<s>[INST] 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
 [/INST] 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. 
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