Function

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  • #29479

    Aniruddha Bardhan
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    Please see the attached file.

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    #29491

    Alpha Beta
    Participant

    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.

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