Rewrite the following differential equation in Sturm-Liouville form $x^2 y''(x)+ 3xy' (x)=-\lambda y(x)$, $x>0$.
Rewrite the following differential equation in Sturm-Liouville form
-
0e^(3 lnx ) y^''+ e^(3 lnx )/x y^'+ (λe^(3 lnx ))/x^2 y=0 is this the final answer of Sturm-Liouville problem...i just want to know the final answer... – 2012-12-10
1 Answers
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}. $