anonymous
  • anonymous
All Calculus students dream :) Find all pair of non constant functions f, g such that (fg)'=f'g' and (fg)''=f''g''
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
i saw this yesterday:)
anonymous
  • anonymous
oh yeah
anonymous
  • anonymous
watchmath had answered:)

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anonymous
  • anonymous
no he didnt give A,B,C
anonymous
  • anonymous
i forgot his solution, but
anonymous
  • anonymous
lets do it now then..:)
anonymous
  • anonymous
we had 2 eqns. (f/f')+(g/g')=1 and (f'/f")+(g'/g")=1
anonymous
  • anonymous
its impossible
anonymous
  • anonymous
we have f'g + fg' = f'g'
anonymous
  • anonymous
and f''g + 2f'g' + fg'' = f''g''
anonymous
  • anonymous
ya so dvide by f'g' both sides...tats same!
nowhereman
  • nowhereman
It's easy, just choose functions with disjoint support. Then either f or g and thus f' or g' likewise are zero.
anonymous
  • anonymous
whats disjoint support
nowhereman
  • nowhereman
Support is the set where they are non-zero. So that just means at every point either of the functions is zero.
anonymous
  • anonymous
f'g + fg' = f'g', f' ( g' - g ) /g' = f
anonymous
  • anonymous
we have to find nonconstant solutions nowhereman!!
anonymous
  • anonymous
f = fg' / ( g' - g)
nowhereman
  • nowhereman
Does f'g' have to be non-constant too?
anonymous
  • anonymous
my bad f' = fg' / ( g' - g)
anonymous
  • anonymous
now, take the derivative of this
anonymous
  • anonymous
cantorset wat r u tryin t say? its f/f'+g/g'=1
anonymous
  • anonymous
now do you get that?
anonymous
  • anonymous
what is the problem the 2 eqns. (f/f')+(g/g')=1 and (f'/f")+(g'/g")=1
anonymous
  • anonymous
saub, we know that (fg)' = f'g + fg' , correct ? so f'g + fg' = f'g'
anonymous
  • anonymous
yes so why are my equations impossible? (f/f')+(g/g')=1 and (f'/f")+(g'/g")=1
anonymous
  • anonymous
(g/g')+(f/f')=1 fg'' + 2g'f' + f''g=f''g'' saubhik: how do you get this? (f'/f")+(g'/g")=1
nowhereman
  • nowhereman
Well 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
  • anonymous
k...amogh u got me... (f/f')+(g/g')=1 is correct and u have to differentiate this and use the 2nd result...
anonymous
  • anonymous
Yes and I get this: fg'' + 2g'f' + f''g = f''g''
anonymous
  • anonymous
ok..so we have to solve: (g/g')+(f/f')=1 and fg'' + 2g'f' + f''g = f''g''
anonymous
  • anonymous
PS:If this question were my dream it wud be my nightmare :P lol
anonymous
  • anonymous
Yea but how did you get this? (f'/f")+(g'/g")=1, sorry if its stupid.
anonymous
  • anonymous
hmmm, ln g(x) + ln f(x) = integral 1 ?
anonymous
  • anonymous
f/f'+g/g'=1, if you take integral of both sides
anonymous
  • anonymous
no tats not right i think!
anonymous
  • anonymous
oh, thats f'/f , woops
anonymous
  • anonymous
integral of f/f' is not ln f!
anonymous
  • anonymous
yes..amogh is correct
anonymous
  • anonymous
Solving these will require deep math cantorset!!
anonymous
  • anonymous
unless and until u start guessing:P
anonymous
  • anonymous
it will be ugly to start solving there will be innumerable veriables and integrals...we have to use the exactness condition..
anonymous
  • anonymous
f=e^{3x/2},g=e^{3x} works . (source:watchmath)
anonymous
  • anonymous
haha
anonymous
  • anonymous
i remember that solution
anonymous
  • anonymous
so can we generalize...
anonymous
  • anonymous
Can you send me the link to the question you posted earlier yesterday?
anonymous
  • anonymous
Thanks
nowhereman
  • nowhereman
I 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/(1-1/G') which can be solved by integration.
anonymous
  • anonymous
youre missing a G there
anonymous
  • anonymous
what is integral(G'/(G'-1)) nowhereman?
anonymous
  • anonymous
nowhereman is not missing a G cantorset..he used 1/(1-1/G') but how to do this integral???
anonymous
  • anonymous
, his G is not the same as g
anonymous
  • anonymous
G = ln g and F = ln f
anonymous
  • anonymous
yups..i know.....but he is right cantorset u r a calculus tutor..how do we do this integral? any ideas?
anonymous
  • anonymous
F'=G'/(G'-1) is correct.
anonymous
  • anonymous
it is integral G ' / ( 1- G' ) ?
anonymous
  • anonymous
F'=G'/(G'-1) i mean
anonymous
  • anonymous
yups..integrate both sides u get F=integral(G'/(G'-1)) what will be the RHS?
anonymous
  • anonymous
first divide that
anonymous
  • anonymous
k. so 1/(1-1/G') then?
