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先解y'-y=0 dy/y=dx ln|y|=x+C y=Ce^x 用常數變易法設y=ue^x y'=u'e^x+ue^x y'-y=u'e^x=2xe^(2x) 即u'=2xe^x 得u=2xe^x-2e^x+C 所以y=(2xe^x-2e^x+C)e^x 代入x=0,y=1,得C=3 所以y=(2xe^x-2e^x+3)e^x
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