# 2nd order ODE - Euler Equation

• Jun 26th 2009, 12:54 AM
utopiaNow
2nd order ODE - Euler Equation
Given:
$\displaystyle 2x^2y'' + 3xy' + (2x^2 - 1)y = 0$

Question:
Find the indicial equation and determine the two singular roots.

I know how to solve it when the equation is of the form:
$\displaystyle x^2y'' + \alpha xy' + \beta y$

However here, $\displaystyle \beta = 2x^2 - 1$ and is not a constant. How am I supposed to find the indicial/characteristic equation here?

In the solutions, they just ignore the $\displaystyle 2x^2$ factor and proceed as if the equation was $\displaystyle 2x^2y'' + 3xy' - y = 0$. I just don't understand how the $\displaystyle 2x^2$ factor can be ignored.

Any suggestions?
• Jul 6th 2009, 09:52 PM
AlephZero
I assume you're looking for a Frobenius solution. Here's a method for finding the indicial equation when the coefficients are polynomials. Divide each term of the DE by the leading polynomial coefficient, and write the DE in the form:

$\displaystyle y^{\prime\prime}+\frac{p_0+p_1x+p_2x^2+\ldots}{x}y ^{\prime}+\frac{q_0+q_1x+q_2x^2+\ldots}{x^2}=0.$

Now your indicial equation is just $\displaystyle r(r-1)+p_0r+q_0=0.$

So in your example, we can rewrite the DE as

$\displaystyle y^{\prime\prime}+\frac{\frac{3}{2}}{x}y^\prime + \frac{-\frac{1}{2}+x^2}{x^2}y=0,$

so that $\displaystyle p_0=3/2$ and $\displaystyle q_0=-1/2.$ Now the indicial equation is $\displaystyle r(r-1)+\frac{3}{2}r-\frac{1}{2}=0,$ or $\displaystyle (r-\frac{1}{2})(r+1)=0.$