Question

Looking for some hints on where to go with an integration problem. I'll post the question and my progress in the comments so I can format things properly.

Answer 1

Question: For any given function f, let
\[I = \int\limits_{}^{}[f'(x)]^2[f(x)]^ndx\]where n is a positive integer. Show tht, if f(x) satisfies
\[f''(x) = kf(x)f'(x)\]
for some constant k, then I can be evaluated in terms of f(x), f'(x), k and n.

Answer 2

From the second condition:\[f(x) = \frac{f''(x)}{kf'(x)}\]\[\int\limits f(x) dx = \frac{1}{k}lnf'(x) + c\]Doesn't immediately appear to be useful.

Answer 3

In the case n=1:
\[I = \int\limits f'(x)[f'(x)f(x)]dx = \int\limits f'(x)\frac{f''(x)}{k}dx = \frac{1}{k}\int\limits f'(x)f''(x)dx\]\[= \frac{1}{k}[f'(x)]^2 - \frac{1}{k}\int\limits f'(x)f''(x)dx\]\[2I = \frac{1}{k}[f'(x)]^2\]\[I = \frac{1}{2k}[f'(x)]^2\]I also tried to consider the case n=2 but I wasn't able to get it into a nice form like this.

Answer 4

i think we need ordinary differential equation humm ill see if it cud work.

Answer 5

I also tried integrating straight off the bat by substitution but I wasn't able to get down to an integral that I could evaluate.

Answer 6

sub \(f'(x) = t\)
then, \(f(x) = t^2/2 + C\)
\(f''(x) dx = dt \)

Answer 7

not really sure, hey ikram help

Answer 8

substitution seems to work, but im not really sure yet... humm

Answer 9

Will see if I can do something with those suggestions!

Answer 10

im really not sure cuz ikram said something about partials :o

Answer 11

*differential equations

Answer 12

humm ordinary , nt partial
in ordinary u assume
f =1 or C
f' =x
f" = x^2
and so
i never used it in case like this :o

Answer 13

I get as far as
\[I = \frac{1}{k}\int\limits t(\frac{t^2}{2}+C)^{n-1}dt\]I was considering looking at some cases of this but it seems to me like the (t^2/2 + C)^(n-1) part is still going to be an issue.

Answer 14

why ?
sub again : \(\large \frac{t^2}{2} +C = u \)

Answer 15

ikram, do u see any goofup in the substitution in my first reply ?

Answer 16

no it works too :)

Answer 17

oww ty :))

Answer 18

np

Answer 19

So u = f'(x) as well?

Answer 20

the integral u reached is also work better :o

Answer 21

\(I = \frac{1}{k}\int\limits t(\frac{t^2}{2}+C)^{n-1}dt\)

yes, this is a good integral

Answer 22

\[I = \frac{1}{k}\int\limits \frac{du}{dt}u^{n-1}dt = \frac{1}{k}\int\limits u^{n-1}du = \frac{1}{nk}u^n + c = \frac{1}{nk}[f(x)]^n + c\]How does that look to you guys? I'm slightly concerned about the absence of any instances of f'(x); did I make a mistake evaluating somewhere?

Answer 23

Still doesn't seem quite right to me, any thoughts @ikram002p @ganeshie8?

Answer 24

\(I = \frac{1}{k}\int\limits t(\frac{t^2}{2}+C)^{n-1}dt\)

we can clearly work it when n = 1,
for n > 1 u can substitute t^2/2 + c = u, then tdt = du
\(I = \frac{1}{k}\int\limits u^{n-1}du\)

Answer 25

btw, u = t^2/2 + c, its not f(x)
u may differentiate it back and see if u get back to the original function

Answer 26

\[I = \frac{1}{nk}(\frac{f'(x)^2}{2}+c_1)^n + c_2\]Anything we can do with this?