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Suppose we have the following functional differential equation:

$$f(a_0+a_1x+a_2f(x))(b_1+b_2f'(x))=c_0+c_1x+c_2f(x)$$

It is easy to see that a linear function: $f(x)=d_0+d_1x$, with appropriate coefficients, will solve this problem. If I am not mistaken, $d_1$ is going to be a root of a 3rd-order polynomial which is a function of $a_1,a_2,b_1,b_2,c_1$, and $c_2$. And after that, it's quite simple to calculate $d_0$ (one only needs to solve a linear equation).

My conjecture is that the solution must be linear (i.e. there are no other solutions), but I was unable to prove it thus far. I was trying the following. If I assume that the solution is a polynomial then, by "counting the degrees", I think I can prove that the solution must be linear. But what if we consider other forms of solutions, say in the form of a ratio of two polynomials? By counting the degrees, I got that the degree of the numerator has to be equal to 1 plus the degree of the denominator. But I don't know how to prove that this ratio will actually simplify to a linear function.

Moreover, what if we consider solutions that aren't a ratio of two polynomials?

Is there someone here who can guide me how to approach this problem?

Thank you,

George

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