Solve $ y'(x)+2y(x)+3y(-x)=0$ I just arrived to an extremely weird ODE and I was wondering how could I solve it: $$
y'(x)+2y(x)+3y(-x)=0.
$$
Actually, I am not even sure if it is possible to prove that it has a unique solution (provided an initial condition). I tried to put it on wolfram alpha but I didn't get anything. My first thought was to use "separation of variables" method but I don't think that it will help because of the term $y(-x)$. Actually I am completely clueless.
Edit: I am actually wondering if this ODE has non-symmetric solutions, that is, neither odd nor even solutions.
 A: Consider the function $B(x)=y(-x)$ and name $A(x)=y(x)$. Then one can show that the following system of equations is satisfied
$$\begin{Bmatrix}A'=-2A-3B\\B'=3A+2B\end{Bmatrix}$$
Adding and subtracting the two equations we obtain that 
$$\begin{Bmatrix}(A+B)'=A-B\\(A-B)'=-5(A+B)\end{Bmatrix}$$
from which we can deduce that both the even and odd parts of $y(x)$ obey the equation
$$y_e''(x)+5y_e(x)=y_o''(x)+5y_o(x)=0$$
Thus we conclude that 
$$y(x)=y_e(x)+y_o(x)=A\cos(\sqrt{5}x)+B\sin(\sqrt{5}x)$$
Substituting into the original equation we find that 
$$(-A\sqrt{5}-B)\sin(\sqrt{5}x)+(B\sqrt{5}+5A)\cos(\sqrt{5}x)=0$$
which implies that the equation is satisfied if $A\sqrt{5}+B=0$ and thus the general solution to the equation reads
$$y(x)=A(\cos(\sqrt{5}x)-\sqrt{5}\sin(\sqrt5x))=y(0)(\cos(\sqrt{5}x)-\sqrt{5}\sin(\sqrt5x))$$
A: Decompose $y(x)$ as a sum of its even and odd part:
$$y(x) = y_e(x) + y_o(x)\quad\text{with}\quad \begin{cases} y_e(x) &= \frac12(y(x) + y(-x))\\ y_o(x) &= \frac12(y(x) - y(-x))\end{cases}$$
The ODE becomes
$$(y_e+y_o)' + 2(y_e+y_o) + 3(y_e-y_o) = 0\tag{*1a}$$
Subsititute $x$ by $-x$ and notice under such a change, $\frac{d}{dx}$ and $y_o$ picks up a minus sign while $y_e$ remains the same. We get
$$-(y_e - y_o)' + 2(y_e - y_o) + 3(y_e + y_o) = 0\tag{*1b}$$
Combined $(*1a)$ and $(*1b)$, we get
$$y'_o + 5y_e = y'_e - y_o = 0\quad\implies\quad y''_e + 5y_e = 0
$$
This implies $$y_e(x) = A \cos(\sqrt{5}x) + B\sin(\sqrt{5}x)$$
for suitably chosen constants $A, B$. Since $y_e(x)$ is even, $B = 0$
and hence $$y_e(x) = A\cos(\sqrt{5}x)
\quad\implies\quad y_o(x) = y'_e(x) = -A\sqrt{5}\sin(\sqrt{5}x)$$
Since $A = y_e(0) = y(0)$, we have
$$y(x) = y(0)(\cos(\sqrt{5}x) - \sqrt{5}\sin(\sqrt{5}x))
$$
Up to an overall scaling factor $A = y(0)$, this solution is unique.
A: Multiply through by $e^{2x}$ and we find $[e^{2x}y(x)]'=-3e^{2x}y(-x)$. Integrating, we have
$y(x)=y(0)e^{-2x}-3\int_0^xe^{2(t-x)}y(-t)\,dt$. Suppose $y(0)=0$ is demanded in that we restrict our attention to the Banach space $X_a=\{f\in C([-a,a]): f(0)=0\}$. Now define 
$$
T(f)(x)=f(0)e^{-2x}-3\int_0^xe^{2(t-x)}f(-t)\,dt=-3\int_0^xe^{2(t-x)}f(-t)\,dt.
$$
Then for any two elements $f$ and $g$, we have
$$
|T(f)(x)-T(g)(x)| \leq ||f-g||_\infty \left[3\int_0^xe^{2(t-x)}\,dt \right]=\frac{3}{2}(1-e^{-2x})||f-g||_\infty .
$$ 
Make $a$ as small as possible so that the coefficient $\alpha=\frac{3}{2}(1-e^{-2a})<1$. Then $T$ is a contraction mapping. Moreover, for any disk $D_r=\{f: ||f||_\infty \leq r\}$, we see that
$$
|T(f)(x)|\leq \alpha r \leq r
$$
so $T$ maps a disk to itself. Therefore, there exists a unique solution on this space. Since $y=0$ solves the problem, it is the only solution.
