TITLE: How to solve this D.E $y''(\frac{x}{2})+y'(\frac{x}{2})+y(x)=x$
QUESTION [5 upvotes]: 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

REPLY [2 votes]: $y''\left(\dfrac{x}{2}\right)+y'\left(\dfrac{x}{2}\right)+y(x)=x$
$y''(x)+y'(x)+y(2x)=2x$
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.