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Dear student put y = 0then you will get f(0) ( 1 + f(x)) = 0 In case of any difficulty, Please feel free to ask again
Dear studentAssume y = 0 to simplify the given equation and determine requiredRegards
Dear student, above answer is wrong. Correct answer is:On putting y=0, we see that f(0)(1+f(x))= 0Since f(x) is non constant, f(0)= 0Now, we can writef(x+y)-f(x)= f(y)(1+f(x))Or Lt y tends to 0 [f(x+y)-f(x)]/y = (1+f(x))*Lt y tends to 0 [f(y)-f(0)]/(y-0)Or f'(x)= f'(0)(1+f(x))= k(1+f(x))This differential equation can be easily solved by integrating as∫f'(x)dx/(1+f(x))= ∫kdxOr ln(1+f(x))= kx+cOr f(x)= Ae^kx-1Put x=0 to get f(0)= A-1Or A=1So, f(x)= e^kx - 1KINDLY APPROVE :))
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