Question

Let the subspace \( H \) of \( P_3(\mathbb{R}) \) be defined as \[ H = \{ p(x) \in P_3(\mathbb{R}) : xp'(x) = 3p(x) \}. \] Then, the dimension of \( H \) is equal to ..............

Hint:

To find the dimension of a subspace defined by a differential equation, solve the equation and determine the number of linearly independent solutions.

Answer 1

Step 1: Analyze the condition for the subspace.
The condition for \( p(x) \in H \) is: \[ xp'(x) = 3p(x). \] This is a first-order linear differential equation for \( p(x) \). We solve it using separation of variables. Step 2: Solve the differential equation.
Rewriting the equation: \[ \frac{p'(x)}{p(x)} = \frac{3}{x}. \] Integrating both sides: \[ \ln |p(x)| = 3 \ln |x| + C \quad \Rightarrow \quad p(x) = Cx^3. \] Step 3: Determine the dimension of \( H \).
Thus, the solutions to the differential equation are of the form \( p(x) = Cx^3 \), where \( C \) is a constant. Therefore, \( H \) is spanned by \( \{ x^3 \} \), so the dimension of \( H \) is 1. Final Answer: \[ \boxed{1}. \]