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

  • $\begingroup$ where did it come from? $\endgroup$ – Calvin Khor May 6 at 17:31
  • $\begingroup$ My friend asked me to solve it $\endgroup$ – user361960 May 6 at 17:33
  • $\begingroup$ I don’t think it’s solvable in elementary functions. $\endgroup$ – 雨が好きな人 May 6 at 17:35
  • 8
    $\begingroup$ One solution is $y = x-1$. This is a functional-differential equation, not an ordinary differential equation. $\endgroup$ – Robert Israel May 6 at 17:41
  • $\begingroup$ Do I understand that its solution will be difficult? $\endgroup$ – 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.




Let $y(x)=u(x)+x-1$ ,

Then $y'(x)=u'(x)+1$


$\therefore u''(x)+u'(x)+1+u(2x)+2x-1=2x$


Let $u(x)=\int_0^\infty e^{-xt}K(t)~dt$ ,

Then $\int_0^\infty t^2e^{-xt}K(t)~dt-\int_0^\infty te^{-xt}K(t)~dt+\int_0^\infty e^{-2xt}K(t)~dt=0$

$\int_0^\infty t^2e^{-xt}K(t)~dt-\int_0^\infty te^{-xt}K(t)~dt+\int_0^\infty\dfrac{1}{2}e^{-xt}K\left(\dfrac{t}{2}\right)dt=0$

$\int_0^\infty e^{-xt}\left((t^2-t)K(t)+\dfrac{1}{2}K\left(\dfrac{t}{2}\right)\right)dt=0$

$\therefore (t^2-t)K(t)+\dfrac{1}{2}K\left(\dfrac{t}{2}\right)=0$





$K(2^t)=\Theta(t)\prod\limits_t\dfrac{1}{2^{t+2}-2^{2t+3}}$ , 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.


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