To determine the dimension of the vector space \( W \) consisting of all bounded real-valued solutions of the differential equation:
\[\frac{d^4y}{dx^4} + 2 \frac{d^2y}{dx^2} + y = 0,\]
we start by finding the characteristic equation of this linear homogeneous differential equation. Assuming a solution of the form \( y = e^{\lambda x} \), we substitute and obtain the characteristic equation:
\[\lambda^4 + 2\lambda^2 + 1 = 0.\]
Letting \( z = \lambda^2 \), we have:
\[z^2 + 2z + 1 = 0.\]
This can be factored as:
\[(z+1)^2 = 0,\]
yielding a repeated root \( z = -1 \). Since \( z = \lambda^2 \), we have:
\[\lambda^2 = -1 \implies \lambda = \pm i.\]
The general solution to the differential equation is based on the roots \(\lambda = i\) and \(\lambda = -i\), leading to solutions such as:
\[y(x) = C_1 \cos(x) + C_2 \sin(x) + C_3 x\cos(x) + C_4 x\sin(x).\]
For \( y(x) \) to be bounded as \( x \to \pm \infty \), the terms involving \( x\cos(x) \) and \( x\sin(x) \) must be excluded because they grow unbounded. Thus, the bounded solutions are:
\[y(x) = C_1 \cos(x) + C_2 \sin(x).\]
Therefore, the dimension of the space of bounded solutions is 2, which matches the given range (2, 2).
Final Conclusion: The dimension of \( W \) is 2.