3y′′+y′−14=0The auxiliary equation is3m2+m−14m1=2,m2=3−7Hence, the general solution isy=Ae2x+Be3−7x∴y′=2Ae2x−37Be3−7xGiven,y(0)=1,y′(0)=−1substituting the given conditions, we have1=A+BB=1−AAlso,−1=2A−37B−3=6A−7+7A13A=4A=134,B=139Hence, the particular solution isy=134e2x+139e−37x
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