Function Home › Forums › Math Olympiad, I.S.I., C.M.I. Entrance › Function Tagged: Function This topic contains 1 reply, has 2 voices, and was last updated by Alpha Beta 2 weeks, 3 days ago. Viewing 2 posts - 1 through 2 (of 2 total) Author Posts June 30, 2019 at 6:50 pm #29479 Aniruddha BardhanParticipant Please see the attached file. This topic was modified 2 weeks, 4 days ago by Aniruddha Bardhan. Attachments:You must be logged in to view attached files. July 1, 2019 at 5:23 am #29491 Alpha BetaParticipant Are you sure it’s not g'(x)-g(x)-2e^x=0 ? Partially differentiating the equation with respect to x and y separately, we get g'(x+y)=g'(x)e^y+g(y)e^x and, g'(x+y)=g(x)e^y+g'(y)e^x so, g'(x)e^y+g(y)e^x=g(x)e^y+g'(y)e^x or, [g'(x)-g(x)]e^y=[g'(y)-g(y)]e^x suppose, for some x, g'(x)=g(x). Then, for all y, 0=[g'(y)-g(y)]e^x. Since, e^x is always positive, g'(y)-g(y)=0, implying g(y)=e^(y+c) for all y, and some constant c. From g'(0)=2, we get g(y)=2e^y for all y. From the equation, 2e^(x+y)=4e^(x+y) for all x,y. Or, 2e^(x+y)=0, which is impossible. Therefore, g'(x)-g(x) is non-zero for all x. or, [g'(y)-g(y)]/[g'(x)-g(x)]=e^y/e^x, for all x,y. So, g'(x)-g(x)=ke^x for all x, where k is a constant. Solving, and imposing the 2 conditions ( g'(0)=2, and g(x+y)=g(x)e^y+g(y)e^x), we get g(x)=2xe^x, implying, g'(x)-g(x)-2e^x=0. Author Posts Viewing 2 posts - 1 through 2 (of 2 total) You must be logged in to reply to this topic.