Legendre's equation

[logistic_guy]

Solve.

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

 

[khansaheb]

Solve.

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

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[logistic_guy]

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

Thank you Sir khan.

 

[logistic_guy]

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

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:

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