# Thread: Trouble with the second root

1. ## Trouble with the second root

Original equation: xy'' + y' + 0. I have found that the roots are both 0, and one of the solutions is:

$y_1=\sum_{m=0}^\infty \frac{(-1)^m a_0}{(m!)^2}x^m$

That solution checks. For y2, I tried using $y_2 = y_1 \int \frac{e^{-\int P(x) dx}}{{y_1}^2}dx$, but using P(x) = 1/x, y2 turned into:

$y_2 = y_1(ln(x) + \frac{x}{2} + \frac{19x^2}{288} + \frac{7x^3}{864} + ...)$

2. ## Re: Trouble with the second root

If the equation is:

$x\frac{d^2y}{dx^2}+\frac{dy}{dx}=0$

then I would simply, rewrite the left side as:

$\frac{d}{dx}\left(x\frac{dy}{dx}\right)=0$

$\int\,d\left(x\frac{dy}{dx}\right)=\int 0\,dx$

$x\frac{dy}{dx}=c_1$

separate variables, losing a trivial solution:

$dy=\frac{c_1}{x}\,dx$

$\int\,dy=c_1\int\frac{1}{x}\,dx$

$y=c_1\ln|x|+c_2$

Obviously, though, I am missing something...

3. ## Re: Trouble with the second root

I mistyped the OP. The differential equation is xy'' + y' + y = 0. (Otherwise, I just set w = y' and solve.)

Everything else is written as intended.

4. ## Re: Trouble with the second root

Originally Posted by phys251
The differential equation is xy'' + y' + y = 0.
$y(x)= c_1 J_0\left(2 \sqrt{x}\right)+2 c_2 Y_0\left(2 \sqrt{x}\right)$