# Problem of the Week #84 - November 4th, 2013

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#### Chris L T521

##### Well-known member
Staff member
Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Consider a differential equation of the form
$A(x)y^{\prime\prime} + B(x)y^{\prime} + C(x)y + \lambda D(x)y = 0.$
Show that you can express this equation in Sturm-Liouville form, given by
$\frac{d}{dx}\left[p(x)\frac{dy}{dx}\right] - q(x)y -\lambda r(x)y = 0.$

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Hint:
First divide every term by $A(x)$ and then multiply the equation by $\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)$.

#### Chris L T521

##### Well-known member
Staff member
This week's problem was correctly answered by Ackbach, MarkFL, and mathbalarka. You can find Mark's solution below.

We are given the following ODE:

$$\displaystyle A(x)y^{\prime\prime} + B(x)y^{\prime} + C(x)y + \lambda D(x)y = 0$$

Dividing through by $A(x)$ we obtain:

$$\displaystyle y^{\prime\prime} + \frac{B(x)}{A(x)}y^{\prime} + \frac{C(x)}{A(x)}y + \lambda \frac{D(x)}{A(x)}y = 0$$

Using the integrating factor:

$$\displaystyle \mu(x)=\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)$$

we obtain:

$$\displaystyle \exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)y^{\prime\prime} + \frac{B(x)}{A(x)}\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)y^{\prime} + \frac{C(x)}{A(x)}\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)y + \lambda \frac{D(x)}{A(x)}\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)y = 0$$

If we use the following definitions:

$$\displaystyle p(x)\equiv \exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)\implies p'(x)=\frac{B(x)}{A(x)}\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)$$

$$\displaystyle q(x)\equiv -\frac{C(x)}{A(x)}\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)$$

$$\displaystyle r(x)\equiv -\frac{D(x)}{A(x)}\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)$$

we may write:

$$\displaystyle p(x)y^{\prime\prime} + p^{\prime}(x)y^{\prime} - q(x)y - \lambda r(x)y = 0$$

Observing that:

$$\displaystyle p(x)y^{\prime\prime} + p^{\prime}(x)y^{\prime}=\frac{d}{dx}\left[p(x)\frac{dy}{dx}\right]$$

we may then write:

$$\displaystyle \frac{d}{dx}\left[p(x)\frac{dy}{dx}\right] - q(x)y -\lambda r(x)y = 0$$

Shown as desired.

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