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 \( y = f(x) \) be the solution of the differential equation\[\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^6 + 4x}{\sqrt{1 - x^2}}, \quad -1 < x < 1\] such that \( f(0) = 0 \). If \[6 \int_{-1/2}^{1/2} f(x)dx = 2\pi - \alpha\] then \( \alpha^2 \) is equal to ______.
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: