# Legendre Equation

From Apostol's Calculus, Vol. II, Section 6.21 #3:

The Legendre equation can be written in the form $$\left[(x^2 -1)y'\right]'-\alpha(\alpha+1)y=0\,,$$ where $\alpha\in\mathbb R$. If $a, b, c$ are constants with $a>b$ and $4c+1>0$, show that a differential equation of the type $$\left[(x-a)(x-b)y'\right]'-cy=0$$ can be transformed to a Legendre equation by a change of variable of the form $x=At+B$ where $A>0$. Determine $A$ and $B$ interms of $a$ and $b$.

It is easy to determine $A=\frac {a-b} 2$, and $B=\frac {a+b} 2$ (which agrees with the answer provided by the book), and we can find that this yields the equation $$\left[(A^2t^2-A^2)y'\right]'-cy=0\implies\left[(t^2-1)y'\right]'-\frac c {A^2} y =0$$ So to finish the proof that this can be considered as a Legendre equation, we must show that $\exists \alpha\in \mathbb R$ such that $\alpha(\alpha+1)=\frac c {A^2}$, however in solving this we find that $\alpha = \frac 1 2 \left(-1 \pm \sqrt{1+4\frac c {A^2}}\right)$. The condition $4c+1>0$ is not enough to guarantee that $\alpha\in\mathbb R$. For instance, we could take $a=\frac 1 4$, $b=0$, and $c=-\frac 1 8$.

Did I make a mistake, or should the condition instead be $16c + (a-b)^2>0$?

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Via $x=At+B$ and the chain rule we have
$$\frac{d}{dx}(\cdot)=\frac{\frac{d}{dt}(\cdot)}{\frac{dx}{dt}}=\frac{1}{A}\frac{d}{dt}(\cdot).$$
$$\frac{d}{dx}\left[(x-a)(x-b)\frac{d}{dx}y\right]=\frac{1}{A}\frac{d}{dt}\left[A^2(t^2-1)\frac{1}{A}\frac{d}{dt}y\right]=\left[(t^2-1)y\,'\right]'$$