Question

Find the Wronskian of two solutions (up to a constant multiple) to the differential equation without solving the equation: (1 - 2?)y" - 2xy + 2y = 0 (this is a well-known equation called Legendre's equation. Its solutions are called Legendre functions). Impossible to determine from the given information O where c is a constant 1-22 O ce 1-22 where c is a constant O cln |1 - 22 where c is a constant

Answer 1

solution:

Given differential equation is

(1-2)' – 2.cy' + 2y = 0

$\Rightarrow y''-\frac{2x}{(1-x^2)}y'+\frac{2}{(1-x^2)}y=0$

compare it with

$\Rightarrow y''+P(x)y'+Q(x)y=0$

we have

$P(x)=-\frac{2x}{1-x^2},Q(x)=\frac{2}{1-x^2}$

Now the wroskian of the given differential equation is given by

$W=ce^{-\int P(x)dx}$

$\Rightarrow W=ce^{\int \frac{2x}{1-x^2}dx}$

$\Rightarrow W=ce^{-ln(1-x^2)}$

$\Rightarrow W=ce^{ln(1-x^2)^{-1}}$

$\Rightarrow W=c(1-x^2)^{-1}$

$\Rightarrow W=\frac{c}{(1-x^2)}$

which is the required wroskian.

Thus, second option is the correct option.

this complete the solution.