Question

Let \( W \) be the vector space (over \( \mathbb{R} \)) consisting of all bounded real-valued solutions of the differential equation \[ \frac{d^4y}{dx^4} + 2 \frac{d^2y}{dx^2} + y = 0. \] Then, the dimension of \( W \) is _________ (in integer).

Hint:

For higher-order linear differential equations, the dimension of the solution space is equal to the number of linearly independent solutions. In this case, the bounded solutions are spanned by \( \cos x \) and \( \sin x \), giving a solution space of dimension 2.

Answer 1

Correct Answer: 2

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.