Question

Reduction of Order:

Professor didn't have time to go over it, but still assigned it for hw. Anyone mind working me through a problem?

Answer 1

\[x^{2}y'' - xy' + 2y = 0; y _{1}= xsin(lnx)\]

Answer 2

And sorry if I reply slow, working a bit. Any help is appreciated still.

Answer 3

reduction of order tends to imply solving a u' equaivalent setup

Answer 4

Yeah, finding a way to makeit linear in u using a sub, correct?

Answer 5

yep

Answer 6

I just dont know the steps in doing that, lol. My book is pathetic :/

Answer 7

Pauls tends to be very readable to me

http://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx

Answer 8

assume y2 = u y1

Answer 9

Right, so y2 = uxsin(lnx)

Answer 10

and y2', y2'' from that gets us?

Answer 11

Oh joy.

y' = u'xsin(lnx) + usin(lnx) + ucos(lnx)

y'' = u''xsin(lnx) + u'sin(lnx) + u'cos(lnx) + u'sin(lnx) + ucos(lnx)/x + u'cos(lnx) - usin(lnx)/x

I know Ill have to simplify that, assuming I did it right. But Yeah, plug it in...

x^2(u''xsin(lnx)+2u'sin(lnx) + 2u'cos(lnx) + (u/x)[cos(lnx) - sin(lnx)) - x(u'xsin(lnx) + usin(lnx) + ucos(lnx)) + 2(uxsin(lnx)

Yeah, thats awful.

Answer 12

:)

