How to solve this D.E $y''(\frac{x}{2})+y'(\frac{x}{2})+y(x)=x$ I know how to slove $y''(x)+y'(x)+y(x)=x$ 
But I couldn't solve this
$$y''(\frac{x}{2})+y'(\frac{x}{2})+y(x)=x$$
any hint to help me?
Thanls
 A: Here is an incomplete attempt that I might be able to rectify later. 
Write the equation as:
$$y''(x)+y'(x)+y(2x) = 2x$$ and let $y(x) = f(x)+x-1.$ Then we get an equation for $f:$
$$f''(x)+f'(x)+f(2x) = 0.$$
Now assume $x\neq 0$ and  By Taylor's theorem:
$$f(2x) = f(x)+xf'(x)+\frac{x^2}{2}f''(x)+x^2h(2x)$$
with $h(t)\to 0$ as $t\to 0.$ Substituting this in the differential equation:
$$\dfrac{f''(x)}{x}+\dfrac{2+2x}{x(2+x^2)}f'(x)+\dfrac{2}{x(2+x^2)}f(x) = -h(2x)\cdot\dfrac{2x}{2+x^2}.$$
Now again by Taylor's, we have $f(x) = f(0)+xf'(0)+o(x^2)$ and $f'(x) = f'(0)+xf''(0)+o(x^2).$ Plug them in and we get:
$$\dfrac{f''(x)}{x}+\dfrac{2+2x}{2+x^2}\left(\dfrac{f'(0)}{x}+f''(0)+o(x)\right)+\dfrac{2}{2+x^2}\left(\dfrac{f(0)}{x}+f'(0)+o(x)\right) = -h(2x)\cdot\dfrac{2x}{2+x^2}.$$
Now, we take the limit $x\to 0$ from both sides and that leaves us with:
$$\lim\limits_{x\to 0}\dfrac{1}{x}\left(f''(x)+\dfrac{2+2x}{2+x^2}f'(0)+ \dfrac{2}{2+x^2}f(0)\right) = 0.$$
The plan of attack is to prove $f''(x) = 0$ and this leaves us only linear functions as possible answer. 
A: $y''\left(\dfrac{x}{2}\right)+y'\left(\dfrac{x}{2}\right)+y(x)=x$
$y''(x)+y'(x)+y(2x)=2x$
Let $y(x)=u(x)+x-1$ ,
Then $y'(x)=u'(x)+1$
$y''(x)=u''(x)$
$\therefore u''(x)+u'(x)+1+u(2x)+2x-1=2x$
$u''(x)+u'(x)+u(2x)=0$
Let $u(x)=\int_0^\infty e^{-xt}K(t)~dt$ ,
Then $\int_0^\infty t^2e^{-xt}K(t)~dt-\int_0^\infty te^{-xt}K(t)~dt+\int_0^\infty e^{-2xt}K(t)~dt=0$
$\int_0^\infty t^2e^{-xt}K(t)~dt-\int_0^\infty te^{-xt}K(t)~dt+\int_0^\infty\dfrac{1}{2}e^{-xt}K\left(\dfrac{t}{2}\right)dt=0$
$\int_0^\infty e^{-xt}\left((t^2-t)K(t)+\dfrac{1}{2}K\left(\dfrac{t}{2}\right)\right)dt=0$
$\therefore (t^2-t)K(t)+\dfrac{1}{2}K\left(\dfrac{t}{2}\right)=0$
$(2t^2-2t)K(t)=-K\left(\dfrac{t}{2}\right)$
$(2^{2t+1}-2^{t+1})K(2^t)=-K(2^{t-1})$
$K(2^t)=\dfrac{K(2^{t-1})}{2^{t+1}-2^{2t+1}}$
$K(2^{t+1})=\dfrac{K(2^t)}{2^{t+2}-2^{2t+3}}$
$K(2^t)=\Theta(t)\prod\limits_t\dfrac{1}{2^{t+2}-2^{2t+3}}$ , where $\Theta(t)$ is an arbitrary periodic function with unit period
$K(t)=\Theta(\log_2t)\left(\prod\limits_t\dfrac{1}{2^{t+2}-2^{2t+3}}\right)(\log_2t)$ , where $\Theta(t)$ is an arbitrary periodic function with unit period
$\therefore u(x)=\int_0^\infty\Theta(\log_2t)e^{-xt}\left(\prod\limits_t\dfrac{1}{2^{t+2}-2^{2t+3}}\right)(\log_2t)~dt$ , where $\Theta(t)$ is an arbitrary periodic function with unit period
But this may be only one of the group of the solution and may be not enough general. I have no idea about the exact number of groups of the solution in the general solution of the functional equation of this type, so I stop here.
