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I am trying to find all functions f satisfying $f'(t)=f(t)+\int_a^bf(t)dt$.

This is a problem from Spivak's Calculus and it is the chapter about Logarithms and Exponential functions. I gave up and read the solution (which I quickly regretted, but at the same time realized that I had not carefully read one very important theorem* in the text) to find that it begins with:

We know $f''(t)=f'(t)$.

How do we know this? Also, in general, how would you have approached this problem? Any solution with your thoughts written out would be very appreciated. I am not interested in the solution, but the thought process behind this.

*For the curious, the theorem was that:

If $f$ is differentiable and $f'(x)=f(x)$ for all $x$ then there is a number $c$ s.t. $f(x)=ce^x$ for all $x$.

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Unfortunately, $t$ is used in two different roles here: an ordinary variable and a "dummy variable" in the integral. It would be better practice, and less confusing, to use a different letter for the dummy variable, e.g. $f'(t) = f(t) + \int_a^b f(s)\ ds$. That makes it more obvious that this term is constant. – Robert Israel Nov 28 '11 at 8:10
Sri, thanks for the edit. I could barely read the question before you intervened :) – The Chaz 2.0 Dec 7 '11 at 0:07
up vote 6 down vote accepted
  1. How do we know that $f''(x) = f'(x)$? Differentiate both sides of $$f'(x) = f(x) + \int_a^b f(t)\,dt.$$ Remember: $\int_a^bf(t)\,dt$ is a number (the net signed area between the graph of $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$). So what is its derivative?

    Why would you do this? Because that integral is somewhat annoying: if you just had $f'(x) = f(x)$, then you would be able to solve the differential equation simply enough (e.g., with the theorem you have). But since all that is standing in our way is a constant that is adding, differentiating should spring to mind: that will get rid of the constant, and just "shift" the problem "one derivative down" (to a relation between $f''(x)$ and $f'(x)$).

  2. Once you know that $f''(x) = f'(x)$, let $g(x) = f'(x)$. Then we have $g'(x)=g(x)$, so the theorem applies to $g(x)$ (exactly what we were hoping for). And you go from there.

Added. As both Didier Piau and Robert Israel point out, it's probably definitely bad practice to use the same letter as both an actual variable and the variable of integration (sometimes called the 'dummy variable'). Though I see from looking at my copy of Spivak that this originated in the text and not with you.

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Sorry but why one would reproduce OP's (mal)practice of using the same unknown in and out of the integral baffles me. Before saying everything else, I would try to explicitly discourage it. – Did Nov 28 '11 at 8:26
@DidierPiau: A fair point; though Spivak does exactly that, so it's not the OP's practice. Problem 26 in Chapter 17 reads: "Find all functions $f(t)$ that satify $f'(t) = f(t) + \int_0^1 f(t)\,dt$." (At least in my Second Edition in Spanish) – Arturo Magidin Nov 28 '11 at 14:13
Yes. But if your edition is faithful to the original in English, you can check that Problem 26 is the one and only (unfortunate) occurrence. Everywhere else, Spivak is correct. Anyway, to me all this has little to do with what we write here and how we choose to write it (and I do not understand the probably in your Addendum). – Did Nov 28 '11 at 18:08

Since $\int_a^bf(t)dt$ is a constant, we denote it as $C$, so we get $f'(t)=f(t)+C$. It is $\frac{df(t)}{dt}=f(t)+C$, i.e. $\frac{df(t)}{f(t)+C}=dt$. Integrate it, we get $ln(f(t)+C)=t+D$, where D is a constant. Thus, $f(t)=e^{t+D}-C$. Let $e^D=K$, we get $f(t)=K{e^{t}}-C$.

Now plug it into the original eqution, we get $Ke^t=Ke^t-C+\int_a^b{(Ke^t-C)}dt$. After some manipulation, we have $K=\frac{b-a+1}{e^b-e^a}C$. Note here we assume that $b \neq a$.

If $b=a$,$C$ is obviously $0$, so the equations can be simplified to $f'(t)=f(t)$. In this case, $f(t)=ke^t$, where k is an arbitrary constant and $k \neq 0$

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@DidierPiau, of course, It my mistake. – NoChance Nov 28 '11 at 9:14
+1, clear answer. – NoChance Nov 28 '11 at 13:03

You know that $f''(t)=f'(t)$ because you can differentiate the equation you are trying to solve.

Since the second term in the right hand side of the equation you are trying to solve is annoying, it is a good idea to try to get rid of it. Since it is constant, then that can be done by differentiating the equality with respect to $t$.

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Yes it's a solved problem. Here's a synopsis that may be a little easier to follow. Start with the equation $$ f^\prime(t) = f(t) + \int_a^bf(t)dt, $$ for $a<b$ and note that the integral on the RHS is $(b-a)$ times the average value of $f$ on $[a,b]$; call this term M, which is constant for a given $f$. Any solution of $f^\prime(t)=f(t)+M$ is differentiable, so the RHS is differentiable, implying that $f^{\prime\prime}=f^\prime$ (and, by recursion, that $f$ is actually infinitely differentiable). This has solution $f^\prime(t)=ce^t$ by a standard argument (the additive constant must be zero) so that $f(t)=ce^t+d$. But then $$ M=\int_a^bf(t)dt=\left[ce^t+dt\right]_a^b=c(e^b-e^a)+d(b-a), $$ giving us a constraint on the constants $c$ and $d$: $$ ce^t=f^\prime(t)=f(t)+M=ce^t+d+M \quad\implies\quad M=-d \quad\implies $$ $$ 0=M+d=c(e^b-e^a)+d(b-a+1)\quad\implies\quad d=-c\frac{e^b-e^a}{1+b-a} $$ so that the general solution is $$ f(t)=c\left[e^t-\frac{e^b-e^a}{1+b-a}\right]. $$ Finally, the "standard argument" referred to above is that if $u^\prime=u$ for some function $u(t)$, then $$ \frac{du}{u}=dt \quad\implies\quad \ln|u|=t+c_1 \quad\implies\quad |u|=e^{t+c_1} \quad\implies\quad u=ce^t $$ where $c=\pm e^{c_1}$ can be any nonzero constant but in fact also zero (which couldn't be inferred from the derivation because we divided by $u$ along the way, so we recover the solution at the end by inspection).

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