Prove that $\sin$ and $\cos$ are the only two functions which satisfy $f'(x) + \int_0^x f(t)\ dt = 0$for all real numbers $x$. Prove that $\sin$ and $\cos$ are the only two functions which satisfy $$f'(x) + \int_0^x f(t)\ dt = 0$$
for all real numbers $x$.
we do not know that $f(x$) is twice differentiable
tried: $$F(x) = \int_0^x f(t)\ dt$$
$$F'(x) = f(x)$$
$$F''(x)= f'(x)$$
so, writing the original formula in terms of $F(x)$ we have
$$F''(x)+F(x)=0$$
$$F(x) = a \cos x + b  \sin x$$ where $a, b$ = $F'(0)$and $F(0)$ respectively
$$= f(0) \cos x + F (0) \sin x$$
am i right up to here? where do i go from here?
 A: All of the previous answers start by taking a derivative (fully justified by copper.hat and others in comments), and then solving a differential equation.
But there is another issue here:  Taking the derivative of such an equation can introduce extraneous solutions.
Here's a stupid example:  The equation $y = 0$ has a unique solution, $f(x) = 0$.  On the other hand, we can differentiate this, getting the differential equation $y' = 0$.  This, now, has solution $f(x) = C$ for any constant.  Of course, this only solves the original equation when $C = 0$.
A similar thing occurs on this problem.  Yes, the general solution to the differential equation everyone has obtained is $f(x) = \alpha \sin x + \beta \cos x$ for some complex coefficients $\alpha$ and $\beta$.
But checking the original condition, with $x=0$ plugged in, we get $\alpha\cos(0) -\beta\sin(0) = 0$, which forces $\alpha = 0$.
So, the final answer is that the solutions all have the form $f(x) = \beta \cos x$.  One can easily check that all of these do in fact solve the problem.
A: $$f'(x)=-\int_0^xf(t){\rm d}t\implies f''(x)+f(x)=0$$
So auxiliary equation is:
$$r^2+1=0\implies r=\pm i\implies f(x)=ae^{ix}+be^{-ix}\implies f(x)=\alpha \sin x+\beta \cos x$$
putting back we get $\alpha=0\implies f(x)=\beta\cos x$
A: Differentiating we get that:
$$f''(x)+f(x)=0$$
Do you have an idea how we could use this?
A: If $f$ is differentiable (which is implied by the fact that
$f'(x) = -\int_0^x f(t) dt$), then it is continuous and
hence $f'$ is differentiable by the fundamental theorem of calculus.
This gives $f''(x) = -f(x)$, and it follows from this that $f$ is, in fact,
smooth.
The system is linear.
The uniqueness theorem for ODEs tells us that there is a unique solution passing through the 'point' $(f(0), f'(0))$. Since $\sin, \cos$ are solutions passing through $(0,1), (1,0)$ respectively, then the solution is
$f(0) \cos + f'(0) \sin$.
Following Jason's astute observation, we note that $f'(0) = 0$.
Here is a marginally different approach:
Again, the system is linear, and $f$ is smooth as above. It is easy to verify that $f(0) \cos$ is a solution. Suppose  $g$ is a solution (with $g(0) = f(0)$), then we would like to verify that $f=g$.
Let $\phi = f-g$ and pick some $0<\lambda <1$ and let $L = \sup_{|x| \le \lambda} |\phi'(x)|$. Since $\phi(0) = 0$, the mean value theorem gives
$|\phi(x)| \le L \lambda$ for $|x| \le \lambda$. The equation gives
$|\phi'(x)| \le |\int_0^x |f(t)| dt| \le L \lambda ^2$ for $|x| \le \lambda$, that is $L \le \lambda^2 L$, which gives $L = 0$. We can repeat this analysis for $x \mapsto \phi(x \pm \lambda)$, etc.,  to show that $\phi = 0$.
A: To specifically address your attempt:
Everything is correct up to (and including) the point where you found that $F(x)$ must have the form $a \cos x + b \sin x$ for some $a$ and $b$. However, the next claim that $a = F'(0)$ and $b = F(0)$ is wrong; it should be the other way around.
The job of finding the corresponding form of $f$ should be straightforward.
However, we're not done yet. As it turns out, the claim in the title is wrong: the function $\sin$ is not a solution to the integro-differential equation, and the solution space is one-dimensional, not two-dimensional. One generally applicable way of finding this out is to substitute the "general form" that you found into the original equation to weed out any non-solutions that may have crept in (see Jason DeVito's answer). In our case, the fact that $F(x)$ must be of the form $a \cos x + b \sin x$ for some $a$ and $b$ does not mean that every function of that form actually works. There was a forgotten constraint $F(0) = 0$ on the equation $F''+F=0$, arising from your definition of $F$ as a definite integral.
