Proof for differentiable functions

Question

Find all positive differentiable functions $f$ that satisfy $$\int_0^x \sin(t) f(t) dt = [f(x)]^2-1$$for all real numbers $x$.

So, it appears that this question has been asked in the past (at this link Find all positive differentiable functions $f$ that satisfy $\int_0^x \sin(t) f(t) dt = [f(x)]^2.$). However, given the hint that this user provided in the previous post, I believe I am still unsure how to go proceed. Also, as the previous poster said, this is the correct problem and I believed people answered about the incorrect problem. My reasoning is that if you differentiate both sides, you can get a function f such that it will only be positive. However, I’m not sure that this I’m fact works. Can someone please help? Thanks. (I’m also sorry about my previous posts - please don’t downvote anymore)

Answer 1

hint

Differentiate both sides to get

$$f(x)\sin(x)=2f(x)f'(x)$$

If $f(x)\ne 0$ then

$$f'(x)=\frac{\sin(x)}{2}$$ at a certain intervall.

$$f(x)=\frac{-\cos(x)}{2}+C$$

plugg it to get $C$.

Answer 2

(I use Fatima's solution)

Usually, we continue as follows: $f(x)=-\frac{\cos x}{2} + C$, plug it in the given equation

$$\int_0^x \sin t \left(-\frac{\cos t}{2} + C\right) dt=\left(-\frac{\cos x}{2} + C\right)^2-1.$$ The integral is easy to compute, but there is a problem with $C$ which is assumed to be a constant. If I am not wrong, we get $$\frac{\cos(2x)}{8}-C\cos x-\frac{1}{8}+C=\frac{(\cos x)^2}{4}-C\cos x +C^2-1.$$

It seems to me that the equation does not have other solution than $f(x)=0.$

Answer 3

@Maam Should be $$\frac{\cos^2(x)}{4}-C\cos x-\frac{1}{4}+C=\frac{(\cos x)^2}{4}-C\cos x +C^2-1.$$

Did you use $u=-\cos(t) $ and integrate $ (\frac{u}{2} + C ) du $ ?

$C^2 -C -\frac{3}{4} $ so the only solution that works to make $f(x)$ positive is $C= \frac {3}{2}$

Note: $f(x) =0 $ is not a solution as its not a positive function.