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

We are given the fourth-order linear homogeneous differential equation: \[ \frac{d^4y}{dx^4} + 2 \frac{d^2y}{dx^2} + y = 0. \] To find the dimension of the vector space \( W \), we first need to solve the characteristic equation associated with the given differential equation. Step 1: Find the characteristic equation
The given differential equation can be rewritten as: \[ y^{(4)} + 2y^{(2)} + y = 0. \] Assuming a solution of the form \( y = e^{rx} \), we substitute this into the differential equation to get the characteristic equation: \[ r^4 + 2r^2 + 1 = 0. \] Step 2: Solve the characteristic equation
We can factor the characteristic equation: \[ (r^2 + 1)^2 = 0. \] This gives a double root \( r = \pm i \), meaning the general solution to the differential equation is: \[ y(x) = c_1 \cos x + c_2 \sin x + c_3 x \cos x + c_4 x \sin x, \] where \( c_1, c_2, c_3, c_4 \) are constants. Step 3: Bounded solutions
For the solutions to be bounded, the terms involving \( x \) (i.e., \( c_3 x \cos x \) and \( c_4 x \sin x \)) must vanish, since these terms grow without bound as \( x \to \infty \). Therefore, we must have \( c_3 = c_4 = 0 \), leaving us with the general solution: \[ y(x) = c_1 \cos x + c_2 \sin x. \] Thus, the space of bounded solutions is spanned by the functions \( \cos x \) and \( \sin x \), so the dimension of the vector space \( W \) is 2. Therefore, the dimension of \( W \) is: \[ \boxed{2}. \]