# 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.

• @DinnoKoluh Oh yes, sorry maybe I forgot to say. I am actually trying to understand if this ODE has nonsymmetric solutions (I mean, non-odd neither even solutions) – Sharik Apr 2 at 21:24
• Oh, okay. One trivial that I see is $y(x) = 0$. – Dinno Koluh Apr 2 at 21:25

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.

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

$$y(x)=A(\cos(\sqrt{5}x)-\sqrt{5}\sin(\sqrt5x))=y(0)(\cos(\sqrt{5}x)-\sqrt{5}\sin(\sqrt5x))$$

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.