Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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... – man Dec 10 '12 at 8:40

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 $$ x\left(x^2 y'' + 3 x y' + \lambda y\right) = x^3 y'' + 3 x^2 y' + \lambda x y = \color{red}{\left(x^3 y'\right)' + \lambda x y = 0}. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.