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.
 A: 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 $\left\{ 1, x^2 - \frac{2x}{3}, x^3 - \frac{x}{2} \right\}$.
A: Integrating the first term on the left-hand side by parts gives
$$ \int_0^1 xp'(x) \, dx = [xp(x)]_0^1 - \int_0^1 p(x) \, dx = p(1)-\int_0^1 p(x) \, dx. $$
Therefore, putting this into the condition and cancelling terms, we find we only need
$$ \int_0^1 p(x) \, dx = p(0), $$
which we can also write as $$ \int_0^1 p(x)-p(0) \, dx = 0 $$. Obviously any constant satisfies this. To find the others, the condition is linear, so it's easy to stick in a general polynomial of degree 3 and see what happens; we find that for $p(x) = ax^3+bx^2+cx+d$, we have
$$ \frac{a}{4}+\frac{b}{3}+\frac{c}{2} = 0. $$
Now just find two linearly independent choices of $a,b,c$, and you'll end up with 3 vectors once you include a constant, which is as you would expect for a 4-dimensional vector space with one constraint.
A: Hint:
Just set $p(x)=ax^3+bx^2+cx+d$, compute both sides in function of $a,b,c,d$. There will result a (single) linear equation in $a, b,c,d$. This shows the space has codimension $1$, i.e. has dimension $4-1=3$. Thus one of the coefficients can be expressed in function of the three others. This defines an isomorphism of $\mathbf R^3$ with  the space of solutions. Take for instance the image under this isomorphism of the canonical basis of $\mathbf R^3$.