\[u''x^3\sin(\ln x)+2u'x^2[\sin(\ln x)+\cos(\ln x)]-ux[\sin(\ln x)-\cos(\ln x)]\\
~~~~~~~~~~~~~~~~~~~~~~-u'x^2[\sin(\ln x)]~~~~~~~~~~~~~~~~~-ux[\sin(\ln x+\cos(\ln x))]\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ +2ux[\sin(\ln x)]
\]

Answer 13

Ill take your word for it. Dunno if you simplifie it more or if I just made a mistake in all that mess. Bit distracted, so forgive me xD

Answer 14

\[u''x^3\sin(\ln x)+u'x^2\sin(\ln x)+2u'x^2\cos(\ln x)]+ux[\cos(\ln x)]\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-ux[\cos(\ln x))]\\\]


\[u''x^3\sin(\ln x)+u'x^2[\sin(\ln x)+2\cos(\ln x)=0 \]

Answer 15

Nice :P

Answer 16

since the term in u drops out, we can "reduce" this by an order: let w=u'

Answer 17

looks like i forgot some brackets and what not in the latex code :) but i think its still readable

Answer 18

looks like i forgot some brackets and what not in the latex code :) but i think its still readable

Answer 19

\[w'x^3\sin(\ln x)+w~x^2[\sin(\ln x)+2\cos(\ln x)]=0\]
\[w'\sin(\ln x)+\frac1xw[\sin(\ln x)+2\cos(\ln x)]=0\]
\[w'+\frac1x[1+2cos(\ln x)]~w=0\]

Answer 20

cotangent in that last one, not cos

Answer 21

The term in u drops out? Obviously not talking about derivatives. So is that just when the u term with no derivatives drops? And you said because of that, does that mean its supposed to happen or did we get lucky with something?

Answer 22

its spose to happen in order to have the u'' and u' parts all alone

Answer 23

since the name of a variable is not important, we can state this in terms of say: u'' = w'; and u' = w to "reduce" the order

Answer 24

So integrating factor is xsin^2(x)

Answer 25

maybe ... with any luck :)

Answer 26

cot, not cos

Answer 27

Yeah, Im pretty sure I did it with cot, lol.

Answer 28

\[w'+\frac1x[1+2\cot(\ln x)]~w=0\]

Answer 29

So, assuming Im right, I would have just :
\[\int\limits_{}^{}(wxsin^{2}x)'= 0 \]
\[wxsin^{2}x = C\]

Im just guessing that its supposed to do that, but im not 100% positive considering its 0 over there.

Answer 30

oops, meant the x in sin to be lnx, lol

Answer 31

\[\int~\frac1x+\frac2x\cot(\ln x)~dx\]
i spose since the lnx, that makes the right part simpler to deal with
\[\ln x+2\ln(\sin(\ln x))\]

Answer 32

Okay, so I got it Im pretty sure, just forgot to keep the angle of sin as lnx.

Answer 33

\[e^{\ln x+2\ln(\sin(\ln x))}w=C\]
etc ... looks like a hoot

Answer 34

you did fine

Answer 35

since w=u', we can determine u; since we know y2 = u y1, we can then determine y2

Answer 36

Okay, cool. so....that means \[\mu'(x) = (wsin^{2}(lnx) + C)\]?

Answer 37

Or no, replace w with u', right

Answer 38

solve for w

Answer 39

yes, replace u' with the results for w

Answer 40

\[\mu' (x)\sin^{2}(lnx) = C\]
\[\mu'(x) = C*\csc^{2}(lnx) \]

Answer 41

Integrate again, right?

Answer 42

\[w=C\frac{\csc^2(\ln x)}{x}\]
\[u'=C\frac{\csc^2(\ln x)}{x}\]

Answer 43

integrate again, yes

Answer 44

Oops, x in there x_x

Answer 45

u = -C*cot(lnx) + C2

Answer 46

looking good; c1 and c2 might clean it up more

Answer 47

Right.
\[y _{2}=(-C _{1}\cot(lnx) + C _{2})xsin(lnx)\]

Answer 48

We need a C1 and C2, though, right?

Answer 49

since c1 is arbitrary, then -c1 is redundant

Answer 50

yes, y'' tells us that there are 2 arbitrary constants

Answer 51

Right, lol. I just didnt want to mess up, so I put it in still xD

Answer 52

Okay, so the book just decided to pick 2 numbers for C1 and C2. Does that mean they can be any 2 numbers?

Answer 53

notice that y1 is incorporated into y2 at the moment, so the full solution would amount to y2

Answer 54

yes; and if we had some initial conditiions we could then specify values of c1 and c2 as well

Answer 55

Yeah, I wasnt sure because usually textbooks would put something liek "answers may vary" in the backbut all of these have specific answers, so i wasnt sure.

Answer 56

\[y = c_1 x \sin(\ln x)+c_2 x \cos(\ln x)\]

Answer 57

So I guess it doesnt matter what C1 and C2 I pick. Okay, cool. SO another thing I wasnt sure what to do with. Basically you do all the substituting and then replace u'' with w', but all I had left was
\[w'e^{2x} = 0\]So how would I get my integrating factor out of that without a regular w term?

Answer 58

theres no need to play with that .... our goal was in determining \(u\):
\[u'=w\]
\[u=\int w\]

Answer 59

Oh O.o Okay, cool, wouldnt have know. Alright, thanks a lot for taking thte time ^_^

Answer 60

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

Ill stop there. In the end, it says y2 is 1 in my book :/ So I want to make sure I did this lead-up stuff correct and then yeah, somehow figure out how the book gets 1, lol.

Answer 61

\[xy'' + y' = 0;~y _{1}=lnx\]
\[y=u~lnx\]
\[y'=u'~lnx+ux^{-1}\]
\[y''=u''~lnx+2u'x^{-1}-ux^{-2}\]
---------------------

\[u''x~lnx+2u'-ux^{-1}+u'~lnx+ux^{-1}=0\]
\[u''x~lnx+(2+lnx)u'=0\]
----------------------
\[w'x~lnx+(2+lnx)w=0\]
\[w'+(\frac{2}{x~lnx}+\frac1x)w=0\]
\[\frac 1w dw=-(\frac{2}{x~lnx}+\frac1x)dx \]
\[ln~w=-2ln(ln~x)-ln~x+C \]
\[w=C~\frac{1}{x~ln^{2}(x)} \]
im up to here

Answer 62

\[u = C_1\frac{1}{ln(x)}+C_2\]

Answer 63

\[y=u lnx\]
\[y=( C_1\frac{1}{ln(x)}+C_2) lnx\]
\[y= C_1\frac{ln(x)}{ln(x)}+C_2 lnx\]