For all $k,l\in\{0,1,2,3\}$ we denote
$$
I_{k,l}=\int\limits_{\mathbb{R}_+}y^{(k)}(x)y^{(l)}(x)d\mu(x)
$$
Since $y(+\infty)=0$, and we are solving linear differetial equation with constant coefficients then $y^{(k)}(+\infty)=0$ for $k\in\{0,1,2,3\}$.
Note the following simple idenities
$$
I_{k,l}=-y_{k-1}y_{l}-I_{k-1,l+1}\quad\text{ for }\quad k\geq 1\\
$$
$$
I_{k+1,k}=-0.5 y_k^2\quad\text{ for }\quad k\geq 0\\
$$
$$
I_{k,l}=I_{l,k}\quad\text{ for }\quad k,l\geq 0\\
$$
They allow us to write down $I$ matrix
$$
I=
\begin{Vmatrix}
I_{0,0} & -0.5 y_0^2 & -y_1y_0-I_{1,1} & -y_2y_0+0.5 y_1^2\\
-0.5 y_0^2 & I_{1,1} & -0.5 y_1^2 & -y_2 y_1-I_{2,2} \\
-y_1y_0-I_{1,1} & -0.5 y_1^2 & I_{2,2} & -0.5 y_2^2 \\
-y_2y_0+0.5 y_1^2 & -y_2 y_1-I_{2,2} & -0.5 y_2^2 & I_{3,3}
\end{Vmatrix}
$$
Now multiply our differential equation by $y^{(k)}(x)$ with $k\in\{0,1,2\}$, and integrate over $\mathbb{R}_+$. Then we get
$$
\sum\limits_{l=0}^3 a_l I_{k,l}=0\quad\text{where}\quad k\in\{0,1,2\}
$$
This is a system of linear equations with unknowns $I_{0,0}$, $I_{1,1}$, $I_{2,2}$. Its solution is very difficult to get by hand but possible
$$
I_{0,0}= -\frac{-a_1 y_0^2-2 a_2 y_1 y_0-2 a_3 y_2 y_0+a_3 y_1^2}{2
a_0}-\frac{a_2 \left(-a_0 a_2 y_0^2-2 a_0 a_3 y_1 y_0-a_2^2 y_1^2-a_1 a_3
y_1^2-a_3^2 y_2^2-2 a_2 a_3 y_1 y_2\right)}{2 a_0 \left(a_1 a_2-a_0
a_3\right)},
$$
$$
I_{1,1}= -\frac{-a_0 a_2 y_0^2-2 a_0 a_3 y_1 y_0-a_2^2 y_1^2-a_1 a_3
y_1^2-a_3^2 y_2^2-2 a_2 a_3 y_1 y_2}{2 \left(a_1 a_2-a_0 a_3\right)},
$$
$$
I_{2,2}=
-\frac{-a_0^2 y_0^2-2 a_0 a_1 y_1 y_0-a_1^2 y_1^2-a_0 a_2 y_1^2-a_1 a_3 y_2^2-2
a_0 a_3 y_1 y_2}{2 \left(a_1 a_2-a_0 a_3\right)}
$$
I must confess, but I used Mathematica to solve this system. Here is the code
i[k_, l_] := -Subscript[y, k - 1] Subscript[y, l] - i[k - 1, l + 1] /;
k > l + 1 && l >= 0
i[k_, l_] := -1/2 Subscript[y, l]^2 /; k == l + 1 && l >= 0
i[k_, l_] := Subscript[i, k, l] /; k == l
i[k_, l_] := i[l, k] /; k < l
n := 3
Solve[
Table[Sum[Subscript[a, l] i[k, l], {l, 0, n}] == 0, {k, 0, n - 1}],
Table[Subscript[i, k, k], {k, 0, n - 1}]]
The answer to your question is $I_{0,0}^{1/2}$.
P.S. This method can be generalized to linear differential equations of arbitrary order.