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anonymous
 5 years ago
All Calculus students dream :) Find all pair of non constant functions f, g such that (fg)'=f'g'
and (fg)''=f''g''
anonymous
 5 years ago
All Calculus students dream :) Find all pair of non constant functions f, g such that (fg)'=f'g' and (fg)''=f''g''

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i saw this yesterday:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0watchmath had answered:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no he didnt give A,B,C

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i forgot his solution, but

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lets do it now then..:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we had 2 eqns. (f/f')+(g/g')=1 and (f'/f")+(g'/g")=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have f'g + fg' = f'g'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and f''g + 2f'g' + fg'' = f''g''

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ya so dvide by f'g' both sides...tats same!

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0It's easy, just choose functions with disjoint support. Then either f or g and thus f' or g' likewise are zero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whats disjoint support

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Support is the set where they are nonzero. So that just means at every point either of the functions is zero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f'g + fg' = f'g', f' ( g'  g ) /g' = f

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have to find nonconstant solutions nowhereman!!

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Does f'g' have to be nonconstant too?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my bad f' = fg' / ( g'  g)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now, take the derivative of this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cantorset wat r u tryin t say? its f/f'+g/g'=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is the problem the 2 eqns. (f/f')+(g/g')=1 and (f'/f")+(g'/g")=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0saub, we know that (fg)' = f'g + fg' , correct ? so f'g + fg' = f'g'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes so why are my equations impossible? (f/f')+(g/g')=1 and (f'/f")+(g'/g")=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(g/g')+(f/f')=1 fg'' + 2g'f' + f''g=f''g'' saubhik: how do you get this? (f'/f")+(g'/g")=1

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Well I would substitute \(F = \ln f\) and \(G = \ln g\) so that you get \(1 = F'(1  1/G')\) and thus \(G' = \int \frac{1}{1\frac{1}{G'}}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k...amogh u got me... (f/f')+(g/g')=1 is correct and u have to differentiate this and use the 2nd result...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes and I get this: fg'' + 2g'f' + f''g = f''g''

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok..so we have to solve: (g/g')+(f/f')=1 and fg'' + 2g'f' + f''g = f''g''

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0PS:If this question were my dream it wud be my nightmare :P lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yea but how did you get this? (f'/f")+(g'/g")=1, sorry if its stupid.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm, ln g(x) + ln f(x) = integral 1 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f/f'+g/g'=1, if you take integral of both sides

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no tats not right i think!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, thats f'/f , woops

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integral of f/f' is not ln f!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes..amogh is correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Solving these will require deep math cantorset!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0unless and until u start guessing:P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it will be ugly to start solving there will be innumerable veriables and integrals...we have to use the exactness condition..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f=e^{3x/2},g=e^{3x} works . (source:watchmath)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i remember that solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so can we generalize...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you send me the link to the question you posted earlier yesterday?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/groups/mathematics/updates/4dda92c95d7a8b0b4fb71986

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0I already told you how to do it. Use the substitution G = ln g and F = ln f or in other terms g = e^G and f = e^F, then for every G you can find an F so that f'g' = (fg)' because the condition can be reformulated to say F' = 1/(11/G') which can be solved by integration.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0youre missing a G there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is integral(G'/(G'1)) nowhereman?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nowhereman is not missing a G cantorset..he used 1/(11/G') but how to do this integral???

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0, his G is not the same as g

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0G = ln g and F = ln f

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yups..i know.....but he is right cantorset u r a calculus tutor..how do we do this integral? any ideas?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0F'=G'/(G'1) is correct.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is integral G ' / ( 1 G' ) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yups..integrate both sides u get F=integral(G'/(G'1)) what will be the RHS?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k. so 1/(11/G') then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0G'/(G'1) = 1 + 1/ G' 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0G'/(G'1) = 1 + 1/ (G' 1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how will u do the later? its the same problem :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm, with respect to x ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its a differential equation...we can differentiate with respect to the variables of the function..i.e. G(x) wrt x and F(x) wrt x...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok...lets go to the main problem after making nowhereman's substitutions we get F'G'=F'+G'..how will u solve this diff eqn?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please dont make fun of me...see that F'G'=F'+G' implies integral(dF(x)*dG(x))=F(x)+G(x) thats an integration with respect to 2 variables!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we need multiple integration here? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok , im lost, how did you get that , capital F' and G'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0see F=lnf and G=lng then use f'g'=(fg)'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then integrate but i dont know the integral of LHS..

