# 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

• where did it come from? – Calvin Khor May 6 at 17:31
• My friend asked me to solve it – user361960 May 6 at 17:33
• I don’t think it’s solvable in elementary functions. – 雨が好きな人 May 6 at 17:35
• One solution is $y = x-1$. This is a functional-differential equation, not an ordinary differential equation. – Robert Israel May 6 at 17:41
• Do I understand that its solution will be difficult? – user361960 May 6 at 17:46

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.

$$y''\left(\dfrac{x}{2}\right)+y'\left(\dfrac{x}{2}\right)+y(x)=x$$

$$y''(x)+y'(x)+y(2x)=2x$$

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.