Let y(x) be a solution problem y"+5y'+6y=0 with the initial conditions y(0)=1; y'(0)=a For what values ​of "a" y(x) remains notnegative for all x>=0?

Question

I tried to solve:
Find the the solutions for $s^2+5s+6=0$
S1=-3 S2=-2
so $ Y(x)=C_1e^{-2x}+C_2e^{-3x}$
use the inicial condition and derivate Y(x) and find $C_1= 3+a$ and $C_2=-a-2$
then I tried to find "a" when is positive $ Y(x)=(3+a)e^{-2x}+(-a-2)e^{-3x}$
but i dont know how to find for all $x\geq0$

Answer 1

Let $x_0\ge 0$.

(Suppose for simplicity $a\neq-3$.)

$$\begin{array}{ll} y(x_0)=0&\Longleftrightarrow& (3+a)\exp(-2x_0)+(-a-2)\exp(-3x_0)=0\\ &\Longleftrightarrow& (3+a)\exp(x_0)+(-a-2)=0\\ &\Longleftrightarrow&\exp(x_0)=\frac{a+2}{a+3} \end{array}$$

The last equation has a positive solution if and only if the quotient $\frac{a+2}{a+3}$ is greater than 1 which is equivalent to $a<-3$.