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Rewrite the following differential equation in Sturm-Liouville form $x^2 y''(x)+ 3xy' (x)=-\lambda y(x)$, $x>0$.

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  • $\begingroup$ e^(3 ln⁡x ) y^''+ e^(3 ln⁡x )/x y^'+ (λe^(3 ln⁡x ))/x^2 y=0 is this the final answer of Sturm-Liouville problem...i just want to know the final answer... $\endgroup$
    – man
    Dec 10, 2012 at 8:40

1 Answer 1

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What you need to do is write the equation as $$ \frac{d}{d x}\left\{p(x) \frac{d y}{d x}\right\} + \big(q(x) + \lambda r(x)\big)y = 0 $$

In order to do that, multiply the ODE by an integrating factor $\mu(x) > 0$, $$ x^2 \mu y'' + 3 x \mu y' + \lambda \mu y = 0 $$

The first term can be written as $$ \frac{d}{dx} \left\{ x^2 \mu \frac{d y}{d x}\right\} - \frac{d y}{d x} \frac{d}{d x}\left(x^2 \mu\right) $$ and then the ODE becomes $$ \left(x^2 \mu y'\right)' + \left[3 x \mu - \left(x^2 \mu\right)'\right]y' + \lambda \mu y = 0 \tag{1} $$ If $$ 3x\mu - (x^2 \mu)' = 0 \tag{2} $$ then the ODE $(1)$ is of the Sturm-Liouville form. Clearly, the solution for $(2)$ is $$ \mu (x) = x $$ which means that, if we multiply the original equation by $x$, then $$ x\left(x^2 y'' + 3 x y' + \lambda y\right) = x^3 y'' + 3 x^2 y' + \lambda x y = \color{red}{\left(x^3 y'\right)' + \lambda x y = 0}. $$

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