Question

ODE Reduction of Order problem below; trying to figure out how to deal with it again once the order has been reduced.

Answer 1

*

Answer 2

Sorry, was just wondering whether it was worth rewriting everything; using a picture instead.

Answer 3

I'm fairly sure that the reduced ODE\[w'+w=0\]is correct, I'm just not really sure how to move from there due to how it's now set up. It's tempting to treat it like a separation of variables problem, but I believe I've set that up wrong.

Answer 4

Here's the algebra error:
\[\cdots-(x+2)(v'+v)+\cdots~~\implies~~\cdots-xv'-xv-2v'-2v+\cdots\]

Answer 5

Yeah, I caught that and fixed it in the bottom, it's just the resulting, correct (I think) DE that I'm concerned with.


\[w'+w=0\]

Answer 6

Tempted to treat it like separation of variables, but not sure how quite to deal with that due to the repeated substitution.

Answer 7

Accounting for that mistake, you should have
\[xv''+(x-2)v'=0\]
Reducing the order with your substitution gives a linear ODE:
\[xw'+(x-2)w=0\]

Answer 8

Following through I get
\[xu'' + (x-2) u' = 0\]
Where I would then substitute

Answer 9

Wait, nevermind, I made *two* different mistakes, then, I guess...my bad, one moment.

Answer 10

Yeah, got the same.

Answer 11

\[w'+\frac{(x-2)}{x}w=0\]\[\mu(x)=e^{\int\limits_{}^{}\frac{(x-2)}{x}dx}\]

Answer 12

\[\mu(x)=e^{1-2\ln(x)}=e^1e^{-2\ln(x)}=\]I'm a little confused here, because I'm inclined to absorb that e^1 into the constant that is arbitrarily decided to be 1 in most cases, but I'm not sure I'm allowed to do that.

Answer 13

\[\int\frac{x-2}{x}\,dx=\int\left(1-\frac{2}{x}\right)\,dx=x-2\ln x\]
So \(\mu(x)=e^{x-2\ln x}\).

Answer 14

Damnit, making a mistake every other part of the problem. Thank you.

Answer 15

In that case, \[\mu(x)=e^xe^{-2\ln(x)}=e^xe^{\ln(x^{-2})}=\frac{e^x}{x^2}\]