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Comments:
<0> One for every string of length n on {x,y}
<1> Cale: Yeah, as you said, we were having trouble with the extra negation with De Morgan's - however, we seem to be having better luck with the distributive law now... thanks :)
<0> nmee: Yeah, De Morgan's law is more useful when you have something that's not in normal form to begin with, I think. You can use it to push negations *inward*
<2> yep
<3> aloha
<4> can anyone here help me with an analytical solution to dy/dx = (1+xy)^2, y(0)=1 ?
<4> solution for y that is
<5> Ugh, that's ugly.
<5> % DSolve[{y'[x]==(1+x y[x])^2,y[0]==1},y[x],x]
<6> |Steve|: {{y[x] -> -(-1 + x)^(-1)}}
<5> Well, there you go: y(x) = -1/(x-1).
<4> thanks
<5> I suppose y(x) = 1/(1-x) is simpler.
<4> any suggestion as to how to derive that?
<4> I need to show some working
<5> Not really. Differential equations are all different.
<5> Someone must have an idea though.
<4> well 1/(1-x) works
<5> Personally, I'd try a power series but since the solution seems so simple, I don't know.
<5> Well, clearly a power series would not be a good choice here.
<7> xy = t ?
<5> % Series[1/(1-x),{x,0,8}]
<6> |Steve|: SeriesData[x, 0, {1, 1, 1, 1, 1, 1, 1, 1, 1}, 0, 9, 1]
<4> haven't done this for a long time, don't really remember anything
<5> The power series is just x+x^2+x^3+...
<5> Oh, geometric series.
<5> Is this for a cl***?
<4> |Steve|: this is for an ***ignment on computation techniques, need to solve it using a few numerical methods and compare that to the analytical solution
<5> Ah. I have no idea about numerical methods.
<4> |Steve|: did the analytical solutions a few years back
<4> I'm fine with the numerical methods
<5> Good. I've never studied those.
<4> just having a bit of difficulty getting the analytical solutions to compare them to
<5> You could ***ume that y(x) = sum a_n x^n
<4> hmm, think I can get away with just using the analytical solution and not writing how I derived it
<5> (1+xy)^2 = 1+2xy+x^2y^2 and then substitute that sum.
<5> You have do differentiate term by term, set like terms equal, solve for the a_n which will all be 1, then say that it's a geometric series.
<5> That actually might not be so bad.
<5> You might have to change a few summation indices.
<7> xy=t 1 + dy/dx = dt/dx
<5> Huh?
<7> umm but y(0) =1 is the knock
<4> I just read the instructions again and they want a "brief report" of "no more than 3 pages"
<4> think I'll just ignore the derivation of the analytical solution
<5> Heh.
<4> thanks though
<4> would have probably taken me a few hours to figure out how to do that
<5> Well, if you have access to mathematica (and you do in this channel), it's pretty easy.
<4> well might as well find the answer to the next problem as well
<4> % DSolve[{y'[x]==(x+y,y[0]==1},y[x],x]
<6> Michael: $Failed
<4> hmm, think I have the syntax a bit wrong
<5> % DSolve[{y'[x]==x+y,y[0]==1},y[x],x]
<6> |Steve|: DSolve[{Derivative[1][y][x] == x + y, y[0] == 1}, y[x], x]
<4> % DSolve[{y'[x]==x+ y[x],y[0]==1},y[x],x]
<6> Michael: {{y[x] -> -1 + 2*E^x - x}}
<5> That one is much easier to do by hand.
<4> probably
<4> dy/dx = x + y
<5> It's a first order linear DE.
<4> I wasn't actually given any initial conditions for this one
<5> Heh, I actually forget how to do first order linear DEs but they're the first ones taught in books.
<4> yeah, I spent about half an hour looking for my book from a few years back
<4> didn't have much luck, although google seems to have a few helpful links
<5> Ah, here we go.
<5> dy/dt + ay = g(t) has the solution: y = e^{-at}\int e^{as}g(s) ds + ce^{-at}.
<4> I seem to have done it
<5> So you have to do simple integration by parts.
<4> y = -1 + Ce^x
<4> where C is a constant that depends on the initial conditions
<5> % Integrate[E^(-s)s,s]
<6> |Steve|: (-1 - s)/E^s
<4> wait, that doesn't seem to work
<5> Multiplying that by e^t (where we replace s with t) gives (-1-t)+ce^t
<5> So in your case, y(x) = -1-x+ce^x.
<4> yeah, that seems right
<4> thanks
<7> (1+x)y = t in the 1st one :( ?
<5> zaphyBeeble: What are you talking about?
<7> from what i remember the methods of PDE were the same as any variable substitution.. just try the combos and u get them
<5> I don't remember that at all.
<7> ok...
<7> (1-x).y = t does seem the right form by the way
<4> well thanks for the help
<4> going to go to sleep, wake up at 4AM and finish this bloody ***ignment
<4> figure out how to code all this crap in VB
<8> can someone elaborate (n-1)^n ?
<8> can anyone hear me?
<9> hello
<8> hi
<9> I have a simple problem, I am new to integration. what is the way to integrate something like x * log(2x) for example (log being based on e)
<9> I saw there is a product rule or something like this..
<10> TwO, they call that "integration by parts"
<9> oh ok
<9> sounds better
<9> :)
<5> int u dv = uv - int v du
<8> we need to know how to expand (n-1)^n . Someone that can help?
<5> So let u = log 2x, dv = x dx, v = x^2/2
<2> ekelund: hmm, why would you have to expand it?
<9> ok... *listening
<5> Now plug into the formula x^2 (log 2x)/2 - int x dx
<5> since du = 1/x
<5> oops, I dropped a factor of 2.
<5> x^2(log 2x)/2 - int x/2 dx
<8> trying to figure out some ordo things
<5> = x^2(log 2x)/2 - x^2/4.
<5> +C
<2> ekelund: you can use the binomial theorem of course.
<5> % Intnegrate[x Log[2x],x]
<6> |Steve|: Intnegrate[x*Log[2*x], x]
<5> % Integrate[x Log[2x],x]
<6> |Steve|: -x^2/4 + (x^2*Log[2*x])/2
<5> Oh yes. I actually got it right. =)
<9> hm will write that on paper.. I see it better that way
<5> That's best.
<8> int-e: thanx will look at that (long time since I used math ;) )
<5> TwO: Any futher questions? I'm going to go to sleep early tonight (5am) so I can try to wake up in time for a lecture by Gordon Moore.
<9> oh thx that should solve my problem for now...
<9> I hope
<9> you know I failed with 1 point missing in math
<9> so tomorrow I will check the exam and maybe if I learn again
<9> I can try to convince my prof..
<9> :(
<5> (It usually helps to study _before_ the exam.)
<9> I nknow
<9> I did
<9> 2 weeks long
<5> ouch
<9> thats my problem =(
<9> I feel bad
<9> :(
<9> *sigh
<9> anyways.. thx for the help man!
<5> You're welcome.
<5> Goodnight #math.
<11> cya
<9> bye
<12> hello
<9> hi
<12> Do you work on what kind of level of math? Primary school? High School? or University?
<9> me?
<12> yes...
<9> university, second semester Computer Science
<9> and you?
<12> I am going to first semestr to University... Computer Science.
<9> ah ok cool
<9> where are you from? :)
<12> Poland
<12> adn you? :)
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