Question

Diff Eq:

I'm just not seeing either my calculation errors or logic errors. I'll post the question and my work for it.

Answer 1

\[xy'' + y' = 0; y _{1}=lnx\]
\[y_{2} = \mu(x)lnx\]
\[y_{2}' = \mu'(x)lnx + \frac{ \mu(x) }{ x }\]
\[y''_{2}=\mu''(x)lnx + \frac{ \mu'(x) }{ x }+\frac{ x \mu'(x)-\mu(x) }{ x^{2} }\]
\[x(\mu''(x)lnx + \frac{\mu'(x)}{x}+\frac{x \mu'(x)- \mu(x)}{x^{2}})+\mu'(x)lnx+ \frac{\mu(x)}{x}=0\]
\[x \mu''(x)lnx + (2+lnx) \mu'(x)\]
\[\mu''(x) = w'(x) = \frac{ dw }{ dx }\]
\[\mu'(x) = w(x)\]
\[(xlnx)\frac{dw}{dx}+(2+lnx)w = 0\]
\[\frac{ dw }{ w }+\frac{ (2+lnx)dx }{ xlnx }=0\]
\[\int\limits_{}^{}\frac{dw}{w}=-\int\limits_{}^{}\frac{ 2+lnx }{ xlnx }dx\]
u = 2+lnx du = (1/x)dx dx = xdu
\[\ln|w| = \int\limits_{}^{}\frac{ u }{ x(u-2) }*xdu\]
\[\ln|w| = -\int\limits_{}^{} 1du+2\int\limits_{}^{}\frac{ 1 }{ u-2 }du\]
\[\ln|w| =-u-2\ln|u-2|+C\]
\[\ln|w| = -2-lnx-2\ln|lnx| + C\]

Hope that pastes properly.

Answer 2

In the end, the book gets y2 to be 1. So before figure out how this mess becomes 1, can someone actually check this mess?

Answer 3

\[\int\limits_{}^{}\frac{dw}{w}=-\int\limits_{}^{}\frac{ 2+lnx }{ xlnx }dx\]
\[\int\limits_{}^{}\frac{dw}{w}=-2\int\frac{1}{xlnx}dx-\int \frac 1xdx\]
\[ln~w=-2~ln(lnx)-lnx+C\]

Answer 4

your messing it up along in there i beleive

Answer 5

I thought of that option, but I thought I could keep it as one fraction and try to u-sub it. I guess I didnt want to deal with 1/(xlnx)

Answer 6

\[\ln|w| = \int\limits_{}^{}\frac{ u }{ (u-2) }du\]
\[\ln|w| = \int\frac{ u-2+2 }{ (u-2) }du\]
\[\ln|w| = \int1+2\frac{1}{u-2}du\]
\[\ln|w| = u+2ln(u-2)+C\]
\[\ln|w| = 2+lnx+2ln(lnx)+C\]

Answer 7

the 2 can be absorbed by the C

Answer 8

Oh, right. I guess it would be something silly.

Answer 9

So wait,t heres no negative?

Answer 10

\[\ln|w| = lnx+2ln(lnx)+C\]
\[w = x~ln^2x+C\]
i might have dropped it someplace .... was kinda wondering that meself

Answer 11

that +C on the end is off too, spose to be a multiplier

Answer 12

i had copied the latex from yours, and YOU dropped it on that one lol

Answer 13

the end of it looks fine now, just absorb that -2 into the arbitrary C and "e" out the sides; the x ln^2x drops to a demoninator

Answer 14

Alright, I think I had something like that earlier before I ruined it, lol. Okay, so gotta integrate it again.....

Oh. My bad, lol. Funny thing was I was trying to be conscious of it x_x I think I corrected it and didnt save it somehow, dunno, haha.

\[C(\frac{ x^{2}(lnx)^{2} }{ 2 }-\frac{ x^{2}(lnx) }{ 2 }+\frac{ x^{2} }{ 4 }) + C_{2}\]

Answer 15

google snapped ....

Answer 16

not good, lol.

Answer 17

yeah, this site is not behaving for me at the moment ....

Answer 18

Not surprised x_xeither way, I tried by parts, lol.

Answer 19

\[\int\frac{1}{x~ln^2(x)}dx\]
\[a=ln(x)\]\[da=\frac1xdx\]
\[\int\frac{1}{a^2}da=-\frac{1}{a}\to -\frac{1}{ln(x)}\]

Answer 20

+C of course