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In response to this input:

DSolve[t[x] == (1/2) Integrate[t[y], {y, x - 1, x + 1}], t[x], x]

Mathematica just returns the same input.

However, it is easy to see that this equation has many solutions - in particular, every constant function $t[x]=c$ is a solution. Why doesn't Mathematica return some solutions? Do I have a mistake in the syntax?

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    $\begingroup$ DSolve[]'s support for integral equations is still somewhat limited, so don't be surprised if some things don't work yet. $\endgroup$ – J. M. is away Oct 15 '17 at 17:04
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Let's look at a few solutions to show why MMA doesn't provide any. Your equation.

inteq = t[x] == (1/2) Integrate[t[y], {y, x - 1, x + 1}]

Start with

t[x_] = a;

inteq
(*True*)

As you pointed out, that works. Now try

t[x_] = a + b x;

inteq//Simplify
(*True*)

Again that works. Try

t[x_] = a + b x + c x^2;

inteq//Simplify
(*c==0*)

Same as before. Try

t[x_] = a + b x + c x^2 + d x^3;

inteq//Simplify
(*c+3 d x==0*)

c = c /. Solve[%, c][[1]]
(*-3 d x*)

t[x]
(*a+b x-2 d x^3*)

inteq//Simplify
(*True*)

Works also. Try

t[x_] = a + b x + c x^2 + d x^3 + e x^4

inteq//Simplify
(*10 e x^2+e==10 d x*)

e = e /. Solve[%, e][[1]]
(*(10 d x)/(10 x^2+1)*)

t[x]
(*a+b x-2 d x^3+(10 d x^5)/(10 x^2+1)*)

inteq//Simplify
(*True*)

And so on for polynomials. It is conceivable that they could go on to many more terms. Now try something different.

t[x_] = a  Sin[x] + b Sin[2 x];

inteq//Simplify
(*Sin[x] (a (Sin[1] - 1) + b (Sin[2] - 2) Cos[x]) == 0*)

a = a /. Solve[%, a][[1]]
(*-((b (Sin[2] - 2) Cos[x])/(Sin[1] - 1))*)

t[x] // Simplify
(*(4 b Sin[1/2]^2 Sin[1] Sin[x] Cos[x])/(Sin[1] - 1)*)

inteq//Simplify
(*True*)

And that works too. With the widely varying possibilities for solution of this integral equation, to me it is unreasonable to expect DSolve to provide a general solution. Even if it were to provide some solutions, it would need much more information as to what the user wants. Even a polynomial option would not be specific enough.

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