What is n in this case?

Question

If f'(x)= [f(x)]² and f(0)=1 then f(6)= 1/n for some integer n

Answer 1

The given equation is f'(x) = [f(x)]².

If we identify f(x) as y, then the equation reads

dy/dx = y²

or

y^-2dy = dx

Integrating, we get

-1/y = x + C

It is given that y = 1, when x is 0. Therefore,

-1/1 = 0 + C ⇒ C = -1

So, we have

-1/y = x - 1

or

y = 1/(1-x)

Reverting back to f(x), we write

f(x) = 1/(1-x)

Setting x = 6, we get

f(6) = 1/(1-6)

f(6) = 1/(-5)

So, n = -5.