For which value of α does the Legendre equation (1-x^2)y''…

For which value of $\alpha$ does the Legendre equation $(1-x^2)y'' - 2xy' + \alpha y = 0$ have a particular solution that is a degree-3 polynomial? Find that solution.

Step 1: Setting up the polynomial ansatz

Seek a genuine degree-3 solution $y=a_0+a_1x+a_2x^2+a_3x^3$ with $a_3\neq0$. Then

\[ y' = a_1+2a_2x+3a_3x^2, \qquad y'' = 2a_2+6a_3x. \]

Step 2: Substituting and collecting by powers of $x$

Substitute into $(1-x^2)y''-2xy'+\alpha y=0$: $(1-x^2)(2a_2+6a_3x) - 2x(a_1+2a_2x+3a_3x^2) + \alpha(a_0+a_1x+a_2x^2+a_3x^3) = 0$.

Expanding and collecting the coefficient of each power of $x$ gives four equations: \[\begin{aligned} x^0:&\quad 2a_2 + \alpha a_0 = 0,\\ x^1:&\quad 6a_3 + (\alpha-2)a_1 = 0,\\ x^2:&\quad (\alpha-6)a_2 = 0,\\ x^3:&\quad (\alpha-12)a_3 = 0. \end{aligned}\]

Step 3: Forcing a genuine degree-3 term

Since a true degree-3 solution requires $a_3\neq0$, the $x^3$ equation $(\alpha-12)a_3=0$ forces

\[ \alpha = 12. \]

Step 4: Solving for the remaining coefficients at $\alpha=12$

With $\alpha=12$, the $x^2$ equation becomes $(12-6)a_2=6a_2=0$, so $a_2=0$. The $x^0$ equation then gives $2(0)+12a_0=0$, so $a_0=0$. The $x^1$ equation gives

\[ 6a_3 + (12-2)a_1 = 6a_3+10a_1 = 0 \;\Longrightarrow\; a_1 = -\frac{3a_3}{5}. \]

Choosing $a_3=5$ for integer coefficients gives $a_1=-3$, $a_2=0$, $a_0=0$, so

\[ y = 5x^3 - 3x = x(5x^2-3). \]

Step 5: Verification

With $y=5x^3-3x$: $y'=15x^2-3$, $y''=30x$. Substituting into $(1-x^2)y''-2xy'+12y$: $(1-x^2)(30x) - 2x(15x^2-3) + 12(5x^3-3x) = 30x-30x^3 - 30x^3+6x + 60x^3-36x = 0$. $\checkmark$