Transforming an equation into a Legendre equation

[InLimbo]

I've been spending way too long trying to figure out how to get started on this problem, any help would be appreciated.

Let a and b be real numbers satisfying a > b and 4c + 1 > 0. Transform the equation
[(x-a)(x-b)u']' -cy
into a Legendre's equation. (I assume that equation is supposed to be suffixed with "= 0" but the above is how the problem was presented to me.)
For reference, a Legendre's equation is of the form (1-x^2)y'' - 2xy' + n(n+1)y = 0.

 

[Deleted member 4993]

I've been spending way too long trying to figure out how to get started on this problem, any help would be appreciated.

Let a and b be real numbers satisfying a > b and 4c + 1 > 0. Transform the equation

[(x-a)(x-b)u']' -cy

into a Legendre's equation. (I assume that equation is supposed to be suffixed with "= 0" but the above is how the problem was presented to me.)
For reference, a Legendre's equation is of the form (1-x^2)y'' - 2xy' + n(n+1)y = 0.

Is that correct?

I am assuming u' means \(\displaystyle \frac{du}{dx}\)

If it is - then what is the relationship between u and y?

 

[HallsofIvy]

Well, that's the question isn't it? Finding the relationship between u, in the given equation, and y, in the standard form for Legendre's equation, would be finding how to convert one to the other!

But what we really need to do is to replace x with a new variable, not u. Since \(\displaystyle (a- x)(b- x)= x^2- (a+ b)x+ ab\), I would recommend "completing the square" so that we have something of the form "\(\displaystyle p^2- x^2\)". Then factor out a \(\displaystyle p\) to have \(\displaystyle p^2(1- x^2/p^2)\) and now make the substitution t= x/p.

 

[Deleted member 4993]

OP has a 'y' in the original equation.

[(x-a)(x-b)u']' -cy

.