Consider a differential equation with a term containing $y(x_0)$, for example $$y'' - 2y' + y = y(x_0)$$ $x_0 \in \mathbb{R}$ is a constant. My question is, does such equations fall under the category of differential equations? I have never studied any equation with such a term. If its a differential equation, then $y(x_0)$ can be considered a constant coefficient?

  • $\begingroup$ There could be multiple terms like $y(x_0), y(x_1),y(x_2)$ too. $\endgroup$
    – Rajesh D
    Aug 11, 2018 at 12:59
  • $\begingroup$ But $y(x_0)$ is assumed to be a constant. $\endgroup$ Aug 11, 2018 at 12:59
  • $\begingroup$ @Dr.SonnhardGraubner : it is not independent of $y$. It depends on the solution. $\endgroup$
    – Rajesh D
    Aug 11, 2018 at 13:00
  • $\begingroup$ Think that $y(x_0) =\int \delta(x-x_0) y(x) dx$. After that you can solve the linear DE using the Laplace Transform. $\endgroup$
    – Cesareo
    Aug 11, 2018 at 13:17
  • $\begingroup$ @Cesareo : differential equations with distributions? Any theory available on this? $\endgroup$
    – Rajesh D
    Aug 11, 2018 at 13:25

1 Answer 1


Although the term $y(x_0)$ depends on the solution $y$ it is still a constant as it doesn't depend on $x$.

Let's try to solve your equation $y'' - 2y' + y = C,$ where $C=y(x_0).$ One solution is $y(x) = C.$ The solutions of the homogeneous equation are $y(x) = (Ax+B) e^x.$ Therefore the general solutions are $$y(x) = (Ax+B) e^x + C.$$

To get $y(x_0) = C$ we must have $A=B=0.$ Thus there is only the constant solution $$y(x) = C = y(x_0).$$

  • 2
    $\begingroup$ You can also have constants $A, B$ with $Ax_0+B=0$ so that the general solution is $$y(x) = A(x-x_0)e^x+B$$ (replaced $C$ with $B$). $\endgroup$
    – Paramanand Singh
    Aug 12, 2018 at 4:23

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