Toronto Math Forum
MAT2442018S => MAT244Tests => Quiz3 => Topic started 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.
$$
(1  x^2)y''  2xy' + \alpha (\alpha + 1)y = 0.
$$

Divide everything by $(1x^2)$ 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 $W = ce^{\intp(x)dx}$ 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 $u = 1x^2$ and $du =2xdx$ we get
$$ ce^{\int{1\over u}du} = ce^{ln(u)+C} = ce^{ln(1x^2)} = c(1x^2)^{1}e^C = {ce^C\over {1x^2}}$$
But $ce^C$ is just some constant, so we can subsume it into just $c$. Simplifying this, we get that the Wronskian is:
$$W = {c\over {1x^2}}$$