1. Reduction of Order

Hi all,
I'm currently doing a problem based on reduction of order. The question goes:
Consider the DE $L[y] = x(1+3x^2)y''+2y'-6xy = 0$. Given that $y_{1} = \frac{1}{x}$, find a second linearly independent solution.

I can see that the solution to this equation is in the form $y_{2} = u(x)y_{1}$. That means $y_{2} = u(x)x^{-1}$. After rearrange the DE to $L[y_{2}] = y''+\frac{2}{ x(1+3x^2)}y'-\frac{6x}{ x(1+3x^2)}y$, I can see that the LDE for $u'$ will be of the form $u''+(\frac{2y'_{1}+\frac{2}{ x(1+3x^2)}y_{1}}{y_{1}})u'=0$. This simplified to the LDE $u''-\frac{6x}{1+3x^2}u'=0$.
To solve for $u'$, I found an integrating factor $e^{\int\frac{6x}{1+3x^2}dx} = \frac{1}{1+3x^2}$. Using this integrating factor, solving for $u'$:
$\frac{d((\frac{-6x}{(1+3x^2)^2})u')}{dx}= 0$
$\frac{-6x}{(1+3x^2)^2})u' = C$
Therefore $u'=-\frac{C(1+3x^2)^2}{6x}$
When integrating to find u, I'm left with this horrible equation:
$u = \frac{-C(4ln(x)+3x^2(3x^2+4))}{24}+D$
I feel this is completely wrong.
I was wondering if someone could look over my work here and point me in the right direction.

2. Originally Posted by Silverflow
Hi all,
I'm currently doing a problem based on reduction of order. The question goes:
Consider the DE $L[y] = x(1+3x^2)y''+2y'-6xy = 0$. Given that $y_{1} = \frac{1}{x}$, find a second linearly independent solution.

I can see that the solution to this equation is in the form $y_{2} = u(x)y_{1}$. That means $y_{2} = u(x)x^{-1}$. After rearrange the DE to $L[y_{2}] = y''+\frac{2}{ x(1+3x^2)}y'-\frac{6x}{ x(1+3x^2)}y$, I can see that the LDE for $u'$ will be of the form $u''+(\frac{2y'_{1}+\frac{2}{ x(1+3x^2)}y_{1}}{y_{1}})u'=0$. This simplified to the LDE $u''-\frac{6x}{1+3x^2}u'=0$.
To solve for $u'$, I found an integrating factor $e^{\int\frac{6x}{1+3x^2}dx} = \frac{1}{1+3x^2}$. Using this integrating factor, solving for $u'$:
$\frac{d((\frac{-6x}{(1+3x^2)^2})u')}{dx}= 0$
$\frac{-6x}{(1+3x^2)^2})u' = C$
Therefore $u'=-\frac{C(1+3x^2)^2}{6x}$
When integrating to find u, I'm left with this horrible equation:
$u = \frac{-C(4ln(x)+3x^2(3x^2+4))}{24}+D$
I feel this is completely wrong.
I was wondering if someone could look over my work here and point me in the right direction.
Process is correct. I don't have time to review the work right now, but what I would suggest is find the general solution via the characteristic equation and compare that to what you have.

3. Originally Posted by Silverflow
Hi all,
I'm currently doing a problem based on reduction of order. The question goes:
Consider the DE $L[y] = x(1+3x^2)y''+2y'-6xy = 0$. Given that $y_{1} = \frac{1}{x}$, find a second linearly independent solution.

I can see that the solution to this equation is in the form $y_{2} = u(x)y_{1}$. That means $y_{2} = u(x)x^{-1}$. After rearrange the DE to $L[y_{2}] = y''+\frac{2}{ x(1+3x^2)}y'-\frac{6x}{ x(1+3x^2)}y$, I can see that the LDE for $u'$ will be of the form $u''+(\frac{2y'_{1}+\frac{2}{ x(1+3x^2)}y_{1}}{y_{1}})u'=0$. This simplified to the LDE $u''-\frac{6x}{1+3x^2}u'=0$.
To solve for $u'$, I found an integrating factor $e^{\int\frac{6x}{1+3x^2}dx} = \frac{1}{1+3x^2}$. Using this integrating factor, solving for $u'$:
$\frac{d((\frac{-6x}{(1+3x^2)^2})u')}{dx}= 0$
$\frac{-6x}{(1+3x^2)^2})u' = C$
Therefore $u'=-\frac{C(1+3x^2)^2}{6x}$
When integrating to find u, I'm left with this horrible equation:
$u = \frac{-C(4ln(x)+3x^2(3x^2+4))}{24}+D$
I feel this is completely wrong.
I was wondering if someone could look over my work here and point me in the right direction.
Well, I just finished my Applied Math exam about an hour ago, let's get right back into it!

Your integrating factor is not correct.

$e^{6 \int \frac{x}{1+3x^2} } = e^{ln(1+3x^2)} = 1+3x^2$

4. Thanks very much, mate.

5. Actually, the correct integrating factor should be:

e^p(x)=e^-ln(1+3x^2)=-1/(1+3x^2)

This is because the p(x) value in this case is negative. Hope this helps!