Answer
The solution to the initial-value problem is
$$
y(x)=-3 e^{x}+3 e^{-\frac{1}{3}x}.
$$
Work Step by Step
$$
3y^{\prime \prime}-2 y^{\prime}- y=0, \quad y(0)=0, \quad y^{\prime}(0)=-4
$$
The given equation is a homogeneous linear equation. So the characteristic equation of the differential equation is
$$
3r^{2}-2r-1 =0
$$
whose roots are
$$
\:r_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:3\left(-1\right)}}{2\cdot \:3}
$$
$\Rightarrow$
$$
r_{1}=1,\:r_{2}=-\frac{1}{3}
$$
Therefore, the general solution of the given differential equation is
$$
y(x)=c_{1}e^{x}+c_{2}e^{-\frac{1}{3}x},
$$
Then
$$
y^{\prime }(x)=c_{1}e^{x}-\frac{1}{3}c_{2}e^{-\frac{1}{3}x},
$$
so
$$
y(0)=0
$$
$\Rightarrow$
$$
y(0)=c_{1}e^{0}+c_{2}e^{0}=c_{1}+c_{2}=0
$$
and
$$
y^{\prime }(0)=-4
$$
$\Rightarrow$
$$
y^{\prime }(0)=c_{1}e^{0}-\frac{1}{3}c_{2}e^{-\frac{1}{3}.0}=c_{1}-\frac{1}{3}c_{2}=-4
$$
giving $c_{1}=-3,\:c_{2}=3$
Thus, the solution to the initial-value problem is
$$
y(x)=-3 e^{x}+3 e^{-\frac{1}{3}x},
$$
