Differentiating an integral and finding f(x)

[supermiedos]

Homework Statement

Find all f(x) satisfying:
∫f dx ∫1/f dx = -1

Homework Equations

The Attempt at a Solution

I solved for ∫1/f dx and differentiated both sides (using the quotient rule for the right side):
∫1/f dx = -1 / ∫f dx
1/f = f / (∫f dx)²
(∫f dx)² = f²
∫fdx = ±f

f = ±f'

Solving the differential equation for f = f ' I get f = ce^x

But when I try to prove if my solution is correct, I got:

∫ce^x dx ∫ 1/(ce^x) dx = -1
(e^x + k₁)(-e^-x + k₂) = -1
And I don't know what to do to get -1 on the left side.

Could you give me a hint please?

 

[SammyS]

Homework Statement

Find all f(x) satisfying:
∫f dx ∫1/f dx = -1

Homework Equations

The Attempt at a Solution

I solved for ∫1/f dx and differentiated both sides (using the quotient rule for the right side):
∫1/f dx = -1 / ∫f dx
1/f = f / (∫f dx)²
(∫f dx)² = f²
∫fdx = ±f

f = ±f '

Wow! It's hard to see the ' on ƒ '

Solving the differential equation for f = f ' I get f = ce^x

But when I try to prove if my solution is correct, I got:

∫ce^x dx ∫ 1/(ce^x) dx = -1
(e^x + k₁)(-e^-x + k₂) = -1
And I don't know what to do to get -1 on the left side.

Could you give me a hint please?

It's been edited. (Sorry for messing up that quote originally.)
It works if k₁ = k₂ = 0 .

Also, where did the constant, c, go in your last line?

 

[supermiedos]

Wow! It's hard to see the ' on ƒ '

Solving the differential equation for f = f ' I get f = ce^x

But when I try to prove if my solution is correct, I got:

∫ce^x dx ∫ 1/(ce^x) dx = -1
(e^x + k₁)(-e^-x + k₂) = -1
And I don't know what to do to get -1 on the left side.

Could you give me a hint please?[/QUOTE]
It works is k₁ = k₂ = 0 .

But is it legal to do that? Giving values to fit the desired result?

Also, where did the constant, c, go in your last line?[/QUOTE]
"c" canceled with the another "c" in the denominator of the second integral.

 

[SammyS]

"c" canceled with the another "c" in the denominator of the second integral.

c does not cancel if k₁, k₂ ≠ 0 .

You're right.

I was missing the fact that you can factor c out of the integrals or equivalently, you can have the integration constants "absorb" c.