Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Live 1-1 coding classes to unleash the creator in your Child
Please solve it. ____________________________________________
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 :))
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -