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:

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

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

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

which is not even close to the answer in the back of the book. Please advise.

Re: Trouble with the second root

If the equation is:

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

then I would simply, rewrite the left side as:

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

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

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

separate variables, losing a trivial solution:

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

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

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

Obviously, though, I am missing something...(Thinking)

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.

Re: Trouble with the second root

Quote:

Originally Posted by

**phys251** The differential equation is xy'' + y' + y = 0.

$\displaystyle y(x)= c_1 J_0\left(2 \sqrt{x}\right)+2 c_2 Y_0\left(2 \sqrt{x}\right)$