Stability of a system in time-domain
Solving the characteristic equation
$$s^2+\frac32 s-1=0\tag{1}$$
gives the following homogeneous solution:
$$y(t)=c_1e^{-2t}+c_2e^{t/2}\tag{2}$$
The requirements of causality and stability are taken into account by choosing at least one of the constants $c_1$ and $c_2$ to be zero for either $t&gt;0$ or $t&lt;0$. If we're looking for a causal solution we know that $y(t)=0$ for $t&lt;0$, i.e.,
$$y_c(t)=\big[c_1e^{-2t}+c_2e^{t/2}\big]u(t)\tag{3}$$
which can be interpreted as choosing $c_1=c_2=0$ in $(2)$ for $t&lt;0$.
For a stable solution, we require $c_2=0$ for $t&gt;0$ because stability means that there can't be exponential growth. For the same reason we require $c_1=0$ for $t&lt;0$:
$$y_s(t)=c_1e^{-2t}u(t)+c_2e^{t/2}u(-t)\tag{4}$$
Eqs $(3)$ and $(4)$ are the general forms of the solutions given the requirements of causality or stability, respectively.
In both cases, the constants $c_1$ and $c_2$ are determined by requiring that $y''+\frac32 y'-y$ equals a Dirac delta impulse with weight $1$.
Note that when calculating the derivatives of $(3)$ and $(4)$ you need to use the product rule:
$$\big[f(t)u(t)\big]'=f'(t)u(t)+f(t)u'(t)=f'(t)u(t)+f(t)\delta(t)\tag{5}$$
Solving such problems using the Laplace transform is indeed much less tedious.

As an example I'll show how to arrive at the values of $c_1$ and $c_2$ for the causal solution $(3)$. Define $f(t)=c_1e^{-2t}+c_2e^{t/2}$. With $y_c(t)=f(t)u(t)$, the derivatives of $y_c(t)$ are
$$\begin{align}y_c'(t)&amp;=f'(t)u(t)+f(t)\delta(t)=f'(t)u(t)+f(0)\delta(t)\\y_c''(t)&amp;=f''(t)u(t)+f'(t)\delta(t)+f(0)\delta'(t)=f''(t)u(t)+f'(0)\delta(t)+f(0)\delta'(t)\end{align}$$
We need to satisfy the differential equation $y_c''(t)+\frac32 y_c'(t)-y_c(t)=\delta(t)$. Consequently, the term $f(0)\delta'(t)$ must vanish:
$$f(0)=c_1+c_2=0$$
Furthermore, the coefficients associated with the Dirac delta impulses must add up to $1$:
$$f'(0)+\frac32 f(0)=1$$
This results in the requirement $$-2c_1+\frac12 c_2+\frac32 (c_1+c_2)=1$$
which leads to $c_1=-\frac25$ and $c_2=\frac25$.
In exactly the same way you can derive the coefficients of the stable solution $(4)$, which results in $c_1=c_2=-\frac25$.