I need to estimate parameter $b$ in following differential equation in the interval $[-3:3]$,
\begin{eqnarray}
y''(x) = b\varphi^2(x)y(x)
\end{eqnarray}
where,
\begin{equation}
\varphi(x) = \frac{1}{2}\left(1 -\sin\left[\frac{\pi}{2} \frac{x}{3} \right] \right)
\end{equation}
subject to boundary conditions,
\begin{eqnarray}
y'(-3) = 0 \\
y'(3) = 1/3
\end{eqnarray}
such that, $y(x)$ has two asymptotes $y=x/3, x\to 3$ and $y=0, x\to -3$.
To estimate the parameter, I proceded as follows, \begin{eqnarray} 0 &=& \int^{+3}_{-3}y''(x) dx -b\int^{+3}_{-3}\varphi^2(x)y(x) dx \nonumber \\ 0 &=& y'(x)\Big{|}^{+3}_{-3} - b\int^{+3}_{-3}\varphi^2(x)y(x) dx \\ 1/3 &=& b\int^{+3}_{-3}\varphi^2(x)y(x) dx \implies \int^{+3}_{-3}\varphi^2(x)y(x) dx = \frac{1}{3b} \end{eqnarray}
So that, for any parameter $b$ above equality holds. Now we find two functions, $y_{min} \le y \le y_{max}$ . \begin{eqnarray} y_{min}(x) &=& 0 \hspace{1cm} -3 < x < 0 \nonumber \\ &=& x/3 \hspace{1cm} 3 > x > 0 \\ y_{max}(x) &=& \frac{1}{6} (x+3) \hspace{1cm} -3 < x < 3 \end{eqnarray}
\begin{eqnarray}
\int^{3}_{-3}\varphi^2(x)y_{min}(x) dx &<& \int^{3}_{-3}\varphi^2(x)y(x) dx < \int^{3}_{-3}\varphi^2(x)y_{max}(x) dx \\
\int^{3}_{-3}\varphi^2(x)y_{min}(x) dx &<& \frac{1}{3b} < \int^{3}_{-3}\varphi^2(x)y_{max}(x) dx \\
0.0305638 &<&\frac{1}{3b} < 0.517073 \\
10.929 &>& b > 0.64467
\end{eqnarray}
Mathematica estimates $b \approx 4$. How can I obtain better estimate without using numerical integration?