Ordinary differential equation for a function with different inputs Find all the functions such that
$$ f'(x)+f(-x) = 0$$
Attempt
I tried using mean value theorem to substitute 
$$a=f'(c) = \frac{f(x)-f(-x)}{2x}$$but this strategy failed when I tried to integrate it over the course of the attempt since $f'(c)$ does depend on $x$.
Can somebody provide me an appropriate substitution to solve this question?
 A: If $f'(x)=-f(-x) $ for all $x$, then 
$f''(x)=f'(-x)= -f(x)$, hence
$$f''(x)+f(x)=0.$$
Thus there are $c_2,c_2 \in \mathbb R$ such that  $f(x)=c_1 \cos x +c_2 \sin x$.
Now use again $f'(x)+f(-x) = 0$ to obtain $c_2=-c_1$.
A: Idea:
Try to differentiate it again, so you get $$f''(x) -f'(-x)=0$$ 
But $f'(-x) = -f(x)$ so we get $$f''(x)+f(x) =0$$ and a solution to this should be something like this $$f(x) =a\cos x+b\sin x$$
A: While taking the derivative of the equation is a quick way to solve this, you can also obtain this answer without assuming that $f$ is twice differentiable -- you're given a first order differential equation for $f$, which in itself only implies that $f$ is continuously differentiable, not more.
The key is to realise that you can write every function (also non-differentiable ones) as a sum of a symmetric and an antisymmetric function:
$$
f(x) = f_s(x) + f_a(x),\tag{1}
$$
where $f_s(-x) = f_s(x)$ and $f_a(-x) = - f_a(x)$. You can easily show this by writing
$$
 f_s(x) = \frac{f(x)+f(-x)}{2}\quad\text{and}\quad f_a(x) = \frac{f(x) - f(-x)}{2}. \tag{2}
$$
It's also important to realise that this splitting is unique.
We can use this symmetric/antisymmetric splitting as follows. We have
$$
 f(-x) = f_s(-x) + f_a(-x) = f_s(x) - f_a(x) \tag{3}
$$
and
$$
 f'(x) = f_s'(x) + f_a'(x). \tag{4}
$$
Now, we use the observation that the derivative of a symmetric function is antisymmetric, and vice versa; you can also derive this fact using $(2)$. Because the symmetric/antisymmetric splitting is unique, we know that the symmetric part of $(3)$ must be equal to minus the symmetric part of $(4)$, and the same holds for the antisymmetric part. So, we have
\begin{align}
f'_a(x) &= - f_s(x) \quad\text{(symmetric parts)},\\
f'_s(x) &= f_a(x) \quad\text{(antisymmetric parts).}
\end{align}
Now we can call $f_s(x) = y(x)$ and $f_a(x) = z(x)$, and we have
\begin{align}
 y' &= z,\\
 z' &= -y,
\end{align}
which a) implies that $y$ and $z$ are indeed twice differentiable, and b) is easy to solve, because it is just the harmonic oscillator. Since $y$ is symmetric and $z$ is antisymmetric, we have
$$
y(x) = y_0 \cos(x)\quad\text{and}\quad z(x) = z_0 \sin(x),
$$
so
$$
 f(x) = y_0 \cos(x) + z_0 \sin(x). \tag{5}
$$
You can now find a relation between $y_0$ and $z_0$ by substituting $(5)$ into the original equation $f'(x) + f(-x) = 0$.
A: Differentiate $$f'(x)+f(-x) = 0$$ to get
$$ f''(x)-f'(-x)=0$$
Since $$f'(-x)=-f(x)$$ we have 
$$ f''(x)+f(x)=0$$
Solving for $f(x)$ we have 
$$ f(x) = C_1 cos(x)+C_2 sin(x)$$
Substituting in  $$f'(-x)=-f(x)$$ We get $$C_1=-C_2$$ 
Thus 
$$ f(x)=C( cos(x)-sin(x))$$
