Polynomial Differential Equation

Using the fact that differentiating a polynomial reduces its degree, we can see that the degrees of the left hand side and the right hand side of the equation can match only if the degree of $f(x)$ is 2.

So let us write

$f(x)=Ax^2+Bx+C$

Substituting into the equation this gives

$Ax^2+Bx+C=2Ax+B+(x+2)^2$

Expanding the bracket and grouping terms with the same power of $x$

$\implies Ax^2+Bx+C=x^2+(2A+4)x+(B+4)$

Two polynomials are equal only if the corresponding coefficients are equal. Applying this principle to the above equation leads us to three simultaneous equations-

$A=1$

$B=2A+4\implies B=6$

$C=B+4\implies C=10$

The required polynomial is therefore $f(x)=x^2+6x+10$ and so

$f(1)=1+6+10= \boxed{17}$

Since the other solutions deal with the Inspiring method to check degrees so I am providing a bit different solution.

$\displaystyle \frac{dy}{dx}-y=-(x+2)^2$

$\displaystyle e^{-x}(\frac{dy}{dx}-y)=-e^{-x}(x+2)^2)$ As $e^{\int -1 dx}$ is the I.F of the linear differential equation.

$\displaystyle ye^{-x}=-\int (x+2)^2e^{-x}dx$

Integrating by parts several times we get , $\displaystyle ye^{-x}=(x+2)^2e^{-x}+2xe^{-x}+6e^{-x}+C$

We eliminate the constant by putting $x=\infty$ and we get $C=0$

$\implies y=(x+2)^2+2x+6=x^2+6x+10$

So, $\boxed{f(1)=17}$

Differentiating multiple times, one can see that $f^{(3)} \ = \ f^{(4)}$ . Since $f$ is a polynomial, we conclude that it must be true that $f^{(3)} \ = \ f^{(4)} \ = \ 0$ , thus our polynomial is of order $2$ . Let $f(x) \ = \ Ax^2 \ + \ Bx \ + \ C$ . we will have:

$Ax^2 \ + \ Bx \ + \ C \ = \ 2Ax \ + \ B \ + \ x^2 \ + \ 4x \ + \ 4 \ = \ Ax^2 \ + \ (2A \ + \ 4)x \ + \ (B \ + \ 4)$

Due to the uniqueness of polynomial expansion, it must be true that $A \ = \ 1$ , implying that $B \ = \ 6$ , which finally implies that $C \ = \ 10$ . We thus have:

$f(x) \ = \ x^2 \ + \ 6x \ + \ 10$

Thus, $f(1) \ = \ 17$ .