Question

What can the functions f and g be if they satisfy this relationship:

(fg)''=(f+g)^2

Answer 1

I tried solving a simpler case, (fg)'=(f+g) and (fg)=(f+g)^0 to see if there was a pattern I could find. Maybe someone can find one.

So I tried then seeing if I could solve h''=h^2 but even this seemed too hard for me, but I found I could change it into a system of equations u'=v, v'=u^2 but being nonlinear I kinda just couldn't plug it into a matrix so bleh.

Then I just guessed for h. h=a*x^n. That gives us: \[\large a*n(n-1)x^{n-2}=a^2x^{2n} \\ \large \frac{n(n-1)}{a}=x^{n+2}\] Since a and n are constants, the right side has to be a constant, so n=-2. This lets us solve for a, which is 6. \[h=\frac{6}{x^2}\] Then if we say \[fg=h \\ f+g=h\] and solve that, we should get an answer to the differential equations above. \[f=\frac{3 \pm \sqrt{9+ 6x^2}}{x^2} \\ g=\frac{3 \mp \sqrt{9+6x^2}}{x^2}\]

So can I just like compute the wronskian of this? I really don't know how that works to be honest or if it applies here since these are different.

The thing that interests me about these types of equations is how they end up being the same, except for an interchanging of addition for multiplication and exponents to derivatives, but they combine the same. \[(fg)''=(f+g)^2\\ f''g+2f'g'+fg''=f^2+2fg+g^2\]

thoughts/ideas/halp? medal/fan/everlasting love/name child of firstborn/etc.

Answer 2

Did you try f and g = e^x ?

Answer 3

nope, good idea though one sec.

Answer 4

I did what you did at the bottom and found that e^x would satisfy this identity.

Answer 5

Wait, what do you mean?

Answer 6

if f and g = e^x or what ever variable you want you get

f'' = e^x g' = e^x f'= e^x g'' =e^x

so,

e^x(e^x) + 2(e^x)(e^x) + e^x(e^x) = e^x(e^x) + 2(e^x)(e^x) + e^x(e^x)

Answer 7

No because you multiply f*g before taking the derivatives. So you'd be getting \[\large (e^{2x})''=(2e^x)^2\]

Answer 8

Yeah you get 4e^2x = 4e^2x

Answer 9

e^x(e^x) + 2(e^x)(e^x) + e^x(e^x) = e^x(e^x) + 2(e^x)(e^x) + e^x(e^x)

4e^2x = 4e^2x

Answer 10

Oh, I guess that works too. Interesting. So are there more solutions then? It seems like there are several solutions here, but not really sure what the most general solution is.

Answer 11

Yeah, I don't know i am sure there is more I tried sin and cos it did not work though.

Answer 12

I'm going to try with variable coefficients like f=ae^bx and g=ce^dx to see if I get anything.

Answer 13

Yeah that is definitely worth a try I just tired e^ix did not work :P

Answer 14

I'm really looking for the general solutions to \[\large \frac{d^n}{dx^n}(fg)=(f+g)^n\] and I guess it seems like making f=g=e^x probably works. But I am pretty sure there are more solutions than that, so it's sort of weird. It's interesting though.

Answer 15

Or even more general, \[\large \frac{d^n}{dx^n}(f_1*f_2*...*f_m) = (f_1+f_2+...+f_m)^n\] but that's just wishful thinking.

Answer 16

Yeah beyond trial and error I don't see solving it as a systems of equations, series solutions or any other clear method :P

Answer 17

I'm out so have fun :D

Answer 18

Alright thanks for playing along haha. =)