anonymous
  • anonymous
G'/(G'-1) = 1 + 1/ G' -1
anonymous
  • anonymous
G'/(G'-1) = 1 + 1/ (G' -1)
anonymous
  • anonymous
how will u do the later? its the same problem :P
anonymous
  • anonymous
hmmm, with respect to x ?
anonymous
  • anonymous
its 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
  • anonymous
ok...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
  • anonymous
dunno
anonymous
  • anonymous
please 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
  • anonymous
we need multiple integration here? :)
anonymous
  • anonymous
am here cantorset..
anonymous
  • anonymous
ok , im lost, how did you get that , capital F' and G'
anonymous
  • anonymous
see F=lnf and G=lng then use f'g'=(fg)'
anonymous
  • anonymous
itll give F'G'=F'+G'
anonymous
  • anonymous
then integrate but i dont know the integral of LHS..
watchmath
  • watchmath
f'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
  • anonymous
yah...
anonymous
  • anonymous
ok , multiply both sides by e ^ int ( f' / (f' - f)
watchmath
  • watchmath
yes :)
anonymous
  • anonymous
what is int ( f' / (f' - f)
anonymous
  • anonymous
keep it like this? we need to simplify...
watchmath
  • watchmath
just keep it like that
anonymous
  • anonymous
g.e ^ int ( f' / (f' - f) =0 is the answer
watchmath
  • watchmath
=constant
anonymous
  • anonymous
yess..
anonymous
  • anonymous
how will u use the 2nd fact watchmath?
anonymous
  • anonymous
this is pretty ugly to differentiate..
watchmath
  • watchmath
just plugin the g. You need fundamental calculus theorem to compute g'.
anonymous
  • anonymous
website keeps frezing on me
anonymous
  • anonymous
ok im a little slow, one sec
anonymous
  • anonymous
watchmath how will u extract the g?
anonymous
  • anonymous
g'- f'/(f'-f)*g =0
anonymous
  • anonymous
k...lot of work though..
anonymous
  • anonymous
but process is clear..
anonymous
  • anonymous
now 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
  • anonymous
g.e^[f'/(f'-f)]=constant.
anonymous
  • anonymous
now since q(x) is zero, that simplifies the numerator, so its g = C / ( e^ int ( f
anonymous
  • anonymous
ln|g|+[f'/(f'-f)]=constant
anonymous
  • anonymous
now cantorset..how will u get g?
anonymous
  • anonymous
, looks like you made a mistake watchman
anonymous
  • anonymous
how ill we use (fg)"=f"g"?
anonymous
  • anonymous
f ' g + f g' - f ' g ' = 0 g' ( f - f ' ) + f ' g = 0 g ' + f' / ( f- f ' ) g = 0
anonymous
  • anonymous
cantorset the integral of what u obtained is ln|g|+[f'/(f'-f)]=constant
anonymous
  • anonymous
g = C / e^ int ( f ' / ( f - f' )
anonymous
  • anonymous
does that make sense so far?
anonymous
  • anonymous
g = C e^ int ( f ' / f' - f)
anonymous
  • anonymous
ln g = C + int f' / ( f ' - f )
anonymous
  • anonymous
saubik im not sure how you got that
anonymous
  • anonymous
watchman, im sure that g = C / e^ int ( f ' / ( f - f' )
anonymous
  • anonymous
or g = C e^ int ( f ' / f' - f)
anonymous
  • anonymous
ya ur right
anonymous
  • anonymous
saubhik, can you show me your reasoning
anonymous
  • anonymous
yep but i simplified it...:) its ln|g|+[f'/(f-f')]=constant
anonymous
  • anonymous
its ok..we are getting the same results..
anonymous
  • anonymous
now how will u plug in g from (fg)"=f"g"
anonymous
  • anonymous
saubhik, our answers dont match, you are missing an integral sign
anonymous
  • anonymous
oops..i meant it sorry :)
anonymous
  • anonymous
oh ok , thats better
anonymous
  • anonymous
ln|g|+int[f'/(f-f')]=constant
anonymous
  • anonymous
i guess you plug this g into the second equation, ?
anonymous
  • anonymous
f'' g + 2 f ' g' + f g'' - f '' g'' = 0
anonymous
  • anonymous
yeah..but its madness...i guess who had this as "dream"!! :)
anonymous
  • anonymous
i think this should be minus
anonymous
  • anonymous
ln|g| - int[f'/(f-f')]=constant
anonymous
  • anonymous
no..i had interchange f and f' so minus is taken care of..
anonymous
  • anonymous
oh ok , i did it a little different
anonymous
  • anonymous
f'' g + 2 f ' g' + f g'' - f '' g'' = 0
anonymous
  • anonymous
this is not a very nice equation to plug g :( <>
anonymous
  • anonymous
nowhere man...how can we integrate F'G'=F'+G' ? using ur substitutions?
nowhereman
  • nowhereman
I 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.
anonymous
  • anonymous
well this is bad:(
nowhereman
  • nowhereman
No, 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/(1-1/H) and you will get H = 1/(1-1/G').
watchmath
  • watchmath
express 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
  • watchmath
Actually 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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