Calculus problem on derivative

Assume $f(x)$ is a polynomial of degree $n$

Degree of $f'(x)$ is $n-1$

Degree of $\left(f'(x)+f(x)\right)=$ Degree of $f(x)$

$\implies$ Degree of $f(x)=$ Degree of $x=1 \hspace{10mm}\small\color{#3D99F6}f'(x)+f(x)=x \text{ (Given)}$

$\begin{aligned}\implies f(x)&=ax+b\\ f'(x)&=a\\ f'(x)+f(x)&=a+ax+b\\ x&=ax+a+b\hspace{10mm}\small\color{#3D99F6}f'(x)+f(x)=x \text{ (Given)}\\ \text{on comparing coefficients we get,}\\ a&=1,\\ b&=-1\\ \implies f(x)&=x-1\\ f(4)&=4-1=\color{#EC7300}\boxed{\color{#333333}3}\end{aligned}$

Outline solution below:

First, given that $f(x)$ is a polynomial, what can its degree be? (The degree of both sides has to be the same)

Next, write a general form for $f(x)$ , and substitute in.

Finally, compare coefficients of the different powers of $x$ on each side to determine the unknown coefficients of $f(x)$ .

Now that you know the coefficients, you can substitute $x=4$ in and see that you do indeed get $f(4)=3$ .

Note that this is a particular solution of the differential equation - the general solution includes an exponential term of the form $Ae^{-x}$ , but we're told to ignore that in this question. In case you want to find more information on these, this is an example of a first-order non-homogeneous linear differential equation with constant coefficients.