Toronto Math Forum

MAT244-2018S => MAT244--Tests => Quiz-3 => Topic started by: Victor Ivrii on February 10, 2018, 05:18:50 PM

Title: Q3-T0601
Post by: Victor Ivrii on February 10, 2018, 05:18:50 PM

Find the Wronskian of two solutions of the given differential equation without solving the equation.
$$
x^2y'' + xy' + (x^2 - \nu^2)y = 0.
$$

Title: Re: Q3-T0601
Post by: Mark Buchanan on February 10, 2018, 05:22:50 PM

Find the Wronskian of two solutions of the given differential equation without solving the equation.
$$x^2y'' + xy' + (x^2-v^2)y = 0$$

Divide everything by $x^2$ to get $y''$ by itself.
$$y'' + {1\over x}y' + {(x^2-v^2)\over x^2}y = 0$$

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 $1\over x$ in this case.  Now we solve the integral:
$$ce^{-\int{1\over x}dx} = ce^{-ln(x)+C} = ce^{ln(x^{-1})+C} = cx^{-1}e^C$$

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