be a differentiable function satisfying f(y).f(x-y) = f(x)  and f '(5) =q and

f(0) / f '(0) = 1 / p then find the value of f(+5)

f(y). f(x-y) = f(x)



put y=x

f(x).f(0) - f(x) =0

f(x)[f(0)-1]=0

as f(x) is not equal to zero

so, f(0) = 1



nw, f(0)/f`(0) = 1/p

so , f`(0) = p



f(y).f(x-y) = f(x)

nw keeping y as const differentitate it

f(y). f`(x-y) = f`(x)

nw put y=x

f(x) . f`(0) = f`(x)

f`(x)/f(x) = p

integrate it



log f(x) = px+c



f(x) = Ce^(px)

f`(x) = pCe^(px)

f`(x) = pf(x)

f`(5) = pf(5)



f(5) = q/p

thats the answer..........



 

An alternative method:

Given that f(x) = f(y) f(x-y)

Put x = y = 0

we get f(0) = f²(0). So, f(0) = 1 or f(0) = 0

But if f(0) = 0, we get f(x) = 0 for all x, and in any case it contradicts  as f'(0) and p are both real numbers.

Hence f(0) = 1

Now let

Since f is differentiable everywhere,

Actually, among functional equations this is counted among the fundamental ones known as Cauchy Functional Equations:

They are

1.f(xy) = f(x) + f(y) for which the general solution as f(x) = f'(1) ln|x|

2. f(x+y) = f(x) + f(y) with the general solution f(x) = cx

3. f(xy) = f(x) f(y) with the general solution f(x) = |x|^c

4. f(x+y) = f(x) f(y) with the general solution f(x) = a^x

It sometimes helps to convert a given functional equation into one of these standard forms as the solutions are already known.

Here we are given that f(x) = f(y) f(y-x).

You can rewrite it as f(y + y-x) = f(y) f(y-x) and recognise that this is in the standard form No.4 given above, which means the function is f(x) = a^x

Now, f'(x) = a^x log a = f(x) log a.

tells us that log a = p

Hence

So you can see how much it simplifies life

Sir,Good explanation.I think you meant to rewrite as

f(y+x-y)=f(y).f(x-y) for it to be in the standard form.

And f(5)=f'(5)/loga=q/p.

I think both these are typo errors but just want to tell them.

thanks for pointing them out. Unfortunately, if I try to edit them now, some huge line breaks will be inserted