Consider the equation (*): (d)/(dx)l[(1-x^2)(dy)/(dx)r] + ay…

Consider the equation $(*)\!: \qquad \frac{d}{dx}\Bigl[(1-x^2)\frac{dy}{dx}\Bigr] + ay = 0, \qquad \text{i.e.} \qquad (1-x^2)y''-2xy'+ay=0$. 1° Show that if $y=f_1+f_2$ ($f_1$ even, $f_2$ odd) solves $(*)$, then $f_1$ and $f_2$ individually solve $(*)$. 2° Find $a$ so that $(*)$ has a degree-3 polynomial solution; give that solution and the general solution.

Step 1: The operator under $x\to-x$

Write $L[y] = (1-x^2)y''-2xy'+ay$, and suppose $L[y]=0$. Define $\tilde y(x):=y(-x)$. Since $\dfrac{d}{dx}\tilde y(x) = -y'(-x)$ and $\dfrac{d^2}{dx^2}\tilde y(x) = y''(-x)$, substituting $x\to-x$ into $L[y]=0$ gives $(1-x^2)y''(-x)+2xy'(-x)+ay(-x)=0$; rewritten in terms of $\tilde y$, the two sign flips (one from $\tilde y'=-y'(-x)$, one from the coefficient $-x\to x$) exactly cancel, reproducing the same operator:

\[ L[\tilde y] = (1-x^2)\tilde y'' - 2x\tilde y' + a\tilde y = 0. \]

So $\tilde y(x)=y(-x)$ is again a solution whenever $y$ is — the equation is invariant under $x\to-x$.

Step 2: Even and odd parts each solve $(*)$

By linearity of $L$, both $L[y+\tilde y]=0$ and $L[y-\tilde y]=0$. Since $y=f_1+f_2$ with $f_1$ even and $f_2$ odd, $\tilde y(x)=y(-x)=f_1(x)-f_2(x)$, so

\[ y+\tilde y = (f_1+f_2)+(f_1-f_2) = 2f_1, \qquad y-\tilde y = (f_1+f_2)-(f_1-f_2) = 2f_2. \]

Hence $L[2f_1]=0$ and $L[2f_2]=0$, and by linearity again, $L[f_1]=0$ and $L[f_2]=0$. $\quad\blacksquare$

Step 3: Seeking a degree-3 polynomial solution

By Part 1°, a degree-3 polynomial solution is automatically odd (its even part would have to solve $(*)$ separately at a lower degree, and the leading behavior forces the ansatz), so try $y=\alpha x^3+\beta x$, with $y'=3\alpha x^2+\beta$, $y''=6\alpha x$. Substituting into $(1-x^2)y''-2xy'+ay=0$:

\[ (1-x^2)(6\alpha x) - 2x(3\alpha x^2+\beta) + a(\alpha x^3+\beta x) = (a\alpha-12\alpha)x^3 + (6\alpha-2\beta+a\beta)x = 0. \]

Step 4: Solving for $a$ and the coefficients

The $x^3$ coefficient gives $\alpha(a-12)=0$; since $\alpha\neq0$ for a genuine degree-3 solution, $a=12$. The $x^1$ coefficient then gives $6\alpha-2\beta+12\beta=0$, i.e. $6\alpha+10\beta=0$, so $\beta=-\tfrac35\alpha$. Choosing $\alpha=5$ clears denominators: $\beta=-3$, giving

\[ a = 12, \qquad y_1 = 5x^3-3x. \]

Substituting $y_1=5x^3-3x$ into $(1-x^2)y''-2xy'+12y$ gives a residual of exactly $0$.\ \checkmark\)

Step 5: The second independent solution

The Legendre equation of order $3$ has a second, linearly independent solution — the Legendre function of the second kind — given for $|x|\lt 1$ by

\[ Q_3(x) = \frac{5x^3-3x}{4}\,\ln\!\frac{1+x}{1-x} - \frac52x^2 + \frac23. \]

Step 6: Verification of both solutions

Direct substitution of $y_1=5x^3-3x$ into $(1-x^2)y''-2xy'+12y$ gives residual $0$, confirmed above. Substituting $Q_3(x)$ into the same operator — differentiating the logarithmic term with the product and chain rules and simplifying the resulting rational and logarithmic pieces — likewise gives residual $0$ identically for $|x|\lt 1$.\ \checkmark\)

Step 7: General solution

\[ y = C_1(5x^3-3x) + C_2\left[\frac{5x^3-3x}{4}\ln\frac{1+x}{1-x} - \frac52x^2+\frac23\right], \qquad |x|\lt 1. \]

\[a=12, \qquad y = C_1(5x^3-3x) + C_2\left[\frac{5x^3-3x}{4}\ln\frac{1+x}{1-x} - \frac52x^2+\frac23\right]\]