Legendre's equation

logistic_guy said:
\(\displaystyle (1 - x^2)y'' - 2xy' + 2y = 0\)
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The general form of Legendre equation is:

\(\displaystyle (1 - x^2)y'' - 2xy' + n(n+1)y = 0\)

Which has the general solution:

\(\displaystyle y(x) = c_1P_n(x) + c_2Q_n(x)\)

Therefore,

\(\displaystyle (1 - x^2)y'' - 2xy' + 2y = 0\)

is a special case where \(\displaystyle n(n+1) = 2\) in this case.

Solving for \(\displaystyle n\) give us:

\(\displaystyle n = 1\) or \(\displaystyle n = -2\)

We know that the indices of Legendre starts at \(\displaystyle n = 0\), so we can safely ignore negative ones.

Then the general solution to \(\displaystyle (1 - x^2)y'' - 2xy' + 2y = 0\) is:

\(\displaystyle y(x) = c_1P_1(x) + c_2Q_1(x)\)

Can we write this solution in different form? The answer is yes!

From accumulated knowledge we know that \(\displaystyle P_1(x) = x\). Now we have two options to find \(\displaystyle Q_1(x)\). The first is that a month ago we learnt a method that gives a second solution if we know the first solution. Or we just become lazy and look at the book or internet to grab the second solution.

I am lazy now so my book says:

\(\displaystyle Q_1(x) = \frac{x}{2}\ln\left|\frac{1 + x}{1 - x}\right| - 1\)

And the general solution becomes:

\(\displaystyle y(x) = c_1x + c_2\left(\frac{x}{2}\ln\left|\frac{1 + x}{1 - x}\right| - 1\right)\)

😉😉
 
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