- anonymous

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

- jamiebookeater

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- anonymous

i saw this yesterday:)

- anonymous

oh yeah

- anonymous

watchmath had answered:)

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## More answers

- anonymous

no he didnt give A,B,C

- anonymous

i forgot his solution, but

- anonymous

lets do it now then..:)

- anonymous

we had 2 eqns. (f/f')+(g/g')=1 and (f'/f")+(g'/g")=1

- anonymous

its impossible

- anonymous

we have f'g + fg' = f'g'

- anonymous

and f''g + 2f'g' + fg'' = f''g''

- anonymous

ya so dvide by f'g' both sides...tats same!

- nowhereman

It's easy, just choose functions with disjoint support. Then either f or g and thus f' or g' likewise are zero.

- anonymous

whats disjoint support

- nowhereman

Support is the set where they are non-zero. So that just means at every point either of the functions is zero.

- anonymous

f'g + fg' = f'g',
f' ( g' - g ) /g' = f

- anonymous

we have to find nonconstant solutions nowhereman!!

- anonymous

f = fg' / ( g' - g)

- nowhereman

Does f'g' have to be non-constant too?

- anonymous

my bad f' = fg' / ( g' - g)

- anonymous

now, take the derivative of this

- anonymous

cantorset wat r u tryin t say? its f/f'+g/g'=1

- anonymous

now do you get that?

- anonymous

what is the problem the 2 eqns. (f/f')+(g/g')=1 and (f'/f")+(g'/g")=1

- anonymous

saub, we know that (fg)' = f'g + fg' , correct ?
so
f'g + fg' = f'g'

- anonymous

yes so why are my equations impossible? (f/f')+(g/g')=1 and (f'/f")+(g'/g")=1

- 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

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

k...amogh u got me... (f/f')+(g/g')=1 is correct and u have to differentiate this and use the 2nd result...

- anonymous

Yes and I get this: fg'' + 2g'f' + f''g = f''g''

- anonymous

ok..so we have to solve: (g/g')+(f/f')=1 and fg'' + 2g'f' + f''g = f''g''

- anonymous

PS:If this question were my dream it wud be my nightmare :P lol

- anonymous

Yea but how did you get this? (f'/f")+(g'/g")=1, sorry if its stupid.

- anonymous

hmmm, ln g(x) + ln f(x) = integral 1 ?

- anonymous

f/f'+g/g'=1, if you take integral of both sides

- anonymous

no tats not right i think!

- anonymous

oh, thats f'/f , woops

- anonymous

integral of f/f' is not ln f!

- anonymous

yes..amogh is correct

- anonymous

Solving these will require deep math cantorset!!

- anonymous

unless and until u start guessing:P

- anonymous

it will be ugly to start solving there will be innumerable veriables and integrals...we have to use the exactness condition..

- anonymous

f=e^{3x/2},g=e^{3x} works .
(source:watchmath)

- anonymous

haha

- anonymous

i remember that solution

- anonymous

so can we generalize...

- anonymous

Can you send me the link to the question you posted earlier yesterday?

- anonymous

Thanks

- 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

youre missing a G there

- anonymous

what is integral(G'/(G'-1)) nowhereman?

- anonymous

nowhereman is not missing a G cantorset..he used 1/(1-1/G')
but how to do this integral???

- anonymous

, his G is not the same as g

- anonymous

G = ln g and F = ln f

- anonymous

yups..i know.....but he is right cantorset u r a calculus tutor..how do we do this integral? any ideas?

- anonymous

F'=G'/(G'-1) is correct.

- anonymous

it is integral G ' / ( 1- G' ) ?

- anonymous

F'=G'/(G'-1) i mean

- anonymous

yups..integrate both sides u get F=integral(G'/(G'-1))
what will be the RHS?

- anonymous

first divide that

- anonymous

k. so 1/(1-1/G') then?

- anonymous

G'/(G'-1) = 1 + 1/ G' -1

- anonymous

G'/(G'-1) = 1 + 1/ (G' -1)

- anonymous

how will u do the later? its the same problem :P

- anonymous

hmmm, with respect to x ?

- 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

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

dunno

- 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

we need multiple integration here? :)

- anonymous

am here cantorset..

- anonymous

ok , im lost, how did you get that , capital F' and G'

- anonymous

see F=lnf and G=lng then use f'g'=(fg)'

- anonymous

itll give F'G'=F'+G'

- anonymous

then integrate but i dont know the integral of LHS..

- 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

yah...

- anonymous

ok , multiply both sides by e ^ int ( f' / (f' - f)

- watchmath

yes :)

- anonymous

what is int ( f' / (f' - f)

- anonymous

keep it like this? we need to simplify...

- watchmath

just keep it like that

- anonymous

g.e ^ int ( f' / (f' - f) =0 is the answer

- watchmath

=constant

- anonymous

yess..

- anonymous

how will u use the 2nd fact watchmath?

- anonymous

this is pretty ugly to differentiate..

- watchmath

just plugin the g. You need fundamental calculus theorem to compute g'.

- anonymous

website keeps frezing on me

- anonymous

ok im a little slow, one sec

- anonymous

watchmath how will u extract the g?

- anonymous

g'- f'/(f'-f)*g =0

- anonymous

k...lot of work though..

- anonymous

but process is clear..

- 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

g.e^[f'/(f'-f)]=constant.

- anonymous

now since q(x) is zero, that simplifies the numerator, so its
g = C / ( e^ int ( f

- anonymous

ln|g|+[f'/(f'-f)]=constant

- anonymous

now cantorset..how will u get g?

- anonymous

, looks like you made a mistake watchman

- anonymous

how ill we use (fg)"=f"g"?

- anonymous

f ' g + f g' - f ' g ' = 0
g' ( f - f ' ) + f ' g = 0
g ' + f' / ( f- f ' ) g = 0

- anonymous

cantorset the integral of what u obtained is ln|g|+[f'/(f'-f)]=constant

- anonymous

g = C / e^ int ( f ' / ( f - f' )

- anonymous

does that make sense so far?

- anonymous

g = C e^ int ( f ' / f' - f)

- anonymous

ln g = C + int f' / ( f ' - f )

- anonymous

saubik im not sure how you got that

- anonymous

watchman, im sure that g = C / e^ int ( f ' / ( f - f' )

- anonymous

or
g = C e^ int ( f ' / f' - f)

- anonymous

ya ur right

- anonymous

saubhik, can you show me your reasoning

- anonymous

yep but i simplified it...:) its ln|g|+[f'/(f-f')]=constant

- anonymous

its ok..we are getting the same results..

- anonymous

now how will u plug in g from (fg)"=f"g"

- anonymous

saubhik, our answers dont match, you are missing an integral sign

- anonymous

oops..i meant it sorry :)

- anonymous

oh ok , thats better

- anonymous

ln|g|+int[f'/(f-f')]=constant

- anonymous

i guess you plug this g into the second equation, ?

- anonymous

f'' g + 2 f ' g' + f g'' - f '' g'' = 0

- anonymous

yeah..but its madness...i guess who had this as "dream"!! :)

- anonymous

i think this should be minus

- anonymous

ln|g| - int[f'/(f-f')]=constant

- anonymous

no..i had interchange f and f' so minus is taken care of..

- anonymous

oh ok , i did it a little different

- anonymous

f'' g + 2 f ' g' + f g'' - f '' g'' = 0

- anonymous

this is not a very nice equation to plug g :(
<

*>*

*
*

- anonymous

nowhere man...how can we integrate F'G'=F'+G' ? using ur substitutions?

- 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

well this is bad:(

- 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

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

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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