# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Quiz-3 => Topic started by: Victor Ivrii on February 10, 2018, 05:20:30 PM

Title: Q3-T0901
Post by: Victor Ivrii on February 10, 2018, 05:20:30 PM
Find the Wronskian of two solutions of the given differential equation without solving the equation.
$$(1 - x^2)y'' - 2xy' + \alpha (\alpha + 1)y = 0.$$
Title: Re: Q3-T0901
Post by: Mark Buchanan on February 10, 2018, 05:39:31 PM
Divide everything by $(1-x^2)$ to get $y''$ by itself.
$$y'' - {2x\over {1 - x^2}}y' + {α(α + 1)\over {1 - x^2}}y = 0$$

Now that it is in the proper form, we can use Abel's theorem of $W = ce^{\int-p(x)dx}$ where c is a constant and $p(x)$ is $- {2x\over {1 - x^2}}$ in this case.  Now we solve the integral:
$$ce^{\int{2x\over {1 - x^2}}dx}$$

Using the substitution $u = 1-x^2$ and $du =-2xdx$ we get
$$ce^{-\int{1\over u}du} = ce^{-ln(u)+C} = ce^{-ln(1-x^2)} = c(1-x^2)^{-1}e^C = {ce^C\over {1-x^2}}$$

But $ce^C$ is just some constant, so we can subsume it into just $c$.  Simplifying this, we get that the Wronskian is:
$$W = {c\over {1-x^2}}$$