MAT244-2018S > Quiz-3

Q3-T0601

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**Victor Ivrii**:

Find the Wronskian of two solutions of the given differential equation without solving the equation.

$$

x^2y'' + xy' + (x^2 - \nu^2)y = 0.

$$

**Mark Buchanan**:

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:

$$W = {c\over x}$$

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