-pi,1],[-2,-2-pi])([xi],[xi_(2)])=([0],[0])-(3)Tc ... See the full answer