Question

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

and (fg)''=f''g''

Answer 1

i saw this yesterday:)

Answer 2

oh yeah

Answer 3

watchmath had answered:)

Answer 4

no he didnt give A,B,C

Answer 5

i forgot his solution, but

Answer 6

lets do it now then..:)

Answer 7

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

Answer 8

its impossible

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

whats disjoint support

Answer 14

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

Answer 15

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

Answer 16

we have to find nonconstant solutions nowhereman!!

Answer 17

f = fg' / ( g' - g)

Answer 18

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

Answer 19

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

Answer 20

now, take the derivative of this

Answer 21

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

Answer 22

now do you get that?

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

(g/g')+(f/f')=1

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

saubhik: how do you get this? (f'/f")+(g'/g")=1

Answer 27

Well I would substitute \(F = \ln f\) and \(G = \ln g\) so that you get \(1 = F'(1 - 1/G')\) and thus \(G' = \int \frac{1}{1-\frac{1}{G'}}\)

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

no tats not right i think!

Answer 36

oh, thats f'/f , woops

Answer 37

integral of f/f' is not ln f!

Answer 38

yes..amogh is correct

Answer 39

Solving these will require deep math cantorset!!

Answer 40

unless and until u start guessing:P

Answer 41

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

Answer 42

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

Answer 43

haha

Answer 44

i remember that solution

Answer 45

so can we generalize...

Answer 46

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

Answer 47

http://openstudy.com/groups/mathematics/updates/4dda92c95d7a8b0b4fb71986

Answer 48

Thanks

Answer 49

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.

Answer 50

youre missing a G there

Answer 51

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

Answer 52

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

Answer 53

, his G is not the same as g

Answer 54

G = ln g and F = ln f

Answer 55

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

Answer 56

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

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

first divide that

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

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

Answer 65

hmmm, with respect to x ?

Answer 66

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...

Answer 67

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?

Answer 68

dunno

Answer 69

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!!!

Answer 70

we need multiple integration here? :)

Answer 71

am here cantorset..

Answer 72

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

Answer 73

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

Answer 74

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

Answer 75

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

Answer 76

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.

Answer 77

yah...

Answer 78

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

Answer 79

yes :)

Answer 80

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

Answer 81

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

Answer 82

just keep it like that

Answer 83

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

Answer 84

=constant

Answer 85

yess..

Answer 86

how will u use the 2nd fact watchmath?

Answer 87

this is pretty ugly to differentiate..

Answer 88

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

Answer 89

website keeps frezing on me

Answer 90

ok im a little slow, one sec

Answer 91

watchmath how will u extract the g?

Answer 92

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

Answer 93

k...lot of work though..

Answer 94

but process is clear..

Answer 95

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)

Answer 96

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

Answer 97

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

Answer 98

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