Originally Posted by

**Silverflow** Hi all,

I'm currently doing a problem based on reduction of order. The question goes:

Consider the DE $\displaystyle L[y] = x(1+3x^2)y''+2y'-6xy = 0$. Given that $\displaystyle y_{1} = \frac{1}{x} $, find a second linearly independent solution.

I can see that the solution to this equation is in the form $\displaystyle y_{2} = u(x)y_{1}$. That means $\displaystyle y_{2} = u(x)x^{-1}$. After rearrange the DE to $\displaystyle L[y_{2}] = y''+\frac{2}{ x(1+3x^2)}y'-\frac{6x}{ x(1+3x^2)}y$, I can see that the LDE for $\displaystyle u'$ will be of the form $\displaystyle u''+(\frac{2y'_{1}+\frac{2}{ x(1+3x^2)}y_{1}}{y_{1}})u'=0$. This simplified to the LDE $\displaystyle u''-\frac{6x}{1+3x^2}u'=0$.

To solve for $\displaystyle u'$, I found an integrating factor $\displaystyle e^{\int\frac{6x}{1+3x^2}dx} = \frac{1}{1+3x^2}$. Using this integrating factor, solving for $\displaystyle u'$:

$\displaystyle \frac{d((\frac{-6x}{(1+3x^2)^2})u')}{dx}= 0$

$\displaystyle \frac{-6x}{(1+3x^2)^2})u' = C$

Therefore $\displaystyle u'=-\frac{C(1+3x^2)^2}{6x}$

When integrating to find u, I'm left with this horrible equation:

$\displaystyle u = \frac{-C(4ln(x)+3x^2(3x^2+4))}{24}+D$

I feel this is completely wrong.

I was wondering if someone could look over my work here and point me in the right direction.

Thank you in advance.