I want to choose constants $a$ and $b$ such that the process $$X_t = t^aP\left(\frac{B_t}{t^b}\right)$$ is a martingale, where $B_t$ is a Brownian Motion and $P(y)$ is a polynomial of degree n.
Thus far, I have used Ito's lemma to determine that the following equation must be true for $X_t$ to be a martingale (This expression is that associated with the drift term):
$$a\cdot P\left(\frac{B_t}{t^b}\right) - b\cdot \frac{B_t}{t^b}\cdot P'\left(\frac{B_t}{t^b}\right) + \frac{1}{2}\cdot \frac{t}{t^{2b}}\cdot P''\left(\frac{B_t}{t^b}\right) = 0 $$
This implies $$a\cdot P(y) - by\cdot P'(y) + \frac{1}{2}\cdot \frac{t}{t^{2b}}\cdot P''(y) = 0$$ where $y = \frac{B_t}{t^b}$.
From there, I realized that this equation must hold for all values of y, which can only happen if $\frac{t}{t^{2b}} = 1 \Longrightarrow b = \frac{1}{2} $. I am, thus, left with the following ODE:
$$ aP(y) - \frac{1}{2}yP'(y)+\frac{1}{2}P''(y) = 0$$
However, I do not know how I should proceed from here to solve for $a$. I know that it should be $\frac{1}{2}$, but I don't understand why. I just know that it has something to do with the highest degree term. I'd appreciate any guidance and explanations.
Also, once I find the ODE with which I have to work, I need to look specifically at a polynomial of degree 2 and determine what its coefficients must be. I tried to do this by setting $P(x)=c_1x^2 + c_2x + c_3$. Then $P'(x) = 2c_1x+c_2$ and $P''(x) = 2c_1$. I plugged this into the tentative ODE above and got: \begin{align*} (c_1x^2 + c_2x + c_3) -x(2c_1x+c_2) + 2c_1 &=0 \\ -c_1x^2 + (2c_1 + c_3) &= 0 \end{align*} Thus, to make the left hand side of the equation equal 0, I thought that I should make all the coefficients also equate to 0. However, if $c_1 = 0$, it's no longer a second-degree polynomial. Am I not thinking about this the right way?
Thus, to summarize my questions are: (1) How do I solve for a? (2) How do I find the coefficients for a polynomial of degree 2?
(1) Let $P$ be an $n$-th degree polynomial, say the highest degree term is $c_n y^n$. Then the $n$-th degree term of $yP'(y)$ is $nc_ny^{n}$ and $P''(y)$ does not have an $n$-th degree term. So the $n$-th term of $$ aP(y) - \frac 12 yP'(y) + \frac 12 P''(y) = 0 $$ is $$ ac_n y^n - \frac 12 n c_n y^n = 0 $$ which gives $a = \frac n2$.
(2) For $n = 2$, $a = 1$, hence the ode reads $$ 0 = (c_2x^2 + c_1x + c_0) - \frac 12x(2c_2x + c_1) + c_2 = \frac 12 c_1 x + (c_0 + c_2) $$ So we must have $c_1 = 0$ and $c_0 = -c_2$, hence $$ P(y) = c_2(y^2 - 1). $$