I am stuck on the following problem.
Consider the initial value problem $16y′′+24y′+9y=0, y(0)=a, y′(0)=−1.$ Find the critical value of a that separates solutions that become negative from those that are always positive for t>0.
I know how to do this if the unknown variable, $a$, is under the $y'(0)$, and $y(0)$ is given. I approached this question in a similar way, getting my solution $Be^{-3/4t}+Cte^{-3/4t}$. I then set B equal to $a$, because $y(0)$ would cancel out the second term. Using the second IC, I get $C = \frac{3}{4}a-1$, or $a = \frac{4}{3}(C+1)$, but the solution requires an exact answer without variables.
Any help is appreciated. Thank you!
After incorporating the first boundary condition, the solution is $$ y(t)=a\exp(-3/4 t)+ct\exp(-3/4t) $$ for $c$ a constant to be determined. The second condition yields $c=3/4a-1$. Right away, we can see $a>0$ is certainly necessary, and $a\geq 4/3$ is sufficient. The latter ends up being necessary as well.
Now, $$ y(t)=a\exp(-3/4 t)+(3/4a-1)t\exp(-3/4t)>0\\ \iff a+(3/4a-1)t>0\\ \iff (1-3/4a)t<a $$ For this to be valid for $t>0$, and $a>0$, we must have $1-3/4a\leq 0$, or $a\geq 4/3$. $4/3$ is thus your critical value.