In this answer, he has three cases $(\lambda = 0, \lambda \lt 0, \lambda \gt 0)$. I understand the first does imply it is linear, hence it isn't consistent with the initial conditions, and looking at $\lambda\lt 0$:
If we have $\lambda \gt 0$, then $F''-\lambda F=0$, which is clearly an appropriate ODE, gives us $m^2 - \lambda m =0$
Gives us solution $m(m - \lambda)$, which has solution $m=0,\lambda$
$A+Be^{\lambda x}=\frac{x}{10}(\pi-x)$ at $x=0$, $A+B=0\implies A=-B$
$x=\pi \implies B(-1+e^{\lambda \pi})=0\implies B=0$ $$0=\frac{x}{10}(\pi-x)$$
Where $f(x)=0$ which is of no interest.$^*$
and then he assumed $\lambda = -\alpha^2$
Question 1: What should be said instead of "Which is of no interest" $^*$
Question 2: Where does $\lambda = -\alpha^2$ get inspired from?
Information: I have to find $u(x,t)$ for the string of length $L=\pi$ when $c^2=1$. $\text{wave equation}: \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}$ - $u(x,0)=\frac x{10}(\pi-x)$ - $u_t(x,0)=0$ ${}$
'Which is of no interest' is what my textbook said, in order to remove the case...$^*$
In regards to question 2, when considering the case when $\lambda<0$, set $\lambda=-\alpha^{2}$ as $\alpha^{2}$ is strictly positive (I'm assuming we're in $\mathbb{R}$) and hence $\lambda$ is strictly negative. Although the poster Semsem didn't do the $\lambda>0$ case, in this particular instance you would set $\lambda=+\alpha^{2}$.