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0f'g+fg'=f'g' (f'f)g'f'g=0 g' f'g/(f'f)=0 Now you can use integrating factor method. The answer will be still in terms of f and f'. But that is ok, we will use the second requirement.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok , multiply both sides by e ^ int ( f' / (f'  f)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is int ( f' / (f'  f)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0keep it like this? we need to simplify...

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0just keep it like that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0g.e ^ int ( f' / (f'  f) =0 is the answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how will u use the 2nd fact watchmath?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is pretty ugly to differentiate..

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0just plugin the g. You need fundamental calculus theorem to compute g'.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0website keeps frezing on me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok im a little slow, one sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0watchmath how will u extract the g?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k...lot of work though..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but process is clear..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now we know , dy/dx + p(x) *y = q(x), the solution is y = [int ( int e^p(x) )*q(x) dx + c ] / e^int p(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0g.e^[f'/(f'f)]=constant.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now since q(x) is zero, that simplifies the numerator, so its g = C / ( e^ int ( f

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lng+[f'/(f'f)]=constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now cantorset..how will u get g?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0, looks like you made a mistake watchman

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how ill we use (fg)"=f"g"?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f ' g + f g'  f ' g ' = 0 g' ( f  f ' ) + f ' g = 0 g ' + f' / ( f f ' ) g = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cantorset the integral of what u obtained is lng+[f'/(f'f)]=constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0g = C / e^ int ( f ' / ( f  f' )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does that make sense so far?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0g = C e^ int ( f ' / f'  f)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln g = C + int f' / ( f '  f )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0saubik im not sure how you got that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0watchman, im sure that g = C / e^ int ( f ' / ( f  f' )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or g = C e^ int ( f ' / f'  f)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0saubhik, can you show me your reasoning

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep but i simplified it...:) its lng+[f'/(ff')]=constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its ok..we are getting the same results..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now how will u plug in g from (fg)"=f"g"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0saubhik, our answers dont match, you are missing an integral sign

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops..i meant it sorry :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lng+int[f'/(ff')]=constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i guess you plug this g into the second equation, ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f'' g + 2 f ' g' + f g''  f '' g'' = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah..but its madness...i guess who had this as "dream"!! :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think this should be minus

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lng  int[f'/(ff')]=constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no..i had interchange f and f' so minus is taken care of..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok , i did it a little different

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f'' g + 2 f ' g' + f g''  f '' g'' = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is not a very nice equation to plug g :( <<i need to go now.. cantorset...will come after smtym.bye>>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nowhere man...how can we integrate F'G'=F'+G' ? using ur substitutions?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0I don't think that the integration can be done for a general G. But to find examples this is a good approach. For example choose F = x^2.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0No, I mean you can't find the integral of the general G itself. So why should you be able to find it of a term involving the general G. In fact you can choose G always in such a way, that the integrand is arbitrary. Set G = integral 1/(11/H) and you will get H = 1/(11/G').

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0express g' and g'' in terms of g. Then you will have an equation in terms of f,f' and f''. It is ugly but you will be able to factor it to get (f''  (f')^2 )(2f'f '')=0. From here we have f=Ae^(Bx) or f=a+be^(cx) (since f andg symmetric, g also of the same form). Then we need to adjust the constants to make it right.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Actually after I replace f,f',f'' by a,b,c it isn't too bad. Also actually we just need algebra to get g'=f'g/(f'f) (no need invoking integral factor)
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