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).

Question

If $y = f(x)$ is the solution of the differential equation $(1 + \sin x) \frac{dy}{dx} + \cos x = 0$, such that $f(0) = 0$, then $f\left(\frac{\pi}{2}\right)$ is:

Answer 1

Step 1: View the differential operator structurally
Instead of immediately solving the equation, rewrite the differential equation using operator notation:

\[ (D^4 + 2D^2 + 1)y = 0, \quad \text{where } D=\frac{d}{dx}. \]

Factor the operator:

\[ D^4 + 2D^2 + 1 = (D^2 + 1)^2. \]

Step 2: Interpret the factorization
The equation

\[ (D^2 + 1)^2 y = 0 \]

means that every solution of $(D^2+1)y=0$ is also a solution, and in addition, generalized solutions arise because the factor is repeated.

The basic equation

\[ (D^2+1)y=0 \]

has solutions:

\[ y=\cos x,\quad y=\sin x. \]

Step 3: Identify which solutions are bounded
Because the factor $(D^2+1)$ is repeated, extra solutions of the form

\[ x\cos x,\quad x\sin x \]

also appear in the full solution space. However, these grow linearly in magnitude as $|x|\to\infty$, and hence are unbounded.

Only $\cos x$ and $\sin x$ remain bounded for all real $x$.

Step 4: Determine the dimension of $W$
The space $W$ of bounded solutions is therefore:

\[ W=\text{span}\{\cos x,\sin x\}. \]

This space has two linearly independent basis functions.

Final Answer:

\[ \boxed{2} \]

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).

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