# Thread: Reduction of Order

1. ## Reduction of Order

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.

2. 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.
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 $\displaystyle L[y] = x(1+3x^2)y''+2y'-6xy = 0$. 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.
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.

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

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!