If $f(x)$ is differentiable, and equals $ x^2+\int_0^x{e^{-t}f(x-t)}dt $ , then find $f(x)$

Substituting $x=0$, we find that $f(0)=0$. Then, substituting $x=t$ gives us $f(t)= t^2+\int_0^t{e^{-t}f(0)}dt = t^2$

However, if we differentiate the given equation, and then solve the resulting differential equation, we get $f(x)=x^3/3+x^2$. It can be verified that this is the actual solution to the functional equation.

My doubt is, why does method $1$ not work?

  • $\begingroup$ $t$ is a dummy variable and we cannot substitute $x$ by $t$. $\endgroup$
    – CY Aries
    Jun 15, 2021 at 7:12
  • $\begingroup$ Can you please elaborate? In what situations can substitutions be made, and why isn't it allowed in this case? $\endgroup$
    – Aspirant
    Jun 15, 2021 at 7:13

1 Answer 1


When you write $$\int_{0}^{x} e^{-t} f(x-t) dt$$ here the integrating variable is $t$, and hence, $x$ is to be treated as a constant. Thus the substitution $t=x$ is meaningless, as the variable cannot be substituted to be a constant during integration.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.