<s>[INST] How do I determine a basis of the vector space of polynomials degree 3 or less that satisfy $\int_{0}^{1}[xp'(x)-p(x)]=p(1)-2p(0)$? I have proven that the set of polynomials of degree 3 or less that satisfy $\int_{0}^{1}[xp'(x)-p(x)]=p(1)-2p(0)$ is indeed a subspace of $\mathscr{P}_{3}(\mathbb{R})$.
Now I need to find a basis. I found that all degree zero polynomials work. I've tried just using the monomials of each degree $(\lambda,0,0,0),(0,\lambda x,0,0),...,(0,0,0,\lambda x^{3})$ and I've found that none of the rest work...  
It doesn't seem right to say that my basis is just one vector, but I'm at a loss on how to find anything else.
 [/INST] Let $p(x) = a + bx + cx^2 + dx^3$. Then the equation reads
$$
\int_0^1 x(b + 2cx + 3d x^2) - (a + bx + cx^2 + dx^3) \mathrm{d} x = a+b+c+d - 2a
$$
Evaluating the left-hand side gives
$$ 
 \frac b2 + \frac{2c}{3} + \frac{3d}{4} - a - \frac{b}{2} - \frac{c}{3} - \frac{d}{4} = a + b + c + d - 2a,
$$
or 
$$
 b = - \frac{2}{3} c - \frac{1}{2} d.
$$
Thus the general form is
$$
 p(x) = a + \left(- \frac{2}{3} c - \frac{1}{2} d \right)x + cx^2 + dx^3 = a + c \left( x^2 - \frac{2}{3} x \right) + d \left( x^3 - \frac{1}{2} x \right)
$$
and we can identify a basis $\left\{ 1, x^2 - \frac{2x}{3}, x^3 - \frac{x}{2} \right\}$.
</s>