Math Help - Method of Reduction of Order question

1. Method of Reduction of Order question

Here is what I'm looking at:

Use the method of reduction of order to find a second solution to

$x^2 y'' + 3xy' + y = 0$

$x > 0$

$y_1 = x^{-1}$

Here is what I did:

$y'' + \frac{3}{x} y' + x^{-2}y = 0$

$p(x) = 3x^{-1}$

$y_2 = \int {\frac{e^{-3ln(x)}}{(x^{-1})^2}} dx$

$y_2 = \int{\frac{x^{-3}}{x^{-2}}} dx$

$y_2 = \int{x^{-1}} dx$

$y_2 = ln|x| + c$

Now, when I plug the original ODE into a solver, the general solution comes up with this:

$c_1x^{-1} + c_2 \frac{ln(x)}{x}$

Does that mean that my solution to $y_2$ is wrong? And if so, where did I make a mistake?

2. Re: Method of Reduction of Order question

Originally Posted by Lancet
Here is what I'm looking at:

Use the method of reduction of order to find a second solution to

$x^2 y'' + 3xy' + y = 0$

$x > 0$

$y_1 = x^{-1}$

Here is what I did:

$y'' + \frac{3}{x} y' + x^{-2}y = 0$

$p(x) = 3x^{-1}$

$y_2 = {\color{red} {y_1}} \int {\frac{e^{-3ln(x)}}{(x^{-1})^2}} dx$
I believe what you're missing is in red above.

3. Re: Method of Reduction of Order question

Originally Posted by Danny
I believe what you're missing is in red above.

Ai! Yes, that would be it, thanks.

A quick question, though - if I put that in, then the final answer I get is this:

$y_2 = \frac{ln|x| + c}{x}$

Which is decidedly different from what's seen here:

$c_1x^{-1} + c_2 \frac{ln(x)}{x}$

What's causing the discrepancy?

4. Re: Method of Reduction of Order question

Originally Posted by Lancet
Ai! Yes, that would be it, thanks.

A quick question, though - if I put that in, then the final answer I get is this:

$y_2 = \frac{ln|x| + c}{x}$

Which is decidedly different from what's seen here:

$c_1x^{-1} + c_2 \frac{ln(x)}{x}$

What's causing the discrepancy?
You don't need the second constant of integration with $y_2$. Note the following

If $y_2 = \dfrac{\ln|x| + k}{x}$ so

$y = c_1 \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln|x| + k}{x}$.

If you expand, you can absorb the $c_2 k$ into the $c_1$.

5. Re: Method of Reduction of Order question

Originally Posted by Danny
You don't need the second constant of integration with $y_2$. Note the following

If $y_2 = \dfrac{\ln|x| + k}{x}$ so

$y = c_1 \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln|x| + k}{x}$.

If you expand, you can absorb the $c_2 k$ into the $c_1$.

If the $k$ were completely seperable, I could understand that. But because it seperates out into $\frac{k}{x}$, I was under the impression that we couldn't just do away with that the way we can with a duplicate $c$ by itself.

6. Re: Method of Reduction of Order question

Originally Posted by Lancet
If the $k$ were completely seperable, I could understand that. But because it seperates out into $\frac{k}{x}$, I was under the impression that we couldn't just do away with that the way we can with a duplicate $c$ by itself.
Post more details on what you mean.

7. Re: Method of Reduction of Order question

Originally Posted by Danny
Post more details on what you mean.
It's like how if you have $2c, c/10, -c, 1/c, e^c$, etc... you can take all of those cases and just revert it back to being a simple +c. But as soon as it's attached to a variable, such as cx, you can't revert that back to a simple c.

If we were looking at:

$y = c_1 \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln|x|k}{x}$

...then this could be rewritten as

$y = c_1 \cdot \dfrac{1}{x} + c_2 \cdot k \cdot \dfrac{\ln|x|}{x}$

...in which case, I can see the two constants being merged. But with what we actually have, seperating out the k would leave it still attached to a variable. And I don't understand how we can merge it with the other constant while it's in that state.

8. Re: Method of Reduction of Order question

Well, let me provide maybe some more details. What you have is

\begin{aligned}y &= c_1 \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln x + k}{x}\\ & = \underbrace{\left(c_1 + c_2 k \right)}_{\text{call this}\; c_1} \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln x}{x}\\&=c_1 \cdot \dfrac{1}{x} + c_2 \cdot \dfrac{\ln |x|}{x}\end{aligned}

9. Re: Method of Reduction of Order question

Okay, I get it now! Wow, definitely a bit trickier than usual.

Thanks for staying with me and helping me understand this!