I'm Such a Meanie!

Let $g(x)=xf(x)$ , so that by the product rule $g'(x)=xf'(x)+f(x)$ . We have $\begin{aligned}g(0)&=0\times f(0)=0\\ g(10)&=10\times 0=0 \end{aligned}$ Hence, by Rolle's theorem , there is some stationary point between 0 and 10, i.e. $g'(c)=0$ for some $c\in(0,10)$ . Hence, $cf'(c)+f(c)=0$ and so $k=0$ .

Notice that this is the only guaranteed value of $k$ : if we take $k\neq0$ then a counterexample is $f(x)=a(x-10)$ for some $a$ , since $g'(x)=a(2x-10)$ and we can choose $a$ small enough so that $|g'(x)|<|k|$ on $(0,10)$ .

Let $f(0) = N$ for any nonzero $N$ . By the Mean-Value Theorem, there exists a point $c \in (0,10)$ such that slope of the tangent line drawn to $f(x)$ at $x=c$ equals the slope of the secant line connecting the endpoints $(x,y) = (0,N)$ and $(10,0)$ . That is $\large f'(c) = \frac{N-0}{0-10} = -\frac{N}{10}$ . Subsituting this value into the given differential equation yields:

$c(-\frac{N}{10}) + f(c) = k \Rightarrow f(c) = k + \frac{Nc}{10}$

Knowing that $0 < f(c) < N$ , we obtain:

$\large 0 < k + \frac{Nc}{10} < N \Rightarrow -\frac{10k}{N} < c < (N-k)(\frac{10}{N})$

and if $c \in (0,10)$ , then we require $\boxed{k=0}.$

Indeed you're a true math meanie! (Did YOU figure it out?)

I just used $y=-x+10$ . I'm a physicist and not a pure mathematician.